Solution of URI 2176 :: Parity

Before seeing the solution make sure that you tried enough. Don’t paste the whole code, just find out the logic. If you stuck in trouble, just inform me on comment.

/**Bismillahir Rahmanir Rahim.**/

#include <stdio.h>
#include <string.h>
int main()
{
    char str[150];
    int i, cnt=0;
    scanf("%s", str);
    for(i=0; str[i]; i++)
        if(str[i]=='1')
            cnt++;
    if(cnt%2 == 0)
        str[i] = '0';
    else
        str[i] = '1';
    str[i+1] = '\0';
    printf("%s\n", str);
    return 0;
}