AcWing 801. 二进制中1的个数

原题链接：https://www.acwing.com/problem/content/803/

思路：

• 采用lowbit()函数
• 采用位运算

#include <iostream>


int bitCount(unsigned int n) {
    n = ((n & 0xAAAAAAAA) >> 1) + (n & ~0xAAAAAAAA);
    n = ((n & 0xCCCCCCCC) >> 2) + (n & ~0xCCCCCCCC);
    n = ((n & 0xF0F0F0F0) >> 4) + (n & ~0xF0F0F0F0);
    n = ((n & 0xFF00FF00) >> 8) + (n & ~0xFF00FF00);
    n = ((n & 0xFFFF0000) >> 16) + (n & ~0xFFFF0000);
    return n;
}


int count(int n) {
    int res = 0;
    while (n != 0) {
        n -= (n & -n);
        res++;
    }
    return res;
}

int main() {
    int n;
    scanf("%d", &n);
    int num[n];
    for (int i = 0; i < n; i++) scanf("%d", &num[i]);
    
    for (int i = 0; i < n; i++) printf("%d ", bitCount(num[i]));
    return 0;
}