Best Cow Fences Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13960   Accepted: 4507 Description 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 题目大意：给一个n个数字，让你求着n个数字中长度大于m的字段的最大平均值。 做这题时，是二分专题训练，所以本题要采用二分法，就是要二分答案，判断答案是否满足条件 这题当时并没有思路，然后看了其他人的博客后才知道如何写出来，二分是很好写的，主要写的是 如何判断这个值是否满足需要的条件。首先求一下数组的前缀和，把数组中的每一个元素先减去二分 出来的那个值，然后求取前缀和，在进行判断。如何判断是关键，要用两个变量minn，maxx， double minn=INF; double maxx=-INF; for(int j=m;j<=n;j++) { if(sum[j-m]maxx) maxx=sum[j]-minn; } 我进行判断二分的条件是，只关注有没有一段区间的和大于0。 #include #include #include #include using namespace std; #define INF 1e10 double field[1000010]; int n,m; double sum[1000010]; bool Is_ok(double x) { for(int i=1;i<=n;i++) { sum[i]=sum[i-1]+(field[i]-x); } double minn=INF; double maxx=-INF; for(int j=m;j<=n;j++) { if(sum[j-m]maxx) maxx=sum[j]-minn; } if(maxx>=0) return true; else return false; } int main() { scanf("%d%d",&n,&m); for(int i=1;i<=n;i++) { scanf("%lf",&field[i]); } double l=0,r=1e6; double mid; double exp=1e-5; while(r-l>exp){ mid=(l+r)/2; if(Is_ok(mid)) l=mid; else r=mid; } cout< 注意倒数第三行代码，强制类型转换我不知道为什么printf过不去！！！！