Best Cow Fences
Time Limit: 1000MS | Memory Limit: 30000K | |
Total Submissions: 12527 | Accepted: 4063 |
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
#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
const int maxn=1e5+10;
int a[maxn],deq[maxn];
ll sum[maxn];
bool check(int i,int j,int x,int y)
{
return (sum[j]-sum[i])*(y-x)>=(sum[y]-sum[x])*(j-i);
}
int main()
{
int n,f;
scanf("%d%d",&n,&f);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
for(int i=1;i<=n;i++)
sum[i]=sum[i-1]+a[i];
int s=0,t=0;
deq[t++]=0;
ll ma=0;
for(int i=f;i<=n;i++)
{
while(s<t-1&&check(deq[s+1],i,deq[s],i))s++;
ma=max(ma,1000*(sum[i]-sum[deq[s]])/(i-deq[s]));
while(s<t-1&&check(deq[t-2],deq[t-1],deq[t-1],i-f+1)) t--;
deq[t++]=i-f+1;
}
printf("%I64d\n",ma);
return 0;
}