Project Euler+ #97 modular exponentation not working - c

I am trying to solve Project Euler+ #97 from Hackerrank. The problem asks to calculate the last 12 digits of A x B ** C + D. My attempt was to use the modular exponentiation mod 10 ** 12 from Wikipedia in order to efficiently calculate the last 12 digits and avoid overflow. However, for all cases aside from the sample 2 x 3 ** 4 + 5 I am getting wrong. According to the constraints there should be no overflow for unsigned long long.
The problem:
Now we want to learn how to calculate some last digits of such big numbers. Let's assume we have a lot of numbers A x B ** C + D and we want to know last 12 digits of these numbers.
Constraints:
1 ≤ T ≤ 500000
1 ≤ A, B, C, D ≤ 10 ** 9
Input: First line contains one integer T - the number of tests.
T lines follow containing 4 integers (A, B, C and D) each.
Output: Output exactly one line containing exactly 12 digits - the last 12 digits of the sum of all results. If the sum is less than 10 ** 12 print corresponding number of leading zeroes then.
My attempt in C
#include <stdio.h>
int main() {
const unsigned long long limit = 1000000000000;
int cases;
for (scanf("%d", &cases); cases; --cases) {
// mult = A, base = B, exp = C, add = D
unsigned long long mult, base, exp, add;
scanf("%llu %llu %llu %llu", &mult, &base, &exp, &add);
base = base % limit;
while (exp) {
if (exp & 1) {
mult = (mult * base) % limit;
}
exp >>= 1;
base = (base * base) % limit;
}
printf("%012llu\n", (mult + add) % limit);
}
return 0;
}

I think you can overflow unsigned long long math (e.g. - modulo 2^64) because your computation of base in your inner loop can get as high as (10^12 - 1)^2 ~= 10^24 ~= 2^79.726, which is much more than 2^64. For example, think about B = 10^6 - 1 and C = 4.
On my MacBook Pro running a 64b version of Mac OS X with clang 8.1.0:
#include <stdio.h>
int main()
{
fprintf(stdout, "sizeof(unsigned long long) = %u\n", (unsigned) sizeof(unsigned long long));
fprintf(stdout, "sizeof(__uint128_t) = %u\n", (unsigned) sizeof(__uint128_t));
fprintf(stdout, "sizeof(long double) = %u\n", (unsigned) sizeof(long double));
return 0;
}
Says:
sizeof(unsigned long long) = 8
sizeof(__uint128_t) = 16
sizeof(long double) = 16
If your platform says 16 or 10 instead for long long, then I think you are in the clear. If it says 8 like mine does, then you need to rework your answer to either force 128b (or 80b) integer math natively or mimic it some other way.
You can try __uint128_t, which is supported by gcc and clang. Otherwise, you'd need to resort to something like long double and fmodl(), which might have enough mantissa bits but might not give exact answers like you want.
Also, you don't accumulate multiple results like the task says. Here's my shot at it, based on your program, but using __uint128_t.
#include <stdio.h>
#include <stdlib.h>
#define BILLION 1000000000
#define TRILLION 1000000000000
int main()
{
const __uint128_t limit = TRILLION;
unsigned long cases = 0;
__uint128_t acc = 0;
if (scanf("%lu", &cases) != 1 || cases == 0 || cases > 500000)
abort();
while (cases-- > 0)
{
unsigned long a, b, c, d;
__uint128_t b2c = 1, bbase;
if (scanf("%lu %lu %lu %lu", &a, &b, &c, &d) != 4 ||
a == 0 || a > BILLION || b == 0 || b > BILLION ||
c == 0 || c > BILLION || d == 0 || d > BILLION)
abort();
for (bbase = b; c > 0; c >>= 1)
{
if ((c & 0x1) != 0)
b2c = (b2c * bbase) % limit; // 64b overflow: ~10^12 * ~10^12 ~= 10^24 > 2^64
bbase = (bbase * bbase) % limit; // same overflow issue as above
}
// can do modulus on acc only once at end of program instead because
// 5 * 10^5 * (10^9 * (10^12 - 1) + 10^9) = 5 * 10^26 < 2^128
acc += a * b2c + d;
}
acc %= limit;
printf("%012llu\n", (unsigned long long) acc);
return 0;
}

Related

Efficient algorithm to calculate the sum of number of base2 digits (number of bits) over an interval of positive integers

Let's say I've been given two integers a, b where a is a positive integer and is smaller than b. I have to find an efficient algorithm that's going to give me the sum of number of base2 digits (number of bits) over the interval [a, b]. For example, in the interval [0, 4] the sum of digits is equal to 9 because 0 = 1 digit, 1 = 1 digit, 2 = 2 digits, 3 = 2 digits and 4 = 3 digits.
My program is capable of calculating this number by using a loop but I'm looking for something more efficient for large numbers. Here are the snippets of my code just to give you an idea:
int numberOfBits(int i) {
if(i == 0) {
return 1;
}
else {
return (int) log2(i) + 1;
}
}
The function above is for calculating the number of digits of one number in the interval.
The code below shows you how I use it in my main function.
for(i = a; i <= b; i++) {
l = l + numberOfBits(i);
}
printf("Digits: %d\n", l);
Ideally I should be able to get the number of digits by using the two values of my interval and using some special algorithm to do that.
