Sum of Digits of a Five Digit Number Discussions | C | HackerRank
We use cookies to ensure you have the best browsing experience on our website. Please read our cookie policy for more information about how we use cookies.
  1. Prepare
  2. C
  3. Conditionals and Loops
  4. Sum of Digits of a Five Digit Number
  5. Discussions

Sum of Digits of a Five Digit Number

Sort by

recency

|

755 Discussions

|

  • inyeobgim
     + 0 comments
    #include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>

int main() {
	
    int n;
    scanf("%d", &n);
    //Complete the code to calculate the sum of the five digits on n.
    
    int digits[5] = {};
    int sum = 0;
    float exp = 10000;
    
    for(int i = 0; i < 6; i++){
        digits[i] = n / exp;
        n = n - digits[i] * exp;
        exp /= 10;
    }
    
    for(int j = 0; j < 6; j++){
        sum += digits[j];
    }
    
    printf("%d", sum);
    
    return 0;
}

  • srijakuruba2005
     + 0 comments

    int main() { int n, res=0; scanf("%d", &n); for(int i=0; i<5; i++) { res+=n%10; n/=10; } printf("%d\n", res); return 0; }

  • westmountdental1
     + 0 comments

    Nice little warm-up! A great way to brush up on basic input handling and digit manipulation. Dental clinic in saskatoon

  • deepthysivan10
     + 0 comments

    int main() {

    int n,i,sum=0;
scanf("%d", &n);
char aray[6];
sprintf(aray, "%d", n);
for(i=0;i<5;i++)
{
    switch(aray[i]){
        case '1':sum=sum+1;;break;
        case '2':sum=sum+2;break;
        case '3':sum=sum+3;break;
        case '4':sum=sum+4;break;
        case '5':sum=sum+5;break;
        case '6':sum=sum+6;break;
        case '7':sum=sum+7;break;
        case '8':sum=sum+8;break;
        case '9':sum=sum+9;break;
    }
}
printf("%d",sum);
//Complete the code to calculate the sum of the five digits on n.
return 0;

    }

  • roshanb1943
     + 0 comments
    #include <stdio.h>

void main() {
        int n, q, result=0;
        scanf("%d", &n);
        for(int i=0; i<5; i++) {
                result+=n%10;
                n/=10;
        }
        printf("%d\n", result);
}