I am trying to print the series but whenever I set the range (input given by me) above 407. I only get the output till 407. However, when I set the range below 407 it gives me the result according to the input I have given. Can anybody tell me what I'm doing wrong?
I used an online compiler (www.onlinegdb.com) to write my code.
Here is the code.
#include<stdio.h>
#include<stdlib.h>
int
main ()
{
int m, n;
printf
("Enter two numbers to find the Armstrong numbers that lie between them.\n");
scanf ("%d%d", &m, &n);
system("clear");
if(m>n)
{
m = m + n;
n = m - n;
m = m - n;
}
for (; m < n; m++)
{
int i = m + 1, r, s = 0, t;
t = i;
while (i > 0)
{
r = i % 10;
s = s + (r * r * r);
i = i / 10;
}
if (t == s)
printf ("%d ", t);
}
return 0;
}
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Try this code!!!
#include <math.h>
#include <stdio.h>
int main() {
int low, high, number, originalNumber, rem, count = 0;
double result = 0.0;
printf("Enter two numbers(intervals): ");
scanf("%d %d", &low, &high);
printf("Armstrong numbers between %d and %d are: ", low, high);
// swap numbers if high < low
if (high < low) {
high += low;
low = high - low;
high -= low;
}
// iterate number from (low + 1) to (high - 1)
// In each iteration, check if number is Armstrong
for (number = low + 1; number < high; ++number) {
originalNumber = number;
// number of digits calculation
while (originalNumber != 0) {
originalNumber /= 10;
++count;
}
originalNumber = number;
// result contains sum of nth power of individual digits
while (originalNumber != 0) {
rem = originalNumber % 10;
result += pow(rem, count);
originalNumber /= 10;
}
// check if number is equal to the sum of nth power of individual digits
if ((int)result == number) {
printf("%d ", number);
}
// resetting the values
count = 0;
result = 0;
}
return 0;
}
Try this code :
#include <stdio.h>
#include <math.h>
int main()
{
int start, end, i, temp1, temp2, remainder, n = 0, result = 0;
printf(“Enter start value and end value : “);
scanf(“%d %d”, &start, &end);
printf(“\nArmstrong numbers between %d an %d are: “, start, end);
for(i = start + 1; i < end; ++i)
{
temp2 = i;
temp1 = i;
while (temp1 != 0)
{
temp1 /= 10;
++n;
}
while (temp2 != 0)
{
remainder = temp2 % 10;
result += pow(remainder, n);
temp2 /= 10;
}
if (result == i) {
printf(“%d “, i);
}
n = 0;
result = 0;
}
printf(“\n”);
return 0;
}
Write a program that examines three variables — x, y, and z — and
prints the largest odd number among them. If none of them are odd, it
should print a message to that effect.
This is my source code:
#include <stdio.h>
int main() {
int x,y,z;
x = 11;
y = 15;
z = 18;
if (x > y && x > z && x % 2 != 0)
printf("%d", x);
else if (y > z && y > x && y % 2 != 0)
printf("%d", y);
else if (z > x && z > y && z % 2 != 0)
printf("%d", z);
else
printf("error");
return 0;
}
The program is compiling and running but is giving the wrong answer. For the above it gives "error" as the output, but the greatest odd number is 15.
You are just printing the largest integer of the three if it is odd. For not to change you code so much and if you doesn't care about the others values, you can set the even variables to -INF. That is:
//INF can be whatever big even value you think that works fine (1e9), 0x3ffffffe, etc
#define INF 1e9
if (x % 2 == 0) x = -INF;
if (y % 2 == 0) y = -INF;
if (z % 2 == 0) z = -INF;
//here the rest of your code
In your example, y is the largest odd number, but it is not the largest number. z is larger, so "y>z" evaluates to false, and the path you want is not taken.
You may have noticed that your code is very repetitive. And the next exercise in the book is likely to be "OK, now do this for ten input numbers", which would be too tedious to contemplate, or "... for any number of input numbers", which just plain can't be done the way you're doing it.