Try this code, i think it gives you what you are needing to calculate the binaries:
int bit(int x)
{
if(!x) return 1;
else
{
int i;
for(i = 0; x; i++, x >>= 1);
return i;
}
}
The main thing to understand here is that the number of digits used to represent a number in binary increases by one with each power of two:
+--------------+---------------+
| number range | binary digits |
+==============+===============+
| 0 - 1 | 1 |
+--------------+---------------+
| 2 - 3 | 2 |
+--------------+---------------+
| 4 - 7 | 3 |
+--------------+---------------+
| 8 - 15 | 4 |
+--------------+---------------+
| 16 - 31 | 5 |
+--------------+---------------+
| 32 - 63 | 6 |
+--------------+---------------+
| ... | ... |
A trivial improvement over your brute force algorithm would then be to figure out how many times this number of digits has increased between the two numbers passed in (given by the base two logarithm) and add up the digits by multiplying the count of numbers that can be represented by the given number of digits (given by the power of two) with the number of digits.
A naive implementation of this algorithm is:
int digits_sum_seq(int a, int b)
{
int sum = 0;
int i = 0;
int log2b = b <= 0 ? 1 : floor(log2(b));
int log2a = a <= 0 ? 1 : floor(log2(a)) + 1;
sum += (pow(2, log2a) - a) * (log2a);
for (i = log2b; i > log2a; i--)
sum += pow(2, i - 1) * i;
sum += (b - pow(2, log2b) + 1) * (log2b + 1);
return sum;
}
It can then be improved by the more efficient versions of the log and pow functions seen in the other answers.
First, we can improve the speed of log2, but that only gives us a fixed factor speed-up and doesn't change the scaling.
Faster log2 adapted from: https://graphics.stanford.edu/~seander/bithacks.html#IntegerLogLookup
The lookup table method takes only about 7 operations to find the log
of a 32-bit value. If extended for 64-bit quantities, it would take
roughly 9 operations. Another operation can be trimmed off by using
four tables, with the possible additions incorporated into each. Using
int table elements may be faster, depending on your architecture.
Second, we must re-think the algorithm. If you know that numbers between N and M have the same number of digits, would you add them up one by one or would you rather do (M-N+1)*numDigits?
But if we have a range where multiple numbers appear what do we do? Let's just find the intervals of same digits, and add sums of those intervals. Implemented below. I think that my findEndLimit could be further optimized with a lookup table.
Code
#include <stdio.h>
#include <limits.h>
#include <time.h>
unsigned int fastLog2(unsigned int v)
{
static const char LogTable256[256] =
{
#define LT(n) n, n, n, n, n, n, n, n, n, n, n, n, n, n, n, n
-1, 0, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
LT(4), LT(5), LT(5), LT(6), LT(6), LT(6), LT(6),
LT(7), LT(7), LT(7), LT(7), LT(7), LT(7), LT(7), LT(7)
};
register unsigned int t, tt; // temporaries
if (tt = v >> 16)
{
return (t = tt >> 8) ? 24 + LogTable256[t] : 16 + LogTable256[tt];
}
else
{
return (t = v >> 8) ? 8 + LogTable256[t] : LogTable256[v];
}
}
unsigned int numberOfBits(unsigned int i)
{
if (i == 0) {
return 1;
}
else {
return fastLog2(i) + 1;
}
}
unsigned int findEndLimit(unsigned int sx, unsigned int ex)
{
unsigned int sy = numberOfBits(sx);
unsigned int ey = numberOfBits(ex);
unsigned int mx;
unsigned int my;
if (sy == ey) // this also means sx == ex
return ex;
// assumes sy < ey
mx = (ex - sx) / 2 + sx; // will eq. sx for sx + 1 == ex
my = numberOfBits(mx);
while (ex - sx != 1) {
mx = (ex - sx) / 2 + sx; // will eq. sx for sx + 1 == ex
my = numberOfBits(mx);
if (my == ey) {
ex = mx;
ey = numberOfBits(ex);
}
else {
sx = mx;
sy = numberOfBits(sx);
}
}
return sx+1;
}
int main(void)
{
unsigned int a, b, m;
unsigned long l;
clock_t start, end;
l = 0;
a = 0;
b = UINT_MAX;
start = clock();
unsigned int i;
for (i = a; i < b; ++i) {
l += numberOfBits(i);
}
if (i == b) {
l += numberOfBits(i);
}
end = clock();
printf("Naive\n");
printf("Digits: %ld; Time: %fs\n",l, ((double)(end-start))/CLOCKS_PER_SEC);
l=0;
start = clock();
do {
m = findEndLimit(a, b);
l += (b-m + 1) * (unsigned long)numberOfBits(b);
b = m-1;
} while (b > a);
l += (b-a+1) * (unsigned long)numberOfBits(b);
end = clock();
printf("Binary search\n");
printf("Digits: %ld; Time: %fs\n",l, ((double)(end-start))/CLOCKS_PER_SEC);
}
Output
From 0 to UINT_MAX
$ ./main
Naive
Digits: 133143986178; Time: 25.722492s
Binary search
Digits: 133143986178; Time: 0.000025s
My findEndLimit can take long time in some edge cases:
From UINT_MAX/16+1 to UINT_MAX/8
$ ./main
Naive
Digits: 7784628224; Time: 1.651067s
Binary search
Digits: 7784628224; Time: 4.921520s
Conceptually, you would need to split the task to two subproblems -
1) find the sum of digits from 0..M, and from 0..N, then subtract.
