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The function should take the address of the integer and modify it by inserting zeros between its digits. For example:
insert_zeros(3) //3
insert_zeros(39) //309
insert_zeros(397) //30907
insert_zeros(3976) //3090706
insert_zeros(39765) //309070605
My code:
#include <stdio.h>
#include <math.h>
void insert_zeros(int* num);
int main() {
int num;
printf("Enter a number:");
scanf("%d", num);
insert_zeros(&num);
printf("Number after inserting zeros: %d", num);
return 0;
}
void insert_zeros(int* num){
int count = 0;
int tmp = *num;
//Count the number of digits in the number
while(tmp != 0){
tmp /= 10;
count++;
}
//calculating the coefficient by which I will divide the number to get its digits one by one
int divider = (int)pow(10, count-1);
int multiplier;
tmp = *num;
*num = 0;
/*
The point at which I'm stuck
Here I tried to calculate the degree for the number 10
(my thought process and calculations are provided below)
*/
(count >= 3)? count += (count/2): count;
//the main loop of assembling the required number
while (count >= 0){
multiplier = (int)pow(10, count); //calculating a multiplier
*num += (tmp / divider) * multiplier; //assembling the required number
tmp %= divider; //removing the first digit of the number
divider /= 10; //decreasing divider
count -= 2; //decreasing the counter,
//which is also a power of the multiplier (witch is 10)
}
}
My idea consists of the following formula:
For number "3" I shold get "30" and it will be:
30 = (3 * 10^1) - the power is a counter for number "3" that equals 1.
For number "39" it will be "309":
309 = (3 * 10^2) + (9 * 10^1)
For number "397" it will be "30907":
30907 = (3 * 10^4) + (9 * 10^2) + (7 * 10^0)
For number "3976" it will be "3090706":
3090706 = (3 * 10^6) + (9 * 10^4) + (7 * 10^2) + (6 * 10^0) - with each iteration power is decreasing by 2
For number "39765" it will be "309070605":
309070605 = (3 * 10^8) + (9 * 10^6) + (7 * 10^4) + (6 * 10^2) + (5 * 10^0)
And so on...
For a 3-digit number, the start power should be 4, for a 4-digit number power should be 6, for a 5-digit it should be 8, for 6-digit it should be 10, etc.
That algorithm works until it takes a 5-digit number. It outputs a number like "30907060" with an extra "0" at the end.
And the main problem is in that piece of code (count >= 3)? count += (count/2): count;, where I tried to calculate the right power for the first iterating through the loop. It should give the right number to which will be added all the following numbers. But it only works until it gets a 5-digit number.
To be honest, so far I don't really understand how it can be realized. I would be very grateful if someone could explain how this can be done.
As noted in comments, your use of scanf is incorrect. You need to pass a pointer as the second argument.
#include <stdio.h>
#include <math.h>
int main(void) {
int num;
scanf("%d", &num);
int num2 = 0;
int power = 0;
while (num > 0) {
num2 += (num % 10) * (int)pow(10, power);
num /= 10;
power += 2;
}
printf("%d\n", num2);
return 0;
}
There's an easy recursive formula for inserting zeros: IZ(n) = 100*IZ(n/10) + n%10.
That gives a very concise solution -- here the test cases are more code than the actual function itself.
#include <stdio.h>
#include <stdint.h>
uint64_t insert_zeros(uint64_t n) {
return n ? (100 * insert_zeros(n / 10) + n % 10) : 0;
}
int main(int argc, char **argv) {
int tc[] = {1, 12, 123, 9854, 12345, 123450};
for (int i = 0; i < sizeof(tc)/sizeof(*tc); i++) {
printf("%d -> %lu\n", tc[i], insert_zeros(tc[i]));
}
}
Output:
1 -> 1
12 -> 102
123 -> 10203
9854 -> 9080504
12345 -> 102030405
123450 -> 10203040500
Adapting some code just posted for another of these silly exercises:
int main() {
int v1 = 12345; // I don't like rekeying data. Here's the 'seed' value.
printf( "Using %d as input\n", v1 );
int stack[8] = { 0 }, spCnt = -1;
// Peel off each 'digit' right-to-left, pushing onto a stack
while( v1 )
stack[ ++spCnt ] = v1%10, v1 /= 10;
if( spCnt == 0 ) // Special case for single digit seed.
v1 = stack[ spCnt ] * 10;
else
// multiply value sofar by 100, and add next digit popped from stack.
while( spCnt >= 0 )
v1 = v1 * 100 + stack[ spCnt-- ];
printf( "%d\n", v1 );
return 0;
}
There's a ceiling to how big a decimal value can be stored in an int. If you want to start to play with strings of digits, that is another matter entirely.
