c program to check digits in number are even -

input , output format:

input consists of number corresponds bill number. bill number 3-digit number , 3 digits in number even.

output consists of string either 'yes' or 'no'. output yes when customer receives prize , no otherwise.

samples:

input     output 565       no 620       yes 66        no         # not 3-digit number (implicit leading zeros not allowed) 002       yes        # 3-digit number 

i have solved problem getting single digit "number" mod 10 , checking if "digit" mod 2 0 or not......

but in case if give input "002", prints "no" instead want should "yes".

code — copied , formatted comment:

#include <stdio.h> #include <stdlib.h> #include <math.h> #include <string.h> int main() {     int num, t, flag, count = 0;     while (num)     {         t = num % 10;         num = num / 10;         if (t % 2 == 0)         {             flag = 1;         }         else         {             flag = 0;         }         count++;     }     if (flag == 1 && count == 3)     {         printf("yes");     }     else     {         printf("no");     }     return 0; } 

you have work strings instead of numbers, otherwise cannot represent 002 value.

recognize digits:

int even(char c) {     switch (c) {         case '0':         case '2':         case '4':         case '6':         case '8':             return 1;         default:             return 0;     } } 

recognize strings digits:

int all_even(char* s) {     while (*s != '\0') {         if (!even(*s)) {             return 0;         }         s++;     }     return 1; } 

return "yes" strings of 3 digits, , return "no" other strings:

char* answer(char* s) {     return (strlen(s) == 3 && all_even(s)) ? "yes" : "no"; }