2) find the floor(log2(x)), because eg for the number 77 the numbers 64,65,...77 all have 6 digits, the next 32 have 5 digits, the next 16 have 4 digits and so on, which makes a geometric progression.
Thus:
int digits(int a) {
if (a == 0) return 1; // should digits(0) be 0 or 1 ?
int b=(int)floor(log2(a)); // use any all-integer calculation hack
int sum = 1 + (b+1) * (a- (1<<b) +1); // added 1, due to digits(0)==1
while (--b)
sum += (b + 1) << b; // shortcut for (b + 1) * (1 << b);
return sum;
}
int digits_range(int a, int b) {
if (a <= 0 || b <= 0) return -1; // formulas work for strictly positive numbers
return digits(b)-digits(a-1);
}
As efficiency depends on the tools available, one approach would be doing it "analog":
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
unsigned long long pow2sum_min(unsigned long long n, long long unsigned m)
{
if (m >= n)
{
return 1;
}
--n;
return (2ULL << n) + pow2sum_min(n, m);
}
#define LN(x) (log2(x)/log2(M_E))
int main(int argc, char** argv)
{
if (2 >= argc)
{
fprintf(stderr, "%s a b\n", argv[0]);
exit(EXIT_FAILURE);
}
long a = atol(argv[1]), b = atol(argv[2]);
if (0L >= a || 0L >= b || b < a)
{
puts("Na ...!");
exit(EXIT_FAILURE);
}
/* Expand intevall to cover full dimensions: */
unsigned long long a_c = pow(2, floor(log2(a)));
unsigned long long b_c = pow(2, floor(log2(b+1)) + 1);
double log2_a_c = log2(a_c);
double log2_b_c = log2(b_c);
unsigned long p2s = pow2sum_min(log2_b_c, log2_a_c) - 1;
/* Integral log2(x) between a_c and b_c: */
double A = ((b_c * (LN(b_c) - 1))
- (a_c * (LN(a_c) - 1)))/LN(2)
+ (b+1 - a);
/* "Integer"-integral - integral of log2(x)'s inverse function (2**x) between log(a_c) and log(b_c): */
double D = p2s - (b_c - a_c)/LN(2);
/* Corrective from a_c/b_c to a/b : */
double C = (log2_b_c - 1)*(b_c - (b+1)) + log2_a_c*(a - a_c);
printf("Total used digits: %lld\n", (long long) ((A - D - C) +.5));
}
:-)
The main thing here is the number and kind of iterations done.
Number is
log(floor(b_c)) - log(floor(a_c))
times
doing one
n - 1 /* Integer decrement */
2**n + s /* One bit-shift and one integer addition */
for each iteration.
Here's an entirely look-up based approach. You don't even need the log2 :)
Algorithm
First we precompute interval limits where the number of bits would change and create a lookup table. In other words we create an array limits[2^n], where limits[i] gives us the biggest integer that can be represented with (i+1) bits. Our array is then {1, 3, 7, ..., 2^n-1}.
Then, when we want to determine the sum of bits for our range, we must first match our range limits a and b with the smallest index for which a <= limits[i] and b <= limits[j] holds, which will then tell us that we need (i+1) bits to represent a, and (j+1) bits to represent b.
If the indexes are the same, then the result is simply (b-a+1)*(i+1), otherwise we must separately get the number of bits from our value to the edge of same number of bits interval, and add up total number of bits for each interval between as well. In any case, simple arithmetic.
Code
#include <stdio.h>
#include <limits.h>
#include <time.h>
unsigned long bitsnumsum(unsigned int a, unsigned int b)
{
// generate lookup table
// limits[i] is the max. number we can represent with (i+1) bits
static const unsigned int limits[32] =
{
#define LTN(n) n*2u-1, n*4u-1, n*8u-1, n*16u-1, n*32u-1, n*64u-1, n*128u-1, n*256u-1
LTN(1),
LTN(256),
LTN(256*256),
LTN(256*256*256)
};
// make it work for any order of arguments
if (b < a) {
unsigned int c = a;
a = b;
b = c;
}
// find interval of a
unsigned int i = 0;
while (a > limits[i]) {
++i;
}
// find interval of b
unsigned int j = i;
while (b > limits[j]) {
++j;
}
// add it all up
unsigned long sum = 0;
if (i == j) {
// a and b in the same range
// conveniently, this also deals with j == 0
// so no danger to do [j-1] below
return (i+1) * (unsigned long)(b - a + 1);
}
else {
// add sum of digits in range [a, limits[i]]
sum += (i+1) * (unsigned long)(limits[i] - a + 1);
// add sum of digits in range [limits[j], b]
sum += (j+1) * (unsigned long)(b - limits[j-1]);
// add sum of digits in range [limits[i], limits[j]]
for (++i; i<j; ++i) {
sum += (i+1) * (unsigned long)(limits[i] - limits[i-1]);
}
return sum;
}
}
int main(void)
{
clock_t start, end;
unsigned int a=0, b=UINT_MAX;
start = clock();
printf("Sum of binary digits for numbers in range "
"[%u, %u]: %lu\n", a, b, bitsnumsum(a, b));
end = clock();
printf("Time: %fs\n", ((double)(end-start))/CLOCKS_PER_SEC);
}
Output
$ ./lookup
Sum of binary digits for numbers in range [0, 4294967295]: 133143986178
Time: 0.000282s
Algorithm
The main idea is to find the n2 = log2(x) rounded down. That is the number of digits in x. Let pow2 = 1 << n2. n2 * (pow2 - x + 1) is the number of digits in the values [x...pow2]. Now find the sun of digits in the powers of 2 from 1 to n2-1
Code
I am certain various simplifications can be made.