EDIT: If this were in Java, this would be a solution, but the problem is in C, which I'm not sure if this can convert to C.
This may be a lot easier if you first convert the integer to a string, then use a for loop to add the zeros, then afterward reconvert to an integer. Example:
int insert_zeros(int num) {
String numString = Integer.toString(num);
String newString = "";
int numStringLength = numString.length();
for (int i = 0; i < numStringLength; i++) {
newString += numString[i];
// Only add a 0 if it's not the last digit (with exception of 1st digit)
if (i < numStringLength - 1 || i == 0) newString += '0';
}
return Integer.parseInt(newString);
}
I think this should give you your desired effect. It's been a little bit since I've worked with Java (I'm currently doing JavaScript), so I hope there's no syntax errors, but the logic should all be correct.
I'm trying to code a program that can tell apart real and fake credit card numbers using Luhn's algorithm in C, which is
Multiply every other digit by 2, starting with the number’s
second-to-last digit, and then add those products’ digits together.
Add the sum to the sum of the digits that weren’t multiplied by 2.
If the total’s last digit is 0 (or, put more formally, if the total
modulo 10 is congruent to 0), the number is valid!
Then I coded something like this (I already declared all the functions at the top and included all the necessary libraries)
//Luhn's Algorithm
int luhn(long z)
{
int c;
return c = (sumall(z)-sumodd(z)) * 2 + sumaodd(z);
}
//sum of digits in odd position starting from the end
int sumodd(long x)
{
int a;
while(x)
{
a = a + x % 10;
x /= 100;
}
return a;
}
//sum of all digits
int sumall(long y)
{
int b;
while(y)
{
b = b + y % 10;
y /= 10;
}
return b;
}
But somehow it always gives out the wrong answer even though there's no error or bug detected. I came to notice that it works fine when my variable z stands alone, but when it's used multiple times in the same line of code with different functions, their values get messed up (in function luhn). I'm writing this to ask for any fix I can make to make my code run correctly as I intended.
I'd appreciate any help as I'm very new to this, and I'm not a native English speaker so I may have messed up some technical terms, but I hope you'd be able to understand my concerns.
sumall is wrong.
It should be sumeven from:
Add the sum to the sum of the digits that weren’t multiplied by 2.
Your sumall is summing all digits instead of the non-odd (i.e. even) digits.
You should do the * 2 inside sumodd as it should not be applied to the other [even] sum. And, it should be applied to the individual digits [vs the total sum].
Let's start with a proper definition from https://en.wikipedia.org/wiki/Luhn_algorithm
The check digit is computed as follows:
If the number already contains the check digit, drop that digit to form the "payload." The check digit is most often the last digit.
With the payload, start from the rightmost digit. Moving left, double the value of every second digit (including the rightmost digit).
Sum the digits of the resulting value in each position (using the original value where a digit did not get doubled in the previous step).
The check digit is calculated by 10 − ( s mod 10 )
Note that if we have a credit card of 9x where x is the check digit, then the payload is 9.
The correct [odd] sum for that digit is: 9 * 2 --> 18 --> 1 + 8 --> 9
But, sumodd(9x) * 2 --> 9 * 2 --> 18
Here's what I came up with:
// digsum -- calculate sum of digits
static inline int
digsum(int digcur)
{
int sum = 0;
for (; digcur != 0; digcur /= 10)
sum += digcur % 10;
return sum;
}
// luhn -- luhn's algorithm using digits array
int
luhn(long z)
{
char digits[16] = { 0 };
// get check digit and remove from "payload"
int check_expected = z % 10;
z /= 10;
// split into digits (we use little-endian)
int digcnt = 0;
for (digcnt = 0; z != 0; ++digcnt, z /= 10)
digits[digcnt] = z % 10;
int sum = 0;
for (int digidx = 0; digidx < digcnt; ++digidx) {
int digcur = digits[digidx];
if ((digidx & 1) == 0)
sum += digsum(digcur * 2);
else
sum += digcur;
}
int check_actual = 10 - (sum % 10);
return (check_actual == check_expected);
}
// luhn -- luhn's algorithm using long directly
int
luhn2(long z)
{
// get check digit and remove from "payload"
int check_expected = z % 10;
z /= 10;
int sum = 0;
for (int digidx = 0; z != 0; ++digidx, z /= 10) {
int digcur = z % 10;
if ((digidx & 1) == 0)
sum += digsum(digcur * 2);
else
sum += digcur;
}
int check_actual = 10 - (sum % 10);
return (check_actual == check_expected);
}
You've invoked undefined behavior by not initializing a few local variables in your functions, for instance you can remove your undefined behaviour in sumodd() by initializing a to zero like so:
//sum of digits in odd position starting from the end
int sumodd(long x)
{
int a = 0; //Initialize
while(x)
{
a += x % 10; //You can "a += b" instead of "a = a + b"
x /= 100;
}
return a;
}
It's also important to note that long is only required to be a minimum of 4-bytes wide, so it is not guaranteed to be wide enough to represent a decimal-16-digit-integer. Using long long solves this problem.
