【二分+贪心】CF1622 C

古城微笑少年丶 2024-03-23 21:59 18阅读 0赞

[Problem - 1622C - Codeforces][]

题意：

![f6cd9c72cb09404e95da429847de7fb5.png][]

![fdc723d055784165bd580acc564e6d5e.png][]

思路：

首先，观察样例可知，肯定是把原本的最小值减到某个值，然后再复制几次

复制的时候肯定是从大到小复制

那把最小值减到哪个值是不确定的，考虑枚举这个值？但是范围太大了

考虑二分答案，我们去二分操作次数，那么问题就是，在操作mid次之内，能不能使它满足条件

一共有两种操作，他们分别占多少不确定，考虑枚举操作1的次数

贡献直接计算即可

注意二分的右边界，取pre\[n\] - k

Code：

#include <bits/stdc++.h>
    
    #define int long long
    
    using i64 = long long;
    
    using namespace std;
    
    constexpr int N = 2e5 + 10;
    constexpr int M = 2e5 + 10;
    
    int n, k;
    int a[N], pre[N];
    
    bool check(int mid) {
        for (int x = max(0ll, mid - n + 1); x <= mid; x ++) {
            int res = 0;
            int y = mid - x;
            int mi = a[n] - x;
            res += (a[n] - mi);
            int sum = 0;
            if (y < n) {
                sum = (pre[y] - pre[0]);
            }else {
                sum = pre[n - 1];
            }
            res += sum - mi * min(y, n - 1);
            if (pre[n] - res <= k) return true;
        }
        return false;
    }
    void solve() {
        cin >> n >> k;
    
        pre[0] = 0;
        for (int i = 1; i <= n; i ++) {
            pre[i] = 0;
            cin >> a[i];
        }
    
        sort(a + 1, a + 1 + n, greater<int>());
    
        for (int i = 1; i <= n; i ++) {
            pre[i] = pre[i - 1] + a[i];
        }
    
        if (pre[n] <= k) {
            cout << 0 << "\n";
            return;
        }
        int l = 0, r = pre[n] - k;
        int ans = 0;
        while(l <= r) {
            int mid = l + r >> 1;
            if (check(mid)) {
                ans = mid;
                r = mid - 1;
            }else {
                l = mid + 1;
            }
        }
        cout << ans << "\n";
    }
    signed main() {
        std::ios::sync_with_stdio(false);
        std::cin.tie(nullptr);
    
        int t = 1;
        std::cin >> t;
        while(t --) {
            solve();
        }
        return 0;
    }

[Problem - 1622C - Codeforces]: https://codeforces.com/problemset/problem/1622/C
[f6cd9c72cb09404e95da429847de7fb5.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/14/fb99cee97934493b828a5641d9532548.png
[fdc723d055784165bd580acc564e6d5e.png]: https://image.dandelioncloud.cn/pgy_files/images/2024/03/14/9ea380677bcd46209f0489dc94f504b6.png