最大均值

ybtoj 二分-1-3

题目大意

给出一个序列，让你求一个长度不小于m的子序列，使其平均值最大

输入样例

10 6
6 
4
2
10
3
8
5
9
4
1

输出样例

6500

数据范围

$1⩽L⩽N⩽10^5$
$A_i⩽2000$

解题思路

二分平均值，然后用数组存下所有数对当前平均值的贡献
然后判断是否有长度大于L的序列贡献值大于0

代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 100010
#define max(x, y) (x)>(y)?(x):(y)
#define min(x, y) (x)<(y)?(x):(y)
using namespace std;
ll n, m;
double l, r, mid, a[N], sum[N];
bool check(double g)
{for (int i = 1; i <= n; ++i)sum[i] = sum[i - 1] + a[i] - g;double ans = -2000.0 * N, minn = 2000.0 * N;for (int i = m; i <= n; ++i)//求长度不小于L的序列贡献值最大是多少{minn = min(minn, sum[i - m]);ans = max(ans, sum[i] - minn);}return ans >= 0;//大于0
}
int main()
{scanf("%lld%lld", &n, &m);for (int i = 1; i <= n; ++i)scanf("%lf", &a[i]);l = 0;r = 2001.0;while(l + 1e-5< r)//二分{mid = (l + r) / 2;if (check(mid)) l = mid;else r = mid;}printf("%lld", (ll)(r*1000));return 0;
}