Alternatively you may find this problem much easier to solve by treating your credit card number as a char[] instead of an integer type altogether, for instance if we assume a 16-digit credit card number:
int luhn(long long z){
char number[16]; //Convert CC number to array of digits and store them here
for(int c = 0; c < 16; ++c){
number[c] = z % 10; //Last digit is at number[0], first digit is at number[15]
z /= 10;
}
int sum = 0;
for(int c = 0; c < 16; c += 2){
sum += number[c] + number[c + 1] * 2; //Sum the even digits and the doubled odd digits
}
return sum;
}
...and you could skip the long long to char[] translation part altogether if you treat the credit card number as an array of digits in the whole program
This expression:
(sumall(z)-sumodd(z)) * 2 + sumall(z);
Should be:
((sumall(z)-sumodd(z)) * 2 + sumodd(z))%10;
Based on your own definition.
But how about:
(sumall(z) * 2 - sumodd(z))%10
If you're trying to be smart and base off sumall(). You don't need to call anything twice.
Also you don't initialise your local variables. You must assign variables values before using them in C.
Also you don't need the local variable c in the luhn() function. It's harmless but unnecessary.
As others mention in a real-world application we can't recommend enough that such 'codes' are held in a character array. The amount of grief caused by people using integer types to represent digit sequence 'codes' and identifiers is vast. Unless a variable represents a numerical quantity of something, don't represent it as an arithmetic type. More issue has been caused in my career by that error than people trying to use double to represent monetary amounts.
#include <stdio.h>
//sum of digits in odd position starting from the end
int sumodd(long x)
{
int a=0;
while(x)
{
a = a + x % 10;
x /= 100;
}
return a;
}
//sum of all digits
int sumall(long y)
{
int b=0;
while(y)
{
b = b + y % 10;
y /= 10;
}
return b;
}
//Luhn's Algorithm
int luhn(long z)
{
return (sumall(z)*2-sumodd(z))%10;
}
int check_luhn(long y,int expect){
int result=luhn(y);
if(result==expect){
return 0;
}
return 1;
}
int check_sumodd(long y,int expect){
int result=sumodd(y);
if(result==expect){
return 0;
}
return 1;
}
int check_sumall(long y,int expect){
int result=sumall(y);
if(result==expect){
return 0;
}
return 1;
}
int main(void) {
int errors=0;
errors+=check_sumall(1,1);
errors+=check_sumall(12,3);
errors+=check_sumall(123456789L,45);
errors+=check_sumall(4273391,4+2+7+3+3+9+1);
errors+=check_sumodd(1,1);
errors+=check_sumodd(91,1);
errors+=check_sumodd(791,8);
errors+=check_sumodd(1213191,1+1+1+1);
errors+=check_sumodd(4273391,15);
errors+=check_luhn(1234567890,((9+7+5+3+1)*2+(0+8+6+4+2))%10);
errors+=check_luhn(9264567897,((9+7+5+6+9)*2+(7+8+6+4+2))%10);
if(errors!=0){
printf("*ERRORS*\n");
}else{
printf("Success\n");
}
return 0;
}
#include <stdio.h>
#include <string.h>
int main() {
int start, end, i, sum = 0, temp, x, y, z;
puts("Please enter the starting number: ");
scanf("%d", &start);
temp = start;
x = start; y = start + 1;
puts("Please enter the last number till you want to add: ");
scanf("%d", &end);
for (i = start; i < end; i++) {
z = x + y;
x = z;
y = y + 1;
}
printf("The sum of the numbers starting from %d to %d is %d", temp, end, z);
}
Can anyone find what's wrong with this code?
It is working fine for less numbers but when I assign large numbers like 1 to 1000000, it doesn't work i.e. it returns strange values.
Case 1:
Please enter the starting number:
1
Please enter the last number till you want to add:
100
The sum of the numbers starting from 1 to 100 is 5050
But in this case -
Case 2:
Please enter the starting number:
1
Please enter the last number till you want to add:
1000000
The sum of the numbers starting from 1 to 1000000 is 1784293664
What's wrong in this case? Thanks in advance.
