#include <bits/stdc++.h>

using namespace std;

int lowbit(int x) {
    return x & (-x);
}

//输出十进制对应的二进制数(模板)
void print(int n) {
    //输出8个数位的二进制数
    for (int i = 7; i >= 0; i--)
        printf("%d", n >> i & 1);
    printf("\n");
}

const int N = 12;
int sum[N];

int main() {
    int x = 28;//二进制 00011100;
    cout << "初始数字：" << x << endl;
    cout << "二进制表示：";
    print(x);
    cout << "仅有末位1：" << lowbit(x) << endl;
    cout << "去掉末位1：" << (x & ~lowbit(x)) << endl;

    //前缀和的lowbit加速技巧
    for (int i = 0; i < 4; i++) sum[1 << i] = 1; //按数位压缩方式存入前缀和数组[前缀和数组下标从1开始]
    cout << "实际的存储情况：" << endl;
    for (int i = 0; i < 9; i++) cout << i << " " << sum[i] << endl;

    //这里补充一下~这个逻辑运算符，它的作用是把一个二进制数每一位取反 即每一位的0变成1,1变成0
    //所以i&~lowbit(i)的意义为去掉一个二进制数的lowbit后的值
    for (int i = 1; i < (1 << 4); i++) {
        sum[i] = sum[i & ~lowbit(i)] + sum[lowbit(i)];
        cout << "i=" << i << "  sum[i]=" << sum[i] << endl;
    }
    //每次去掉最后一个1
    for (int j = x; j; j -= lowbit(j))//枚举可能的上一次状态
        cout << j << endl;
    return 0;
}