Untested code. Will review later.
// Let us use unsigned for everything.
unsigned ulog2(unsigned value) {
unsigned result = 0;
if (0xFFFF0000u & value) {
value >>= 16; result += 16;
}
if (0xFF00u & value) {
value >>= 8; result += 8;
}
if (0xF0u & value) {
value >>= 4; result += 4;
}
if (0xCu & value) {
value >>= 2; result += 2;
}
if (0x2 & value) {
value >>= 1; result += 1;
}
return result;
}
unsigned bit_count_helper(unsigned x) {
if (x == 0) {
return 1;
}
unsigned n2 = ulog2(x);
unsigned pow2 = 1u << n;
unsigned sum = n2 * (pow2 - x + 1u); // value from pow2 to x
while (n2 > 0) {
// ... + 5*16 + 4*8 + 3*4 + 2*2 + 1*1
pow2 /= 2;
sum += n2 * pow2;
}
return sum;
}
unsigned bit_count(unsigned a, unsigned b) {
assert(a < b);
return bit_count_helper(b - 1) - bit_count_helper(a);
}
For this problem your solution is the simplest, the one called "naive" where you look for every element in the sequence or in your case interval for check something or execute operations.
Naive Algorithm
Assuming that a and b are positive integers with b greater than a let's call the dimension/size of the interval [a,b], n = (b-a).
Having our number of elements n and using some notations of algorithms (like big-O notation link), the worst case cost is O(n*(numberOfBits_cost)).
From this we can see that we can speed up our algorithm by using a faster algorithm for computing numberOfBits() or we need to find a way to not look at every element of the interval that costs us n operations.
Intuition
Now looking at a possible interval [6,14] you can see that for 6 and 7 we need 3 digits, with 4 need for 8,9,10,11,12,13,14. This results in calling numberOfBits() for every number that use the same number of digits to be represented, while the following multiplication operation would be faster:
(number_in_subinterval)*digitsForThisInterval
((14-8)+1)*4 = 28
((7-6)+1)*3 = 6
So we reduced the looping on 9 elements with 9 operations to only 2.
So writing a function that use this intuition will give us a more efficient in time, not necessarily in memory, algorithm. Using your numberOfBits() function I have created this solution:
int intuitionSol(int a, int b){
int digitsForA = numberOfBits(a);
int digitsForB = numberOfBits(b);
if(digitsForA != digitsForB){
//because a or b can be that isn't the first or last element of the
// interval that a specific number of digit can rappresent there is a need
// to execute some correction operation before on a and b
int tmp = pow(2,digitsForA) - a;
int result = tmp*digitsForA; //will containt the final result that will be returned
int i;
for(i = digitsForA + 1; i < digitsForB; i++){
int interval_elements = pow(2,i) - pow(2,i-1);
result = result + ((interval_elements) * i);
//printf("NumOfElem: %i for %i digits; sum:= %i\n", interval_elements, i, result);
}
int tmp1 = ((b + 1) - pow(2,digitsForB-1));
result = result + tmp1*digitsForB;
return result;
}
else {
int elements = (b - a) + 1;
return elements * digitsForA; // or digitsForB
}
}
Let's look at the cost, this algorithm costs is the cost of doing correction operation on a and b plus the most expensive one that of the for-loop. In my solution however I'm not looping over all elements but only on numberOfBits(b)-numberOfBits(a) that in the worst case, when [0,n], become log(n)-1 thats equivalent to O(log n).
To resume we passed from a linear operations cost O(n) to a logartmic one O(log n) in the worst case. Look on this diagram the diferinces between the two.
Note
When I talk about interval or sub-interval I refer to the interval of elements that use the same number of digits to represent the number in binary.
Following there are some output of my tests with the last one that shows the difference:
Considered interval is [0,4]
YourSol: 9 in time: 0.000015s
IntuitionSol: 9 in time: 0.000007s
Considered interval is [0,0]
YourSol: 1 in time: 0.000005s
IntuitionSol: 1 in time: 0.000005s
Considered interval is [4,7]
YourSol: 12 in time: 0.000016s
IntuitionSol: 12 in time: 0.000005s
Considered interval is [2,123456]
YourSol: 1967697 in time: 0.005010s
IntuitionSol: 1967697 in time: 0.000015s

Finding the last 10 digits of 2^n

So i'm supposed to find out the last 10 digits of 2^n(0<=n<=100) where n is the input. I found a method to handle large numbers but the program fails when n>64. Any leads on how to go about with this would be appreciated.