You are using int which only have value range -32,768 to 32,767 or -2,147,483,648 to 2,147,483,647 for 2 byte and 4 byte respectively.
You should use unsigned long long for your code which have a value range 0 to 18,446,744,073,709,551,615.
It's a case of data overflow.
For your example of computing from 1 to 1000000, the sum of consecutive numbers from 1 to x is x(x+1)/2, which is 500.000.500.000, a value largely above the range of an int (2.147.483 647).
Consider using long long int, and don't forget that regardless of the byte size of your variable up to a certain value you'll overflow anyway.
For large numbers you must use unsigned long long int instead of int. Here int means integer which is up to 3 or 4 digit numbers but when you enter 5 digit numbers or more digit numbers it gives you strange values.
Problem in finding sum of consecutive numbers
OP's code repeatedly adds and attempts a sum greater than INT_MAX leading to incorrect results.
Use a wider type for the computation of the sum such as long long or intmax_t.
Performance: Do not use a loop. The sum of integers [0...N] is N*(N+1)/2
Instead:
//for (i = start; i < end; i++) {
// z = x + y;
// x = z;
// y = y + 1;
//}
// Sum 0 to start
long long sum_start = ((long long) start + 1) * start / 2; // or (start + 1LL)*start/2;
// Sum 0 to end
long long sum_end = ((long long) end + 1) * end / 2;
// vvvv specifier change
printf("The sum of the numbers starting from %d to %d is %lld",
temp, end, sum_end - (sum_start - start));
The range of type int is not sufficient to represent the value of the sum for large values of start and stop (for example the sum for 0 to 65536 exceeds 2147483647). The variables x, y and z should be defined with type long long which is guaranteed to represent values up to 9223372036854775807 (9.22E18) or unsigned long up to 18446744073709551615 (1.84E19).
Furthermore your algorithm to compute the sum is convoluted, with unnecessary or unused intermediary variables. Simplicity is king!
Here is a modified version:
#include <stdio.h>
#include <string.h>
int main() {
int start, end, i;
long long sum;
puts("Please enter the starting number: ");
if (scanf("%d", &start) != 1)
return 1;
puts("Please enter the last number till you want to add: ");
if (scanf("%d", &end) != 1)
return 1;
sum = 0;
for (i = start; i <= end; i++) {
sum = sum + i;
}
printf("The sum of the numbers from %d to %d is %lld\n", start, end, sum);
return 0;
}
For positive start and stop values, there is a direct formula to compute the result. You could replace the for loop with this expression:
if (start <= stop) {
// take the sum of integers from 0 to stop: stop * (stop + 1) / 2
sum = (long long)stop * (stop + 1) / 2;
// subtract the sum of integers from 0 to start - 1: (start - 1) * start / 2
sum -= (long long)start * (start - 1) / 2;
} else {
sum = 0;
}
which simplifies as:
if (start <= stop) {
sum = ((long long)stop + start)(stop - start + 1) / 2;
} else {
sum = 0;
}
If start and/or stop can be negative, the sum can still be computed directly as:
if (start <= stop) {
// take the sum of integers from 0 to stop - start:
sum = (long long)(start - stop) * ((long long)stop - start + 1) / 2;
// add the initial value (stop - start + 1) times
sum += ((long long)stop - start + 1) * start;
} else {
sum = 0;
}
Which simplifies as:
if (start <= stop) {
sum = ((long long)stop + start)((long long)stop - start + 1) / 2;
} else {
sum = 0;
}
Sorry, but don't you know the following way to compute the sum?
long sum_from_a_to_b(long a, long b)
{
int n;
int acc = 0;
for (n = a; n <= b; n++)
acc += n;
return acc;
}
by the way, there's a better formula, which requires no loop:
#include <assert.h>
/* assumed that a <= b */
long sum_from_a_to_b(long a, long b)
{
assert(a <= b);
return b*(b + 1)/2 - a*(a - 1)/2;
}
In the second case, you have used a number that is greater than the allowed range for integer data type(it's from -32768 to 32767 for a 16-bit architecture and from -2,147,483,648 to 2,147,483,647 for a 32-bit architecture.) and that's why it leads to integer overflow issue. To fix this problem, you could use other bigger data types such as long int or long long int but I recommend you use their unsigned data types i.e unsigned long int or unsigned long long int if you're not going to use negative numbers because the range of unsigned data types begins from 0.
