Best Cow Fences POJ

Farmer John's farm consists of a long row of N (1 <= N <= 100,000)fields. Each field contains a certain number of cows, 1 <= ncows <= 2000.

FJ wants to build a fence around a contiguous group of these fields in order to maximize the average number of cows per field within that block. The block must contain at least F (1 <= F <= N) fields, where F given as input.

Calculate the fence placement that maximizes the average, given the constraint.

Input

* Line 1: Two space-separated integers, N and F.

* Lines 2..N+1: Each line contains a single integer, the number of cows in a field. Line 2 gives the number of cows in field 1,line 3 gives the number in field 2, and so on.

Output

* Line 1: A single integer that is 1000 times the maximal average.Do not perform rounding, just print the integer that is 1000*ncows/nfields.

Sample Input

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

Sample Output

6500


题意：n块田地，每块田地有cows【i】头牛，求出一个长度不小于F的子段，使子段牛的平均数最大。
思路：我们令 avr = sum【i，j】/（i-j+1）   
那么这题就是 求是avr的最大值，我们二分枚举ans，判断 avr 是否不小于 ans，即avr >= ans,为了不维护（i-j+1）的值，变形成sum【i，j】 - ans*（i-j+1） >= 0，
我们把原数组cows【i】-ans，那么Sum【i，j】 = sum【i，j】-ans（i-j+1）。
现在关键就是取得一个  max{Sum【i，j】}，对于Sum【i，j】，我们可以用前缀和相减的方式求得，sum（i）-min（sum（j））, 0 <= j <= i-F。

坑点：注意最后答应的是二分的R值，我打印L值Wa的怀疑人生.而且题目要求精度是1e-4，当我们枚举的精度不小于题目要求精度的时候L和R值都是OK的
我感觉是如果两个值转换为整数不同的话，R值转换的整数值在L~R区间内的，而l转换的值是小于l的，如果两个值转换的值相同，打印那个都行，且整数肯定小于L。

其实像是最大值最小，最小值最大，都可以用二分解决，答案是单调的，这题还有种用凸包方法写的，以后再填坑吧。

#include<cstdio>
#include<iostream>
using namespace std;

const int maxn = 1e5+5;
int n,f;
int cows[maxn];
const double eps = 1e-8;

bool solve(double x,int f)
{
    double fcows[n+1];
    double sum[n+1];
    for(int i=1;i<=n;i++)fcows[i] = cows[i] - x;
    for(int i=1;i<=n;i++)sum[i] = sum[i-1]+fcows[i];
    double ans = -1e10,minn = 1e10;
    for(int i=f;i<=n;i++)
    {
        minn = min(minn,sum[i-f]);
        ans = max(ans,sum[i]-minn);
    }
    return ans >= 0;
}
int main()
{
    scanf("%d%d",&n,&f);
    double low=0;
    double high = 0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&cows[i]);
        high += cows[i];
    }
    while(low + eps < high)
    {
        double mid = (low+high)/2;
        if(solve(mid,f))low = mid;
        else high = mid;
    }
    printf("%d
",(int)(1000*high)); 
}