So instead you should be thinking about a code structure like this:
int current_largest_odd_number = 0; /* note: zero is even */
for (;;) {
int n = read_a_number();
if (n == -1) break; /* read_a_number returns -1 on EOF */
if (is_even(n)) continue;
if (n > current_largest_odd_number)
current_largest_odd_number = n;
}
if (current_largest_odd_number > 0)
printf("largest odd number: %d\n", current_largest_odd_number);
The code is not printing the largest odd number, but is printing the largest number if it happens to be odd.
You need to keep track of the largest odd number you've found and then print that.
For example:
int found = false, largest = 0;
if (x%2 != 0) {
found = 1;
largest = x;
}
if (y%2 != 0) {
found = 1;
if (y > largest) {
largest = y;
}
}
if (z%2 != 0) {
found = 1;
if (z > largest) {
largest = z;
}
}
if (found) {
printf("%d", largest);
} else {
printf("error");
}
Your code prints the largest number if and only if is is odd.
You could instead test each number in turn and update a pointer to the maximum odd value if any.
#include <stdio.h>
int main() {
int x = 11;
int y = 15;
int z = 18;
int *largest = NULL;
if (x % 2 != 0) {
largest = &x;
}
if (y % 2 != 0) {
if (!largest || y > *largest) {
largest = &y;
}
}
if (z % 2 != 0) {
if (!largest || z > *largest) {
largest = &z;
}
}
if (largest) {
printf("%d\n", *largest);
} else {
printf("error\n");
}
return 0;
}
Here is an alternative approach with an indicator instead of a pointer:
#include <stdio.h>
int main() {
int x = 11;
int y = 15;
int z = 18;
int largest = 0;
int found = 0;
if (x % 2 != 0) {
largest = x;
found = 1;
}
if (y % 2 != 0) {
if (!found || y > largest) {
largest = y;
found = 1;
}
}
if (z % 2 != 0) {
if (!found || z > largest) {
largest = z;
found = 1;
}
}
if (found) {
printf("%d\n", largest);
} else {
printf("error\n");
}
return 0;
}
My version examines all possible three integer numbers and prints the largest
odd number (As an absolute beginner.)
x = int(input("number x = "))
y = int(input("number y = "))
z = int(input("number z = "))
/*Ask the user to input three variables*/
ans=0
if x%2==0 and y%2==0 and z%2==0:
print("There are no odd numbers.")
else:
if x<=y<=z and z%2!=0:
ans=z
elif y%2!=0:
ans=y
elif x%2!=0:
ans=x
if x<=y>=z and y%2!=0:
ans=y
elif x>=z and x%2!=0:
ans=x
elif z%2!=0:
ans=z
if x>=y<=z and x>=z and x%2!=0:
ans=x
elif z%2!=0:
ans=z
elif y%2!=0:
ans=y
if x>=y>=z and x%2!=0:
ans=x
elif y%2!=0:
ans=y
elif z%2!=0:
ans=z
print("The largest odd number is "+str(ans))
#include <stdio.h>
int main() {
int rangeValue;
int x;
printf("Please input a number to be considered in the range:\n");
scanf("%d", &rangeValue);
while (rangeValue != 1) {
x = rangeValue;
if ((x % 2) == 0) {
x = x / 2;
printf("%d,", x);
} else {
x = (3 * x) + 1;
printf("%d,", x);
}
rangeValue--;
}
return 0;
}
My goal is to do the Collatz sequence of every number from 1 to the number I give to rangeValue. I expected this to work. Can anyone help me make it work?
You are mixing the range of sequences to print, the maximum number of iterations and the current number in the sequence.