I am making a library management in C for practice. Now, in studentEntry I need to generate a long int studentID in which every digit is non-zero. So, I am using this function:
long int generateStudentID(){
srand(time(NULL));
long int n = 0;
do
{
n = rand() % 10;
}while(n == 0);
int i;
for(i = 1; i < 10; i++)
{
n *= 10;
n += rand() % 10;
}
if(n < 0)
n = n * (-1); //StudentID will be positive
return n;
}
output
Name : khushit
phone No. : 987546321
active : 1
login : 0
StudentID : 2038393052
Wanted to add another student?(y/n)
I wanted to remove all zeros from it. Moreover, when I run the program the first time the random number will be the same as above, and second time random number is same as past runs like e.g:-
program run 1
StudentID : 2038393052
StudentID : 3436731238
program run 2
StudentID : 2038393052
StudentID : 3436731238
What do I need to fix these problems?
You can either do as gchen suggested and run a small loop that continues until the result is not zero (just like you did for the first digit) or accept a small bias and use rand() % 9 + 1.
The problem with the similar sequences has its reason with the coarse resolution of time(). If you run the second call of the function to fast after the first you get the same seed. You might read this description as proposed by user3386109 in the comments.
A nine-digit student ID with no zeros in the number can be generated by:
long generateStudentID(void)
{
long n = 0;
for (int i = 0; i < 9; i++)
n = n * 10 + (rand() % 9) + 1;
return n;
}
This generates a random digit between 1 and 9 by generating a digit between 0 and 8 with (rand() % 9) and adding 1. There's no need to for loops to avoid zeros.
Note that this does not call srand() — you should only call srand() once in a given program (under normal circumstances). Since a long must be at least 32 bits and a 9-digit number only requires 30 bits, there cannot be overflow to worry about.
It's possible to argue that the result is slightly biassed in favour of smaller digits. You could use a function call to eliminate that bias:
int unbiassed_random_int(int max)
{
int limit = RAND_MAX - RAND_MAX % max;
int value;
while ((value = rand()) >= limit)
;
return value % max;
}
If RAND_MAX is 32767 and max is 9, RAND_MAX % 9 is 7. If you don't ignore the values from 32760 upwards, you are more likely to get a digit in the range 0..7 than you are to get an 8 — there are 3642 ways to each of 0..7 and only 3641 ways to get 8. The difference is not large; it is smaller if RAND_MAX is bigger. For the purposes on hand, such refinement is not necessary.
Slightly modify the order of your original function should perform the trick. Instead of removing 0s, just do not add 0s.
long int generateStudentID(){
srand(time(NULL));
long int n = 0;
for(int i = 0; i < 10; i++)
{
long int m = 0;
do
{
m = rand() % 10;
}while(m == 0);
n *= 10;
n += m;
}
//Not needed as n won't be negative
//if(n < 0)
//n = n * (-1); //StudentID will be positive
return n;
}
I have to find the sum of the first 4 digits, the sum of the last 4 digits and compare them (of all the numbers betweem m and n). But when I submit my solution online there's a problem with the time limit.
Here's my code:
#include <stdio.h>
int main()
{
int M, N, res = 0, cnt, first4, second4, sum1, sum2;
scanf("%d", &M);
scanf("%d", &N);
for(cnt = M; cnt <= N; cnt++)
{
first4 = cnt % 10000;
sum1 = first4 % 10 + (first4 / 10) % 10 + (first4 / 100) % 10 + (first4 / 1000) % 10;
second4 = cnt / 10000;
sum2 = second4 % 10 + (second4 / 10) % 10 + (second4 / 100) % 10 + (second4 / 1000) % 10;
if(sum1 == sum2)
res++;
}
printf("%d", res);
return 0;
}
I'm trying to find a more efficient way to do this.
Finally, if you are still interested, there is a much faster way to do this.
Your task doesn't specifically require you to calculate the sums for all the numbers,
it only asks for the number of some special numbers.
In such cases optimization techniques like memoization or dynamic programming come really handy.
In this case, when you have the first four digits of some number (let them be 1234),
you calculate their sum (in this case 10) and you immediately know,
what is the sum of the other four digits supposed to be.
Any 4-digit number, that yields sum 10 can now be the other half to create a valid number.
Therefore total number of valid numbers beginning with 1234 is exactly the number of all four digit numbers that give the sum 10.
Now consider another number, say 3412. This number has also sum equal to 10,
therefore any right-side that completes 1234 also completes 3412.
What this means is that the number of valid numbers beginning with 3412 is the same
as the number of valid numbers beginning with 1234, which is in turn the same as the total number of valid numbers, where the first half yields the sum 10.
