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.
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.
* 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】给出n个数,在这n个数里面找到一些连续的数,这些数的数量大于等于m,并且他们的平均值在这n个数里面最大
【思路】先把n个数的最大最小值确定,然后二分枚举平均值,对于每一个连续数,只要他们减去平均值大于0,就调制上限制,不然调整下限制,.......
#include<iostream> #include<stdio.h> #include<string.h> using namespace std; const int inf=0x3f3f3f3f; const int N=100005; double a[N],sum[N]; int n,m; int check(double k) { double tmp=sum[m-1]-(m-1)*k; for(int i=m;i<=n;i++) { tmp+=a[i]-k; tmp=max(tmp,sum[i]-sum[i-m]-m*k);//新选m个数和前面的数比较,取最大值 if(tmp>-1e-6) return 1; } return 0; } int main() { while(~scanf("%d%d",&n,&m)) { sum[0]=0; double maxn=0,minx=inf; for(int i=1;i<=n;i++) { scanf("%lf",&a[i]); if(a[i]>maxn) maxn=a[i]; if(a[i]<minx) minx=a[i]; sum[i]=sum[i-1]+a[i]; } while(maxn-minx>=1e-6) { double mid=(maxn+minx)/2; if(check(mid)) minx=mid; else maxn=mid; } int ans=1000*maxn; printf("%d ",ans); } return 0; }