43 lines
1.4 KiB
C++
43 lines
1.4 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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int lowbit(int x) {
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return x & (-x);
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}
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//输出十进制对应的二进制数(模板)
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void print(int n) {
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//输出8个数位的二进制数
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for (int i = 7; i >= 0; i--)
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printf("%d", n >> i & 1);
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printf("\n");
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}
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const int N = 12;
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int sum[N];
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int main() {
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int x = 28;//二进制 00011100;
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cout << "初始数字：" << x << endl;
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cout << "二进制表示：";
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print(x);
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cout << "仅有末位1：" << lowbit(x) << endl;
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cout << "去掉末位1：" << (x & ~lowbit(x)) << endl;
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//前缀和的lowbit加速技巧
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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]
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cout << "实际的存储情况：" << endl;
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for (int i = 0; i < 9; i++) cout << i << " " << sum[i] << endl;
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//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
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//所以i&~lowbit(i)的意义为去掉一个二进制数的lowbit后的值
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for (int i = 1; i < (1 << 4); i++) {
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sum[i] = sum[i & ~lowbit(i)] + sum[lowbit(i)];
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cout << "i=" << i << " sum[i]=" << sum[i] << endl;
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}
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//每次去掉最后一个1
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for (int j = x; j; j -= lowbit(j))//枚举可能的上一次状态
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cout << j << endl;
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return 0;
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} |