Therefore if we precompute for each i the number of four digit numbers
that yield the sum i, we would know for each first four digits the exact number of
combinations of last four digits that complete a valid number,
without having to iterate over all 10000 of them.
The following implementation of this algorithm
Precomputes number of different ending halves for each sum of the beginning half
Splits the [M,N] interval in three subintervals, because in the first and the last beginning not every ending is possible
This algorithm runs quadratically faster than the naive implementation (for sufficiently big N-M).
#include <string.h>
int sum_digits(int number) {
return number%10 + (number/10)%10 + (number/100)%10 + (number/1000)%10;
}
int count(int M, int N) {
if (M > N) return 0;
int ret = 0;
int tmp = 0;
// for each i from 0 to 36 precompute number of ways we can get this sum
// out of a four-digit number
int A[37];
memset(A, 0, 37*4);
for (int i = 0; i <= 9999; ++i) {
++A[sum_digits(i)];
}
// nearest multiple of 10000 greater than M
int near_M = ((M+9999)/10000)*10000;
// nearest multiple of 10000 less than N
int near_N = (N/10000)*10000;
// count all numbers up to first multiple of 10000
tmp = sum_digits(M/10000);
if (near_M <= N) {
for (int i = M; i < near_M; ++i) {
if (tmp == sum_digits(i % 10000)) {
++ret;
}
}
}
// count all numbers between the 10000 multiples, use the precomputed values
for (int i = near_M / 10000; i < near_N / 10000; ++i) {
ret += A[sum_digits(i)];
}
// count all numbers after the last multiple of 10000
tmp = sum_digits(N / 10000);
if (near_N >= M) {
for (int i = near_N; i <= N; ++i) {
if (tmp == sum_digits(i % 10000)) {
++ret;
}
}
}
// special case when there are no multiples of 10000 between M and N
if (near_M > near_N) {
for (int i = M; i <= N; ++i) {
if (sum_digits(i / 10000) == sum_digits(i % 10000)) {
++ret;
}
}
}
return ret;
}
EDIT: I fixed the bugs mentioned in the comments.
I don't know if this would be significantly faster or not, but you might try breaking the number into two 4 digit numbers, then use a table lookup to get the sums. That way there's only one division operation instead of eight.
You can pre-compute the table of 10000 sums so it gets compiled in so there's no runtime cost at all.
Another slightly more complicated, but probably much faster, approach that can be used is have a table or map of 10000 elements that's the reverse of the sum lookup table where you can map the sum to the set of four digit numbers that would produce that sum. That way, when you have to find the result for a particular range 10000 number range, it's a simple lookup on the sum of the most significant four digits. For example, to find the result for the range 12340000 - 12349999, you could use a binary search on the reverse lookup table to quickly find how many numbers in the range 0 - 9999 have the sum 10 (1 + 2 + 3 + 4).
Again - this reverse sum lookup table can be pre-computed and compiled in as a static array.
In this way, the results for complete 10000 number ranges are performed with a couple binary searches. Any partial ranges can also be handled with the reverse lookup table with slightly more complication due to having to ignore matches that are from out of the range of interest. But that complication only has to happen at most twice for your whole set of subranges.
This would reduce the complexity of the algorithm from O(N*N) to O(N log N) (I think).
update:
Here's some timings I got (Win32-x86, using VS 2013 (MSVC 12) with release build default options):
range range
start end count time
================================================
alg1(10000000, 99999999): 4379055, 1.854 seconds
alg2(10000000, 99999999): 4379055, 0.049 seconds
alg3(10000000, 99999999): 4379055, 0.001 seconds
with:
alg1() is the original code from the question
alg2() is my first cut suggestion (lookup precomputed sums)
alg3() is the second suggestion (binary search lookup of sum matches using a table sorted by sums)
I'm actually surprised at the difference between alg1() to alg2()
You are going about this the wrong way. A little bit of cleverness is worth a lot of horsepower. You should not be comparing the first and last four digits of every number.
First - notice that the first four digits will change very slowly - so for sure you can have a loop of 10000 of the last four digits without re-computing the first sum.
Second - the sum of digits repeats itself every 9th number (until you get overflow). This is the basis of the "number is divisible by 9 if sum of digits is divisible by 9". example:
1234 - sum = 10
1234 + 9 = 1243 - sum is still 10
What this means is that the following will work pretty well (pseudo code):
take first 4 digits of M, find sum (call it A)
find sum of last four digits of M (call it B)
subtract: C = (A - B)
If C < 9:
D = C%9
first valid number is [A][B+D]. Then step by 9, until...