Here is how to fix the code:
#include <stdio.h>
int main(void) {
int rangeValue;
printf("Please input a number to be considered in the range:\n");
if (scanf("%d", &rangeValue) != 1)
return 1;
// iterate for all numbers upto rangeValue
for (int n = 1; n <= rangeValue; n++) {
printf("%d", n);
for (long long x = n; x != 1; ) {
if ((x % 2) == 0) {
x = x / 2;
} else {
x = (3 * x) + 1;
}
printf(",%lld", x);
}
printf("\n");
}
return 0;
}
Write a program that will find the largest number smaller than N that is totally different from a given number X. One number is totally different from other only if it doesn't contain any of the digits from the other number. N and X are read from standard input. The problem should be solved without the use of arrays.
Example Input 1: 400 897
Example Output 1: 366
Example Input 2: 1000 1236498
Example Output 2:777
No it's not homework, it was on one of the midterms and it's been killing me. I though about taking the first numbers last digit with %10 then taking the second numbers digit with %10 comparing them but...I just can't get it to work...I ended up with an endless loop...I just don't understand how to get every digit of the numbers and compare them to the other number.
#include <stdio.h>
int main () {
int N, X, num_N, num_X, i, lastDigit_N, lastDigit_X, flag, smaller_than_N;
scanf("%d%d", &N, &X);
smaller_than_N = N - 1;
for (i = smaller_than_N; i > 0; i--) {
num_N = i;
num_X = X;
flag = 0;
while (num_N > 0) {
lastDigit_N = num_N % 10;
while (num_X > 0) {
lastDigit_X = num_X % 10;
if (lastDigit_N == lastDigit_X) {
break;
}
else {
flag = 1;
}
num_X /= 10;
}
num_N /= 10;
}
if(flag) {
printf("%d", i);
break;
}
}
return 0;
}
You could build a bitmask for your numbers showing the digits which are contained.
uint16_t num2bitmask(int number)
{
uint16_t result = 0;
while (number) {
int digit = number % 10;
number /= 10;
result |= (1 << digit);
}
return result;
}
With this function, you can create your bitmask for X and then iterate from N-1 down to 1 until you find a value which doesn't have any bits in common with the other value.
If you have a number with digits d_1, d_2, ..., d_n, and you're allowed to use digits in the set D, then possible solutions look like:
d_1, ..., d_{i-1}, max(d in D | d < d_i), max(d in D), ..., max(d in D).
That is, the digits are the same up to some point, then the next digit is as large as possible while being below the input digit, then the rest are just as large as possible.
Not all these "solutions" will be valid, but if you iterate through them in reverse order (there's exactly n for an input number of size n), the first valid one you find is the answer.
Some code, including tests:
#include <stdio.h>
int digit_length(int a) {
int r = 0;
while (a) {
a /= 10;
r += 1;
}
return r;
}
int get_digit(int a, int k) {
while (k--) a /= 10;
return a % 10;
}
int largest_different(int a, int b) {
int lena = digit_length(a);
int invalid = b ? 0 : 1;
for (; b; b /= 10) invalid |= 1 << (b % 10);
int max_valid = 9;
while (max_valid >= 0 && (invalid & (1 << max_valid)))
max_valid--;
if (max_valid == -1) return -1;
for (int i = 0; i < lena; i++) {
int d = get_digit(a, i) - 1;
while (d >= 0 && (invalid & (1 << d)))d--;
if (d < 0) continue;
int solution = 0;
for (int k = lena - 1; k >= 0; k--) {
solution *= 10;
solution += (k < i ? max_valid : k > i ? get_digit(a, k) : d);
}
return solution;
}
return -1;
}
int main(int argc, char *argv[]) {
struct {int n; int x; int want;} examples[] = {
{400, 897, 366},
{1000, 1236498, 777},
{998, 123, 997},
};
int error = 0;
for (int i = 0; i < sizeof(examples) / sizeof(*examples); i++) {
int got = largest_different(examples[i].n, examples[i].x);
if (got != examples[i].want) {
error = 1;
printf("largest_different(%d, %d) = %d, want %d\n",
examples[i].n, examples[i].x, got, examples[i].want);
}
}
return error;
}
There's not always a solution. In that case, the function returns -1.
