题解 | #查找输入整数二进制中1的个数#

#include <stdio.h>
#include <string.h>
//多项式分解
// x= an 2^n+...a2 2^2+ a1 2^1+ a0 2^0
// x%2 =a0 x/=2
int main(){
int n;
while(scanf("%d",&n)!=EOF){
    int count=0;
    while(n!=0){
        count += n%2;
        n /=2;
    }
    printf("%d\n",count);
    
}


return 0;
}