[HDOJ4349]Xiao Ming's Hope

题目连接：http://acm.hdu.edu.cn/showproblem.php?pid=4349

数出n的二进制有x位1，再求2的x次幂就可以，数据不大不用优化

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

typedef long long LL;

LL solve(int x){
    int cnt = 0;
    while(x){
      if(x % 2)
          cnt++;
      x >>= 1;
    }
    LL sum = 1;
    for(int i = 1; i <= cnt; ++i) {
        sum *= 2;
    }
    return sum;
}

int main(){
    int n;
    while(scanf("%d",&n) != EOF){
       LL ans = solve(n);
       printf("%d
", ans);
    }
    return 0;
}