I would like to know how I can find the length of an integer in C.
For instance:
1 => 1
25 => 2
12512 => 5
0 => 1
and so on.
How can I do this in C?
C:
You could take the base-10 log of the absolute value of the number, round it down, and add one. This works for positive and negative numbers that aren't 0, and avoids having to use any string conversion functions.
The log10, abs, and floor functions are provided by math.h. For example:
int nDigits = floor(log10(abs(the_integer))) + 1;
You should wrap this in a clause ensuring that the_integer != 0, since log10(0) returns -HUGE_VAL according to man 3 log.
Additionally, you may want to add one to the final result if the input is negative, if you're interested in the length of the number including its negative sign.
Java:
int nDigits = Math.floor(Math.log10(Math.abs(the_integer))) + 1;
N.B. The floating-point nature of the calculations involved in this method may cause it to be slower than a more direct approach. See the comments for Kangkan's answer for some discussion of efficiency.
If you're interested in a fast and very simple solution, the following might be quickest (this depends on the probability distribution of the numbers in question):
int lenHelper(unsigned x) {
if (x >= 1000000000) return 10;
if (x >= 100000000) return 9;
if (x >= 10000000) return 8;
if (x >= 1000000) return 7;
if (x >= 100000) return 6;
if (x >= 10000) return 5;
if (x >= 1000) return 4;
if (x >= 100) return 3;
if (x >= 10) return 2;
return 1;
}
int printLen(int x) {
return x < 0 ? lenHelper(-x) + 1 : lenHelper(x);
}
While it might not win prizes for the most ingenious solution, it's trivial to understand and also trivial to execute - so it's fast.
On a Q6600 using MSC I benchmarked this with the following loop:
int res = 0;
for(int i = -2000000000; i < 2000000000; i += 200) res += printLen(i);
This solution takes 0.062s, the second-fastest solution by Pete Kirkham using a smart-logarithm approach takes 0.115s - almost twice as long. However, for numbers around 10000 and below, the smart-log is faster.
At the expense of some clarity, you can more reliably beat smart-log (at least, on a Q6600):
int lenHelper(unsigned x) {
// this is either a fun exercise in optimization
// or it's extremely premature optimization.
if(x >= 100000) {
if(x >= 10000000) {
if(x >= 1000000000) return 10;
if(x >= 100000000) return 9;
return 8;
}
if(x >= 1000000) return 7;
return 6;
} else {
if(x >= 1000) {
if(x >= 10000) return 5;
return 4;
} else {
if(x >= 100) return 3;
if(x >= 10) return 2;
return 1;
}
}
}
This solution is still 0.062s on large numbers, and degrades to around 0.09s for smaller numbers - faster in both cases than the smart-log approach. (gcc makes faster code; 0.052 for this solution and 0.09s for the smart-log approach).
int get_int_len (int value){
int l=1;
while(value>9){ l++; value/=10; }
return l;
}
and second one will work for negative numbers too:
int get_int_len_with_negative_too (int value){
int l=!value;
while(value){ l++; value/=10; }
return l;
}
You can write a function like this:
unsigned numDigits(const unsigned n) {
if (n < 10) return 1;
return 1 + numDigits(n / 10);
}
length of n:
length = ( i==0 ) ? 1 : (int)log10(n)+1;
The number of digits of an integer x is equal to 1 + log10(x). So you can do this:
#include <math.h>
#include <stdio.h>
int main()
{
int x;
scanf("%d", &x);
printf("x has %d digits\n", 1 + (int)log10(x));
}
Or you can run a loop to count the digits yourself: do integer division by 10 until the number is 0:
int numDigits = 0;
do
{
++numDigits;
x = x / 10;
} while ( x );
You have to be a bit careful to return 1 if the integer is 0 in the first solution and you might also want to treat negative integers (work with -x if x < 0).
