CF231C To Add or Not to Add 题解 双指针

题目链接:https://www.luogu.com.cn/problem/CF231C

解题思路:双指针。
题解有时间补上。

实现代码如下:

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
int n;
long long k, a[maxn], cost, tmp_cost, cnt, ans;
int main() {
    cin >> n >> k;
    for (int i = 1; i <= n; i ++) cin >> a[i];
    sort(a+1, a+1+n);
    int j = 1;
    cnt = 1; ans = a[1];
    for (int i = 2; i <= n; i ++) {
        tmp_cost += (a[i] - a[i-1]) * (i-j);
        while (tmp_cost > k) {
            tmp_cost -= a[i]-a[j];
            j ++;
        }
        if (i-j+1 > cnt) {
            cnt = i-j+1;
            ans = a[i];
        }
    }
    cout << cnt << " " << ans << endl;
    return 0;
}

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转载自www.cnblogs.com/quanjun/p/12740808.html
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