POJ2018总结

Best Cow Fences

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

我的代码

#include<stdio.h>

int n,f,sum,num;
int fields[100010];
double temp;

int main()
{
double temp1,temp2,ans;
scanf("%d%d",&n,&f);
for (int i=0;i<n;i++)
{
scanf("%d",&fields[i]);
}
sum=0;num=0;
for (int i=0;i<f;i++)
{
num++;
sum+=fields[i];
}
temp=sum;
ans=1.0*sum/num;
for (int i=f;i<n;i++)
{
temp1=1.0*(sum+fields[i])/(num+1);
temp+=fields[i];
temp-=fields[i-f];
temp2=1.0*temp/f;

if (temp1>temp2){
sum+=fields[i];
num++;
if (temp1>ans) ans=temp1;
}else{
sum=temp;
num=f;
if (temp2>ans) ans=temp2;
}
}
printf("%d\n",(int)(ans*1000));
return 0;
}

反思

Do not perform rounding, just print the integer that is 1000*ncows/nfields.