A correct snprintf implementation:
int count = snprintf(NULL, 0, "%i", x);
The most efficient way could possibly be to use a fast logarithm based approach, similar to those used to determine the highest bit set in an integer.
size_t printed_length ( int32_t x )
{
size_t count = x < 0 ? 2 : 1;
if ( x < 0 ) x = -x;
if ( x >= 100000000 ) {
count += 8;
x /= 100000000;
}
if ( x >= 10000 ) {
count += 4;
x /= 10000;
}
if ( x >= 100 ) {
count += 2;
x /= 100;
}
if ( x >= 10 )
++count;
return count;
}
This (possibly premature) optimisation takes 0.65s for 20 million calls on my netbook; iterative division like zed_0xff has takes 1.6s, recursive division like Kangkan takes 1.8s, and using floating point functions (Jordan Lewis' code) takes a whopping 6.6s. Using snprintf takes 11.5s, but will give you the size that snprintf requires for any format, not just integers. Jordan reports that the ordering of the timings are not maintained on his processor, which does floating point faster than mine.
The easiest is probably to ask snprintf for the printed length:
#include <stdio.h>
size_t printed_length ( int x )
{
return snprintf ( NULL, 0, "%d", x );
}
int main ()
{
int x[] = { 1, 25, 12512, 0, -15 };
for ( int i = 0; i < sizeof ( x ) / sizeof ( x[0] ); ++i )
printf ( "%d -> %d\n", x[i], printed_length ( x[i] ) );
return 0;
}
Yes, using sprintf.
int num;
scanf("%d",&num);
char testing[100];
sprintf(testing,"%d",num);
int length = strlen(testing);
Alternatively, you can do this mathematically using the log10 function.
int num;
scanf("%d",&num);
int length;
if (num == 0) {
length = 1;
} else {
length = log10(fabs(num)) + 1;
if (num < 0) length++;
}
int digits=1;
while (x>=10){
x/=10;
digits++;
}
return digits;
sprintf(s, "%d", n);
length_of_int = strlen(s);
You may use this -
(data_type)log10(variable_name)+1
ex:
len = (int)log10(number)+1;
In this problem , i've used some arithmetic solution . Thanks :)
int main(void)
{
int n, x = 10, i = 1;
scanf("%d", &n);
while(n / x > 0)
{
x*=10;
i++;
}
printf("the number contains %d digits\n", i);
return 0;
}
Quite simple
int main() {
int num = 123;
char buf[50];
// convert 123 to string [buf]
itoa(num, buf, 10);
// print our string
printf("%s\n", strlen (buf));
return 0;
}
keep dividing by ten until you get zero, then just output the number of divisions.
int intLen(int x)
{
if(!x) return 1;
int i;
for(i=0; x!=0; ++i)
{
x /= 10;
}
return i;
}
This goes for both negative and positive intigers
int get_len(int n)
{
if(n == 0)
return 1;
if(n < 0)
{
n = n * (-1); // if negative
}
return log10(n) + 1;
}
Same logic goes for loop
int get_len(int n)
{
if(n == 0)
return 1;
int len = 0;
if(n < 0)
n = n * (-1);
while(n > 1)
{
n /= 10;
len++;
}
return len;
}
Why don't you cast your integer to String and get length like this :
int data = 123;
int data_len = String(data).length();
For simple programs...
int num = 456, length=0 // or read value from the user to num
while(num>0){
num=num/10;
length++;
}
Use another variable to retain the initial num value.