#include<stdio.h>
#include<math.h>
/* Iterative Function to calculate (x^y)%p in O(log y) */
int power(long long int x, long long int y, long long int p)
{
long long int res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0) {
// If y is odd, multiply x with result
if (y & 1)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
// C function to print last 10 digits of a^b
void printLastDigits(long long int a,long long int b)
{
long long int temp = pow(10,10);
// Calling modular exponentiation
temp = power(a, b, temp);
if (temp)
printf("%d",temp);
}
int main()
{
long long int n;
scanf("%d",&n);
printLastDigits(2,n);
return 0;
}
You don't need to worry about the 'high' bits, since multiplication by 2 left shifts them out of range of the lower part of the product you're interesting in. Just be sure you're using the unsigned long long type (of at least 64 bits) to hold integral types that are wide enough, e.g.,
#include <inttypes.h>
#include <stdio.h>
void low_digits (unsigned int n)
{
unsigned long long base = 2, modulus = 10000000000ULL;
for (unsigned int i = 1; i <= n; i++)
{
fprintf(stdout, "2^%u mod 10^10 = %llu\n", i, base);
base = (base * 2) % modulus;
}
}
You can test 2^1000 with a bignum calculator:
10715086071862673209484250490600018105614048117055336074437503883703\
51051124936122493198378815695858127594672917553146825187145285692314\
04359845775746985748039345677748242309854210746050623711418779541821\
53046474983581941267398767559165543946077062914571196477686542167660\
429831652624386837205668069376
while n = 1000 above yields: 5668069376
Others have noted, that this is a naive method, and modular exponentiation is far more efficient for sufficiently large values of (n). Unfortunately, this is going to require products that exceed the range of an unsigned 64-bit value, so unless you're prepared to implement [hi64][lo64] multi-precision mul / mod operations, it's probably beyond the scope of your task.
Fortunately, later versions of gcc and clang do provide an extended 128 bit integral type:
#include <inttypes.h>
#include <stdio.h>
void low_digits (unsigned int n)
{
unsigned long long base = 2, modulus = 10000000000ULL;
__extension__ unsigned __int128 u = 1, w = base;
while (n != 0)
{
if ((n & 0x1) != 0)
u = (u * w) % modulus; /* (mul-reduce) */
if ((n >>= 1) != 0)
w = (w * w) % modulus; /* (sqr-reduce) */
}
base = (unsigned long long) u;
fprintf(stdout, "2^%u mod 10^10 = %llu\n", n, base);
}
The following uses strings to perform the multiplication:
void lastdigits(char digits[11], int n)
{
int i, j, x, carry;
for (i=0; i<n;i++) {
for (j=9, carry=0; j>=0; j--) {
x= digits[j]-'0';
x *= 2;
x += carry;
if (x>9) {carry= 1; x -= 10;}
else carry= 0;
digits[j]= x+'0';
}
}
}
void test(void)
{
char digits[11];
strcpy(digits,"0000000001");
lastdigits(digits,10);
printf("%s\n",digits);
strcpy(digits,"0000000001");
lastdigits(digits,20);
printf("%s\n",digits);
strcpy(digits,"0000000001");
lastdigits(digits,100);
printf("%s\n",digits);
}
Output:
0000001024
0001048576
6703205376
Since the other answers you've received don't actually show what you're doing wrong:
x = (x * x) % p;
You're assuming that x * x still fits in long long int. But if x is 0x100000000 (4294967296, for 10 decimal digits) and long long int is 64 bits, then it will not fit.
Either:
You need a way to accurately multiply two arbitrary 10-digit numbers. The result may have 20 digits and may not fit in long long int or even unsigned long long int. This means you'd need to use some bigint library or implement something like that yourself.
Or:
You need to avoid multiplying multiple possibly-10-digit numbers.
The answer you've accepted opts for simple repeated multiplication by 2. This is sufficient for your problem now, but beware that this does significantly increase the complexity if you want to allow very large exponents.
Let's say you are finding the last digit of 2^n, you just need to consider last digit and ignore every other digit
1. 2*2 = 4
2. 4*2 = 8
3. 8*2 = 16 (ignore last-but-one digit i.e 1)
4. 6*2 = 12 (ignore last-but-one digit i.e 1)
5. 2*2 = 4
6. 4*2 = 8
7. 8*2 = 16 (ignore last-but-one digit i.e 1)
8. 6*2 = 12 (ignore last-but-one digit i.e 1)
9. 2*2 = 4
... n-1 iterations
To find the last 2 digits of 2^n, ignore all digits except last 2 digits.
1. 2*2 = 4
2. 4*2 = 8
3. 8*2 = 16
4. 16*2 = 32
5. 32*2 = 64
6. 64*2 = 128 (Consider last 2 digits)
7. 28*2 = 56
8. 56*2 = 112 (Consider last 2 digits)
9. 12*2 = 24
... n-1 iterations
Similarly, to find the last 10 digits of 2^n, consider just last 10 digits at each iteration and repeat it for n-1 iterations.
