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.
输入
* 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.
输出
* 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.
解题思路： 题目意思找长度大于等于的F的连续的平均值最大 我是从1e-6~1e6 开始二分的 二分一直是每个数减去平均值求前缀和 从长度为L的地方开始判断 减去i-L之前的最小值看是不是大于0即可
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <algorithm> 5 #define eps 1e-6 6 using namespace std; 7 const double INF=0x3f3f3f3f*1.0; 8 int n,m; 9 double arr[100005]; 10 double sum[100005]; 11 12 bool check(double num){ 13 double minn=INF; 14 for(int i=1;i<=n;i++){ 15 sum[i]=sum[i-1]+arr[i]-num; 16 } 17 double ans; 18 for(int i=m;i<=n;i++){ ///看长度大于等于m的有没有大于等于0的 19 minn=min(minn,sum[i-m]); //i-m 之前的最小值 20 ans=sum[i]-m; // 21 if(ans>=0) return true; //这个平均值是可以达到的 22 } 23 return false; 24 } 25 26 void Calculation(){ 27 double left=1e-6; 28 double right=1e6; 29 while(left+eps<right){ 30 double mid=(left+right)/2; 31 if(check(mid)) left=mid; 32 else right=mid; 33 } 34 printf("%d ",(int)(right*1000)); 35 } 36 37 38 int main(){ 39 scanf("%d%d",&n,&m); 40 for(int i=1;i<=n;i++) scanf("%lf",&arr[i]); 41 Calculation(); 42 return 0; 43 }