【二分】【边界判定】

https://ac.nowcoder.com/acm/contest/22353/G
注意点：check中，不仅要判断用的joker数是否大于joker牌的数量，还要判断组成套数是否小于用的joker数量，

原文链接：https://blog.csdn.net/a_forever_dream/article/details/106548941

#include<bits/stdc++.h>typedef long long ll;
using namespace std;bool check(ll goal, ll* a, int n, int m) {ll joker_count = 0;for(int i = 0; i < n; i++) {if(a[i] < goal) {joker_count += goal - a[i];}}return joker_count <= m && joker_count <= goal; // 只需要判断 joker 数量是否超过了 m
}int main() {int n, m;cin >> n >> m;ll a[50];ll total_cards = 0;  // 记录所有牌的总数// 读取每种牌的数量并计算总牌数for(int i = 0; i < n; i++) {cin >> a[i];total_cards += a[i];}ll low = 1; // low 最小从 1 开始ll high = total_cards + m; // high 设置为总牌数加上 Joker 数量ll ans = 0;while(low <= high) {ll mid = (low + high) / 2; // 中间值if(check(mid, a, n, m)) {ans = mid;  // 更新答案为 mid，因为 mid 满足条件low = mid + 1; // 继续尝试更大的值} else {high = mid - 1; // 尝试更小的值}}cout << ans << endl;return 0;
}