Note:
With this approach, the biggest number you'll get during the calculation can be of 11 digits ~ 10^11, while for a 64-bit machine the max value is ~ 2^64 = ~ 10^18

Why does x * x is computed wrong?

I am trying find the nth number in the array with first element 3 and next elements built after the following rule: v[i] = (v[i - 1] * v[i - 1] / (i + 2) + v[i - 1] * i + i + 1) % 666013. Where v is the array. I think my code is working, but I have the next problem. For n = 7 first y = 600198, and at the next step y = 3755353636, but I expected y at that step to be 360237639204. Why does that happen? I am using Visual Studio 2017 on Windows 7 ultimate x64.
#define _CRT_SECURE_NO_DEPRECATE
#include "stdafx.h"
#include <stdlib.h>
#include <stdio.h>
#include <stdbool.h>
int main()
{
FILE *input;
if ((input = fopen("C:\\Users\\HP\\Documents\\Visual Studio 2017\\Projects\\ConsoleApplication2\\hex.in", "r")) == NULL)
{
perror("Error opening hex.in\n");
return 1;
}
FILE *output;
if ((output = fopen("hex.out", "w+")) == NULL)
{
perror("Error opening hex.out\n");
return 1;
}
int n;
fscanf_s(input, "%d", &n);
int i = 1;
unsigned long x = 3;
unsigned long y = 8;
bool found = false;
while(!found)
{
if (i == n)
{
found = true;
fprintf(output, "%d", x);
}
i++;
x = y;
y = x * x;//for i = 7 I expect y to be 360237639204 after this step
y /= (i + 2);
y += x * i;
y += i + 1;
y %= 666013;
}
fclose(input);
fclose(output);
return 0;
}
The Visual Studio C++ compiler still have long (and unsigned long too of course) as a 32-bit datatype. Use long long if you want a (at least) 64-bit integer datatype.
You have integer overflow, try to use 64-bit variable.
unsigned long can accept number up to 2^32-1 which is less than you have expected and that's why you have overflow.
Use 64-bit variable instead:
unsigned long long x = 3;
A 32-bit integer can represent 2^32-1 (around 4 billion) values. The result you desire is around 360 billion. So you can roughly imagine it as having wrapped around 360 / 4 = 90 times and was left with a remainder of around 3 billion, the result you're actually seeing.
Use a 64-bit integer instead, as others have stated.
unsigned long long x = 3;
unsigned long long y = 8;

Decimal to octal in C

I have just begun teaching myself C out of K.N King's C Programming: A Modern Approach (2ndEdn).
I'm enjoying it, but am hoping to post the odd question here for advice if appropriate because unfortunately I don't have a tutor and some bits raise more questions then they answer!
I'm doing a question on taking an integer entered and displaying it in octal. It says there is an easy way to do it, but that comes later in the book. I have come up with the following:
// Convert a number to octal
int n, n2, n3, n4, n5, n6;
printf("Enter a number between 0 and 32767: ");
scanf("%d", &n);
n6 = n % 8;
n5 = (n / 8) % 8;
n4 = ((n / 8) / 8) % 8;
n3 = (((n / 8) / 8) / 8) % 8;
n2 = ((((n / 8) / 8) / 8) / 8) % 8;
printf("%d%d%d%d%d", n2, n3, n4, n5, n6);
It works OK, but I'm not good at math and was wondering if there is a more efficient way of doing this or have I done it the only way possible...
If anyone else has the book it's Q4 p.71.
Thanks for your time.
Andrew
P.S I did look in the search engine but couldn't find anything that was doing it this 'slower' way!
Everyone is right in saying that there's a built-in way to do that with printf. But what about doing it yourself?
The first thing that came to mind is that one octal digit is exactly three bits. Therefore you can do the conversion this way:
Loop while n != 0
Isolate the leftmost 3 bits of n into d and print d
Shift n 3 bits to the left
The code is trivial, but I 'm not providing it so you can do it yourself (you will need to be familiar with the bitwise and shift operators in order to do it).
The easy way is probably to use printf()'s %o format specifier:
scanf("%d", &n);
printf("%o", n);
Others have posted the real, production code answer, and now I see from your comments that you haven't done loops yet. Perhaps your book is trying to teach you about recursion:
void print_oct(int n)
{
if (n != 0) {
print_oct(n / 8);
printf("%d", n % 8);
}
}
This works for n > 0.
With loops you can roll up your five very similar lines like this:
for (int d = 8 * 8 * 8 * 8; d > 0; d /= 8)
printf("%d", n / d % 8);
printf("\n");
d will start at 8 * 8 * 8 * 8, which is the divisor you use for n2 and then step through 8 * 8 * 8, 8 * 8, 8 and finally 1, which is the divisor for n6, printing each digit along the way.
A good compiler will actually optimize this by unrolling it back into five lines, so you'll get almost the same thing you started with. The advantage of writing it as a loop is that you can't make a mistake in just one of the lines.
The compiler will also take care of replacing divisions by 8 with shifts by 3 bits. Both give the same result in binary, but the latter is faster.