You need to think a bit about the "until", and also about what to do when C >= 9. This means you need to find a zero in B and replace it with a 9, then repeat the above.
If you want to do nothing else, then see that you don't need to re-compute the sum of digits that did not change. In general when you add 1 to a number, the sum of digits increases by 1 (unless there is carry - then it decreases by 9; and that happens every 9th, 99th (twice -> sum drops by 18), 999th (drop by 27), etc.
I hope this helps you think about the problem differently.
I am going to try an approach which doesn't make use of the lookup table (even though I know that the second one should be faster) to investigate how much we can speedup just optimizing calculus. This algorithm can be used where stack is an important resource...
Let's work on the idea that divisions and modulus are slow, for example in cortex R4 a 32 bit division requires up to 16 loops while a multiplication can be done in a single loop, with older ARMs things can be even worse.
This basic idea will try to get rid of them using digit arrays instead of integers. To keep it simple let's show an implementation using printf before a pseudo optimized version.
void main() {
int count=0;
int nmax;
char num[9]={0};
int n;
printf( "Insert number1 ");
scanf( "%d", &nm );
printf( "Insert number2 ");
scanf( "%d", &nmax );
while( nm <= nmax ) {
int sumup=0, sumdown=0;
sprintf( num, "%d", nm );
for( n=0; n<4; n++ ) {
sumup += num[n] -'0'; // subtracting '0' is not necessary (see below)
sumdown += num[7-n]-'0'; // subtracting '0' is not necessary (see below)
}
if( sumup == sumdown ) {
/* whatever */
count++;
}
nm++;
}
}
You may want to check that the string is a valid number using strtol before calling the for loop and the length of the string using strlen. I set here fixed values as you required (I assume length always 8).
The downside of the shown algorithm is the sprintf for any loop that may do thing worse... So we apply two major changes
we use [0-9] instead of ['0';'9']
we drop the sprintf for a faster solution which takes in account that we need to format a digit string starting from the previous number (n-1)
Finally the pseudo optimized algorithm should look something like the one shown below in which all divisions and modules are removed (apart from the first number) and bytes are used instead of ASCII.
void pseudo_optimized() {
int count=0;
int nmax,nm;
char num[9]={0};
int sumup=0, sumdown=0;
int n,i;
printf( "Insert number1 ");
scanf( "%d", &nm );
printf( "Insert number2 ");
scanf( "%d", &nmax );
n = nm;
for( i=7; i>=0; i-- ) {
num[i]=n%10;
n/=10;
}
while( nm <= nmax ) {
sumup = num[0] + num[1] + num[2] + num[3];
sumdown = num[7] + num[6] + num[5] + num[4];
if( sumup == sumdown ) {
/* whatever */
count++;
}
nm++;
/* Following loop is a faster sprintf replacement and
* it will exit at the first value 9 times on 10
*/
for( i=7; i>=0; i-- ) {
if( num[i] == 9 ) {
num[i]=0;
} else {
num[i] += 1;
break;
}
}
}
}
Original algo on my vm 5.500000 s, this algo 0.950000 s tested for [00000000=>99999999]
The weak point of this algorithm is that it uses sum of digits (which are not necessary and a for...loop that can be unrolled.
* update *
further optimization. The sums of digits are not necessary.... thinking about it I could improve the algorithm in the following way:
int optimized() {
int nmax=99999999,
int nm=0;
clock_t time1, time2;
char num[9]={0};
int sumup=0, sumdown=0;
int n,i;
int count=0;
n = nm;
time1 = clock();
for( i=7; i>=0; i-- ) {
num[i]=n%10;
n/=10;
}
sumup = num[0] + num[1] + num[2] + num[3];
sumdown = num[7] + num[6] + num[5] + num[4];
while( nm <= nmax ) {
if( sumup == sumdown ) {
count++;
}
nm++;
for( i=7; i>=0; i-- ) {
if( num[i] == 9 ) {
num[i]=0;
if( i>3 )
sumdown-=9;
else
sumup-=9;
} else {
num[i] += 1;
if( i>3 )
sumdown++;
else
sumup++;
break;
}
}
}
time2 = clock();
printf( "Final-now %d %f\n", count, ((float)time2 - (float)time1) / 1000000);
return 0;
}
with this we arrive to 0.760000 s which is 3 times slower than the result achieved on the same machine using lookup tables.
