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位运算

视频讲解

🎥 视频讲解

位运算

在计算机中，所有数据都是以二进制形式存储的。位运算就是直接对二进制位进行操作的运算。

例题1

自建OJ：统计二进制中1的个数

参考代码

参考实现 
#include <bits/stdc++.h>
using namespace std;

int main() {
    int n;
    cin >> n;

    int ans = 0;
    for (int i = 30; i >= 0; i--) {
        if ((n >> i) & 1) {
            ans++;
        }
    }

    cout << ans;
    return 0;
}

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();

        int ans = 0;
        for (int i = 30; i >= 0; i--) {
            if (((n >> i) & 1) == 1) {
                ans++;
            }
        }

        System.out.println(ans);
    }
}

n = int(input())
ans = 0

for i in range(30, -1, -1):
    if (n >> i) & 1:
        ans += 1

print(ans)

例题2

自建OJ：位运算的操作

参考实现

参考实现 
#include <bits/stdc++.h>
using namespace std;

int n, q;

int main() {
    cin >> n >> q;
    for (int i = 1; i <= q; i++) {
        int x;
        cin >> x;

        if (x == 1) {
            int k;
            cin >> k;
            cout << ((n >> k) & 1) << endl;
        }

        if (x == 2) {
            int l, r;
            cin >> l >> r;
            int p = (1 << (r + 1)) - 1;
            int t = (1 << l) - 1;
            p -= t;
            cout << (n ^ p) << endl;
            n ^= p;
        }

        if (x == 3) {
            int l, r;
            cin >> l >> r;
            int p = (1 << (r + 1)) - 1;
            int t = (1 << l) - 1;
            p -= t;
            cout << (n | p) << endl;
            n |= p;
        }

        if (x == 4) {
            int l, r;
            cin >> l >> r;
            int p = (1 << (r + 1)) - 1;
            int t = (1 << l) - 1;
            p -= t;
            p = ~p;
            cout << (n & p) << endl;
            n &= p;
        }

        if (x == 5) {
            cout << (n & -n) << endl;
        }
    }
    return 0;
}

import java.util.Scanner;

public class Main {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int q = sc.nextInt();

        while (q-- > 0) {
            int x = sc.nextInt();

            if (x == 1) {
                int k = sc.nextInt();
                System.out.println((n >> k) & 1);
            }

            if (x == 2) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                int p = (1 << (r + 1)) - 1;
                int t = (1 << l) - 1;
                p -= t;
                System.out.println(n ^ p);
                n ^= p;
            }

            if (x == 3) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                int p = (1 << (r + 1)) - 1;
                int t = (1 << l) - 1;
                p -= t;
                System.out.println(n | p);
                n |= p;
            }

            if (x == 4) {
                int l = sc.nextInt();
                int r = sc.nextInt();
                int p = (1 << (r + 1)) - 1;
                int t = (1 << l) - 1;
                p -= t;
                p = ~p;
                System.out.println(n & p);
                n &= p;
            }

            if (x == 5) {
                System.out.println(n & -n);
            }
        }
    }
}

n, q = map(int, input().split())

for _ in range(q):
    x = int(input().split()[0])

    if x == 1:
        k = int(input())
        print((n >> k) & 1)

    if x == 2:
        l, r = map(int, input().split())
        p = (1 << (r + 1)) - 1
        t = (1 << l) - 1
        p -= t
        print(n ^ p)
        n ^= p

    if x == 3:
        l, r = map(int, input().split())
        p = (1 << (r + 1)) - 1
        t = (1 << l) - 1
        p -= t
        print(n | p)
        n |= p

    if x == 4:
        l, r = map(int, input().split())
        p = (1 << (r + 1)) - 1
        t = (1 << l) - 1
        p -= t
        p = ~p
        print(n & p)
        n &= p

    if x == 5:
        print(n & -n)

练习题单

位运算