/* Converts a positive base_10 into base_b */
int DecimalToBase(int n, int b)
{
int rslt=0, digitPos=1;
while (n)
{
rslt += (n%b)*digitPos;
n /= b;
digitPos *= 10;
}
return rslt;
}
Use %o format specifier inside printf
printf("Enter a number between 0 and 32767: ");
scanf("%d", &n);
printf("%o", n);
Since only basics are introduced you don't want (at least at this point) to use functions, loops, bitwise operators, %o format specifier and all that stuff. Here is my basic solution:
int n, d1, d2, d3, d4, d5, o;
printf("Enter a number between 0 and 32767: ");
scanf("%d", &n);
d5 = n % 8;
n /= 8;
d4 = n % 8;
n /= 8;
d3 = n % 8;
n /= 8;
d2 = n % 8;
n /= 8;
d1 = n % 8;
o = 10000 * d1 + 1000 * d2 + 100 * d3 + 10 * d4 + d5;
printf("In octal, your number is: %.5d\n", o);
Note that since n is not needed in output, you can modify (divide) it for every step (thus saving divides, which are computationally and relatively expensive). You are safe up to 32767 (in octal: 77777), as 32768 (8*8*8*8*8 = 8^5 = (2^3)^5 = 2^15) is the first number, that requires six digits in octal: 100000.
This o variable is not really needed, morever it will not work when int is signed 16-bit (on some ancient system), so from this point it's better to just print separate digits.
Existing answers aren't clean enough for my liking. Here's mine:
#include <stdio.h>
#define OCTALBASE 8
#define OCTALSIZE 8
int main(int argc, char **argv) {
int indecimal = 1337;
char output[OCTALSIZE + 1];
output[OCTALSIZE] = '\0';
int outindex = OCTALSIZE;
int outdigit = 0;
int outvalue = indecimal;
while (--outindex >= 0) {
outdigit = outvalue % OCTALBASE;
if (outvalue > 0 || outdigit > 0)
{ output[outindex] = '0' + outdigit; }
else { output[outindex] = ' '; }
outvalue /= OCTALBASE;
}
fprintf(stdout, "{ DEC: %8d, OCT: %s }\n", indecimal, output);
fflush(stdout);
return 0;
}
Result:
{ DEC: 1337, OCT: 2471 }
Convert Decimal to Octal in C Language
#include<stdio.h>
#include<conio.h>
void main()
{
A:
long int n,n1,m=1,rem,ans=0;
clrscr();
printf("\nEnter Your Decimal No :: ");
scanf("%ld",&n);
n1=n;
while(n>0)
{
rem=n%8;
ans=(rem*m)+ans;
n=n/8;
m=m*10;
}
printf("\nYour Decimal No is :: %ld",n1);
printf("\nConvert into Octal No is :: %ld",ans);
printf("\n\nPress 0 to Continue...");
if(getch()=='0')
goto A;
printf("\n\n\n\tThank You");
getch();
}

Subtraction without minus sign in C

How can I subtract two integers in C without the - operator?
int a = 34;
int b = 50;
You can convert b to negative value using negation and adding 1:
int c = a + (~b + 1);
printf("%d\n", c);
-16
This is two's complement sign negation. Processor is doing it when you use '-' operator when you want to negate value or subtrackt it.
Converting float is simpler. Just negate first bit (shoosh gave you example how to do this).
EDIT:
Ok, guys. I give up. Here is my compiler independent version:
#include <stdio.h>
unsigned int adder(unsigned int a, unsigned int b) {
unsigned int loop = 1;
unsigned int sum = 0;
unsigned int ai, bi, ci;
while (loop) {
ai = a & loop;
bi = b & loop;
ci = sum & loop;
sum = sum ^ ai ^ bi; // add i-th bit of a and b, and add carry bit stored in sum i-th bit
loop = loop << 1;
if ((ai&bi)|(ci&ai)|(ci&bi)) sum = sum^loop; // add carry bit
}
return sum;
}
unsigned int sub(unsigned int a, unsigned int b) {
return adder(a, adder(~b, 1)); // add negation + 1 (two's complement here)
}
int main() {
unsigned int a = 35;
unsigned int b = 40;
printf("%u - %u = %d\n", a, b, sub(a, b)); // printf function isn't compiler independent here
return 0;
}
I'm using unsigned int so that any compiler will treat it the same.
If you want to subtract negative values, then do it that way:
unsgined int negative15 = adder(~15, 1);
Now we are completly independent of signed values conventions. In my approach result all ints will be stored as two's complement - so you have to be careful with bigger ints (they have to start with 0 bit).
Pontus is right, 2's complement is not mandated by the C standard (even if it is the de facto hardware standard). +1 for Phil's creative answers; here's another approach to getting -1 without using the standard library or the -- operator.
C mandates three possible representations, so you can sniff which is in operation and get a different -1 for each:
negation= ~1;
if (negation+1==0) /* one's complement arithmetic */
minusone= ~1;
else if (negation+2==0) /* two's complement arithmetic */
minusone= ~0;
else /* sign-and-magnitude arithmetic */
minusone= ~0x7FFFFFFE;
r= a+b*minusone;
The value 0x7FFFFFFFE would depend on the width (number of ‘value bits’) of the type of integer you were interested in; if unspecified, you have more work to find that out!
