python/TangDou/LuoGuBook/P1036_Prepare_2.cpp

#include <bits/stdc++.h>

using namespace std;

//手工版本的检查二进制中1的个数的办法
int OneCount(int n) {
    int cnt = 0;
    for (int i = 31; i >= 0; i--)
        if ((n >> i) & 1) cnt++;//固定1，右移i大法好！

    return cnt;
}

int OneCount2(int n) {
    int cnt = 0;
    for (int i = 31; i >= 0; i--)
        if (n & (1 << i)) cnt++;//固定n，左移1大法好！
    return cnt;
}

int main() {
    int n;
    cin >> n;
    int cnt = __builtin_popcount(n);
    cout << "__builtin_popcount: cnt=" << cnt << endl;

    int cnt2 = OneCount(n);
    cout << "OneCount: cnt2=" << cnt2 << endl;

    int cnt3 = OneCount2(n);
    cout << "OneCount2: cnt3=" << cnt3 << endl;

    if (cnt == 1) cout << "是2的整数次幂！";
    else cout << "不是2的整数次幂！";
    return 0;
}