Codeforces Round 973 (Div. 2) - D题

wx200411022024-09-21 12:12

传送门：Problem - D - Codeforces

题目大意：

思路：

尽量要最大值变小，最小值变大

即求最大值的最小和最小值的最大 -> 二分答案

AC代码：

代码有注释

cpp 复制代码

#include<bits/stdc++.h>
using namespace std;
#define int long long
void solve()
{
    int n; cin >> n;
    vector<int> a(n + 1), b(n + 1);
    for (int i = 1; i <= n; i++) cin >> a[i];
    auto check1 = [&](int limit)
        {
            // limit 此时就是 最大值的最小值
            // 经过操作后，若 b[i] <= limit 就是ok的，否则就放弃这个值
            // 最大值最小
            for (int i = 1; i <= n; i++) b[i] = a[i];
            for (int i = 1; i < n; i++)
            {
                // b[i] 超过 limit ，就要减小 b[i]
                if (b[i] > limit)
                {
                    b[i + 1] += (b[i] - limit);
                    b[i] = limit;
                }
            }
            for (int i = 1; i <= n; i++)
            {
                if (b[i] > limit) return false;
            }
            return true;
        };
    int left = 0; int right = 1e12;
    while (right > left)
    {
        int mid = left + right >> 1;
        if (check1(mid))right = mid;
        else left = mid + 1;
    }
    int ans = left;
    auto check2 = [&](int limit)
        {
            // 最小值最大
            // limit 就是最小值的最大值
            for (int i = 1; i <= n; i++) b[i] = a[i];
            for (int i = 1; i < n; i++)
            {
             
                if (b[i] > limit)
                {
                    b[i + 1] += (b[i] - limit);
                    b[i] = limit;
                }
            }
            int mn = 2e18;
            for (int i = 1; i <= n; i++) mn = min(mn, b[i]);
            // 经过操作后，mn 仍大于 limit ，则可以继续增大limit
            if (mn >= limit)return true;
            else return false;
        };
    left = 0; right = 1e12;
    while (right > left)
    {
        int mid = left + right + 1 >> 1;
        if (check2(mid))left = mid;
        else right = mid - 1;
    }
    cout << ans - left << endl;
}
signed main()
{
    int tt; cin >> tt;
    while (tt--)solve();
    return 0;
}