+ No bit setting
+ Language independent
+ Can be adjusted for different number types (int, float, etc)
- Almost certainly not your C homework answer (which is likely to be about bits)
Expand a-b:
a-b = a + (-b)
= a + (-1).b
Manufacture -1:
float: pi = asin(1.0);
(with minusone_flt = sin(3.0/2.0*pi);
math.h) or = cos(pi)
or = log10(0.1)
complex: minusone_cpx = (0,1)**2; // i squared
integer: minusone_int = 0; minusone_int--; // or convert one of the floats above
+ No bit setting
+ Language independent
+ Independent of number type (int, float, etc)
- Requires a>b (ie positive result)
- Almost certainly not your C homework answer (which is likely to be about bits)
a - b = c
restricting ourselves to the number space 0 <= c < (a+b):
(a - b) mod(a+b) = c mod(a+b)
a mod(a+b) - b mod(a+b) = c mod(a+b)
simplifying the second term:
(-b).mod(a+b) = (a+b-b).mod(a+b)
= a.mod(a+b)
substituting:
a.mod(a+b) + a.mod(a+b) = c.mod(a+b)
2a.mod(a+b) = c.mod(a+b)
if b>a, then b-a>0, so:
c.mod(a+b) = c
c = 2a.mod(a+b)
So, if a is always greater than b, then this would work.
Given that encoding integers to support two's complement is not mandated in C, iterate until done. If they want you to jump through flaming hoops, no need to be efficient about it!
int subtract(int a, int b)
{
if ( b < 0 )
return a+abs(b);
while (b-- > 0)
--a;
return a;
}
Silly question... probably silly interview!
For subtracting in C two integers you only need:
int subtract(int a, int b)
{
return a + (~b) + 1;
}
I don't believe that there is a simple an elegant solution for float or double numbers like for integers. So you can transform your float numbers in arrays and apply an algorithm similar with one simulated here
If you want to do it for floats, start from a positive number and change its sign bit like so:
float f = 3;
*(int*)&f |= 0x80000000;
// now f is -3.
float m = 4 + f;
// m = 1
You can also do this for doubles using the appropriate 64 bit integer. in visual studio this is __int64 for instance.
I suppose this
b - a = ~( a + ~b)
Assembly (accumulator) style:
int result = a;
result -= b;
As the question asked for integers not ints, you could implement a small interpreter than uses Church numerals.
Create a lookup table for every possible case of int-int!
Not tested. Without using 2's complement:
#include <stdlib.h>
#include <stdio.h>
int sillyNegate(int x) {
if (x <= 0)
return abs(x);
else {
// setlocale(LC_ALL, "C"); // if necessary.
char buffer[256];
snprintf(buffer, 255, "%c%d", 0x2d, x);
sscanf(buffer, "%d", &x);
return x;
}
}
Assuming the length of an int is much less than 255, and the snprintf/sscanf round-trip won't produce any unspecified behavior (right? right?).
The subtraction can be computed using a - b == a + (-b).
Alternative:
#include <math.h>
int moreSillyNegate(int x) {
return x * ilogb(0.5); // ilogb(0.5) == -1;
}
This would work using integer overflow:
#include<limits.h>
int subtractWithoutMinusSign(int a, int b){
return a + (b * (INT_MAX + INT_MAX + 1));
}
This also works for floats (assuming you make a float version…)
For the maximum range of any data type , one's complement provide the negative value decreased by 1 to any corresponding value. ex:
~1 --------> -2
~2---------> -3
and so on... I will show you this observation using little code snippet
#include<stdio.h>
int main()
{
int a , b;
a=10;
b=~a; // b-----> -11
printf("%d\n",a+~b+1);// equivalent to a-b
return 0;
}
Output: 0
Note : This is valid only for the range of data type. means for int data type this rule will be applicable only for the value of range[-2,147,483,648 to 2,147,483,647].
Thankyou .....May this help you
Iff:
The Minuend is greater or equal to 0, or
The Subtrahend is greater or equal to 0, or
The Subtrahend and the Minuend are less than 0
multiply the Minuend by -1 and add the result to the Subtrahend:
SUB + (MIN * -1)
Else multiply the Minuend by 1 and add the result to the Subtrahend.
SUB + (MIN * 1)
Example (Try it online):
#include <stdio.h>
int subtract (int a, int b)
{
if ( a >= 0 || b >= 0 || ( a < 0 && b < 0 ) )
{
return a + (b * -1);
}
return a + (b * 1);
}
int main (void)
{
int x = -1;
int y = -5;
printf("%d - %d = %d", x, y, subtract(x, y) );
}
Output:
-1 - -5 = 4
int num1, num2, count = 0;
Console.WriteLine("Enter two numebrs");
num1 = int.Parse(Console.ReadLine());
num2 = int.Parse(Console.ReadLine());
if (num1 < num2)
{
num1 = num1 + num2;
num2 = num1 - num2;
num1 = num1 - num2;
}
for (; num2 < num1; num2++)
{
count++;
}
Console.WriteLine("The diferrence is " + count);
void main()
{
int a=5;
int b=7;
while(b--)a--;
printf("sud=%d",a);
}

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