In my opinion the shortest and easiest solution would be:
int length , n;
printf("Enter a number: ");
scanf("%d", &n);
length = 0;
while (n > 0) {
n = n / 10;
length++;
}
printf("Length of the number: %d", length);
My way:
Divide as long as number is no more divisible by 10:
u8 NumberOfDigits(u32 number)
{
u8 i = 1;
while (number /= 10) i++;
return i;
}
I don't know how fast is it in compared with other propositions..
int intlen(int integer){
int a;
for(a = 1; integer /= 10; a++);
return a;
}
A more verbose way would be to use this function.
int length(int n)
{
bool stop;
int nDigits = 0;
int dividend = 1;
do
{
stop = false;
if (n > dividend)
{
nDigits = nDigits + 1;
dividend = dividend * 10;
}
else {
stop = true;
}
}
while (stop == false);
return nDigits;
}
int returnIntLength(int value){
int counter = 0;
if(value < 0)
{
counter++;
value = -value;
}
else if(value == 0)
return 1;
while(value > 0){
value /= 10;
counter++;
}
return counter;
}
I think this method is well suited for this task:
value and answers:
-50 -> 3 //it will count - as one character as well if you dont want to count
minus then remove counter++ from 5th line.
566666 -> 6
0 -> 1
505 -> 3
Solution
Use the limit where the integer length changes, in the case of the decimal it is a power of 10, and thus use a counter for each verification that the specified integer has not exceeded the limit.
With the math.h dependency:
#include <math.h>
int count_digits_of_integer(unsigned int integer) {
int count = 1;
while(1) {
int limit = pow(10, count);
if(integer < limit) break;
count++;
}
return count;
}
Without dependency:
int int_pow(int base, int exponent) {
int potency = base;
for(int i = 1; i < exponent; i++) potency *= base;
return potency;
}
int count_digits_of_integer(unsigned int integer) {
int count = 1;
while(1) {
int limit = int_pow(10, count);
if(integer < limit) break;
count++;
}
return count;
}
Implementation
#include <stdio.h>
// Copy and paste the solution code here
int main() {
printf("%i -> (%i digits)\n", 0, count_digits_of_integer(0));
printf("%i -> (%i digits)\n", 12, count_digits_of_integer(12));
printf("%i -> (%i digits)\n", 34569, count_digits_of_integer(34569));
printf("%i -> (%i digits)\n", 1234, count_digits_of_integer(1234));
printf("%i -> (%i digits)\n", 3980000, count_digits_of_integer(3980000));
printf("%i -> (%i digits)\n", 100, count_digits_of_integer(100));
printf("%i -> (%i digits)\n", 9, count_digits_of_integer(9));
printf("%i -> (%i digits)\n", 385784, count_digits_of_integer(385784));
return 0;
}
Output:
0 -> (1 digits)
12 -> (2 digits)
34569 -> (5 digits)
1234 -> (4 digits)
3980000 -> (7 digits)
100 -> (3 digits)
9 -> (1 digits)
385784 -> (6 digits)
Hmm, maybe like this...?
#define _LEN(x) (sizeof(#x)/sizeof(char)-1)
You can also use this function to find the length of an integer:
int countlength(int number)
{
static int count = 0;
if (number > 0)
{
count++;
number /= 10;
countlength(number);
}
return count;
}
I think I got the most efficient way to find the length of an integer
its a very simple and elegant way
here it is:
int PEMath::LengthOfNum(int Num)
{
int count = 1; //count starts at one because its the minumum amount of digits posible
if (Num < 0)
{
Num *= (-1);
}
for(int i = 10; i <= Num; i*=10)
{
count++;
}
return count;
// this loop will loop until the number "i" is bigger then "Num"
// if "i" is less then "Num" multiply "i" by 10 and increase count
// when the loop ends the number of count is the length of "Num".
}
int main(void){
unsigned int n, size=0;
printf("get the int:");
scanf("%u",&n);
/*the magic*/
for(int i = 1; n >= i; i*=10){
size++;
}
printf("the value is: %u \n", n);
printf("the size is: %u \n", size);
return 0;
}
#include <stdio.h>
int main(void){
int c = 12388884;
printf("length of integer is: %d",printf("%d",c));
return 0;
}