* update* Optimized and unrolled:
int optimized_unrolled(int nm, int nmax) {
char num[9]={0};
int sumup=0, sumdown=0;
int n,i;
int count=0;
n = nm;
for( i=7; i>=0; i-- ) {
num[i]=n%10;
n/=10;
}
sumup = num[0] + num[1] + num[2] + num[3];
sumdown = num[7] + num[6] + num[5] + num[4];
while( nm <= nmax ) {
if( sumup == sumdown ) {
count++;
}
nm++;
if( num[7] == 9 ) {
num[7]=0;
if( num[6] == 9 ) {
num[6]=0;
if( num[5] == 9 ) {
num[5]=0;
if( num[4] == 9 ) {
num[4]=0;
sumdown=0;
if( num[3] == 9 ) {
num[3]=0;
if( num[2] == 9 ) {
num[2]=0;
if( num[1] == 9 ) {
num[1]=0;
num[0]++;
sumup-=26;
} else {
num[1]++;
sumup-=17;
}
} else {
num[2]++;
sumup-=8;
}
} else {
num[3]++;
sumup++;
}
} else {
num[4]++;
sumdown-=26;
}
} else {
num[5]++;
sumdown-=17;
}
} else {
num[6]++;
sumdown-=8;
}
} else {
num[7]++;
sumdown++;
}
}
return count;
}
Unrolling vectors improves the speed of about 50%. The algorithm costs now 0.36000 s, by the way it makes use of the stack a bit more than the previous solution (as some 'if' statements may result in a push, so it cannot be always used). The result is comparable with Alg2#Michael Burr on the same machine, [Alg3-Alg5]#Michael Burr are a lot faster where stack isn't a concern.
Note all test where performed on a intel VMS. I will try to run all those algos on a ARM device if I will have time.
#include <stdio.h>
int main(){
int M, N;
scanf("%d", &M);
scanf("%d", &N);
static int table[10000] = {0,1,2,3,4,5,6,7,8,9};
{
register int i=0,i1,i2,i3,i4;
for(i1=0;i1<10;++i1)
for(i2=0;i2<10;++i2)
for(i3=0;i3<10;++i3)
for(i4=0;i4<10;++i4)
table[i++]=table[i1]+table[i2]+table[i3]+table[i4];
}
register int cnt = M, second4 = M % 10000;
int res = 0, first4 = M / 10000, sum1=table[first4];
for(; cnt <= N; ++cnt){
if(sum1 == table[second4])
++res;
if(++second4>9999){
second4 -=10000;
if(++first4>9999)break;
sum1 = table[first4];
}
}
printf("%d", res);
return 0;
}
If you know that the numbers are fixed like that, then you can you substring functions to get the components and compare them. Otherwise, your modulator operations are contributing unnecessary time.
i found faster algorithm:
#include <stdio.h>
#include <ctime>
int main()
{
clock_t time1, time2;
int M, N, res = 0, cnt, first4, second4, sum1, sum2,last4_ofM,first4_ofM,last4_ofN,first4_ofN,j;
scanf("%d", &M);
scanf("%d", &N);
time1 = clock();
for(cnt = M; cnt <= N; cnt++)
{
first4 = cnt % 10000;
sum1 = first4 % 10 + (first4 / 10) % 10 + (first4 / 100) % 10 + (first4 / 1000) % 10;
second4 = cnt / 10000;
sum2 = second4 % 10 + (second4 / 10) % 10 + (second4 / 100) % 10 + (second4 / 1000) % 10;
if(sum1 == sum2)
res++;
}
time2 = clock();
printf("%d\n", res);
printf("first algorithm time: %f\n",((float)time2 - (float)time1) / 1000000.0F );
res=0;
time1 = clock();
first4_ofM = M / 10000;
last4_ofM = M % 10000;
first4_ofN = N / 10000;
last4_ofN = N % 10000;
for(int i = first4_ofM; i <= first4_ofN; i++)
{
sum1 = i % 10 + (i / 10) % 10 + (i / 100) % 10 + (i / 1000) % 10;
if ( i == first4_ofM )
j = last4_ofM;
else
j = 0;
while ( j <= 9999)
{
sum2 = j % 10 + (j / 10) % 10 + (j / 100) % 10 + (j / 1000) % 10;
if(sum1 == sum2)
res++;
if ( i == first4_ofN && j == last4_ofN ) break;
j++;
}
}
time2 = clock();
printf("%d\n", res);
printf("second algorithm time: %f\n",((float)time2 - (float)time1) / 1000000.0F );
return 0;
}
i just dont need to count sum of the first four digits all the time the number in changed. I need to count it one time per 10000 iterations. In worst case output is:
10000000
99999999
4379055
first algorithm time: 5.160000
4379055
second algorithm time: 2.240000
about half the better result.