二分是基础算法之一，常用于答案有单调性的题目，或者穷举会超时的题目

int search(int l, int r) {
    while (l + 1 < r) {
        int mid = l + (r - l) >> 1; // 防溢出
        if (check(mid)) l = mid; else r = mid;
    }
    return l;
}

例题

洛谷 - P2440 经典二分例题

code

// URL: https://www.luogu.com.cn/problem/P2440
// Memory Limit: 128 MB
// Time Limit: 1000 ms

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

#define debug(args...) fprintf(stderr, ##args)

#define int ll
using ll = long long; 
using ull = unsigned long long;

const int inf = 0x3f3f3f3f;
const int mod = 998244353; // 998244383, 1e9 + 7
const double eps = 1e-6;
const int N = 1e5 + 7;
const int M = 2e6 + 7;

namespace ckb2008 {
    int n, k, l, r = inf, a[N];
    
    bool check(int x) {
        int y = 0;
        for (int i = 1; i <= n; ++i) y += a[i] / x;
        return y >= k;
    }
    
    signed main() {
        scanf("%d %d", &n, &k);
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
        while (l + 1 < r) {
            int mid = (l + r) >> 1;
            if (check(mid)) l = mid;
            else r = mid;
        }
        printf("%d", l);

        return 0;
    }
}

signed main() { return ckb2008::main(); }

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