Best Cow Fences 二分
问题 K: Best Cow Fences
时间限制: 1 Sec 内存限制: 128 MB
提交: 25 解决: 11
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题目描述
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.
样例输入
10 6 6 4 2 10 3 8 5 9 4 1
样例输出
6500 求最大平均数,长度不小于k的连续子串 如果先减去平均值,就转变为求一个长度不小于k且非负的连续子串
利用前缀和相减的形式计算
double minn=INF,ans=-INF;
for(int i=k; i<=n; i++){
minn=min(minn,sum[i-k]);
ans=max(ans,sum[i]-minn);
}
因为一段数的平均数不会大于这些数里的最大值,不会小于最小值,所以利用二分来计算平均值 代码:
#include <bits/stdc++.h>
using namespace std;
const int maxx=1e5+100;
const int INF=1e10;
const int MOD=1e9+7;
typedef long long ll;
double a[maxx];
double b[maxx],sum[maxx];
int main()
{
int n,k;
double l=INF,r=0;
scanf("%d%d",&n,&k);
for(int i=1; i<=n; i++){
scanf("%lf",&a[i]);
l=min(l,a[i]);
r=max(r,a[i]);
}
//l=-1e6;
//r=1e6;
while(r-l>1e-5){
double mid=(l+r)/2;
for(int i=1; i<=n; i++) b[i]=a[i]-mid;
for(int i=1; i<=n; i++) sum[i]=sum[i-1]+b[i];
double minn=INF,ans=-INF;
for(int i=k; i<=n; i++){
minn=min(minn,sum[i-k]);
ans=max(ans,sum[i]-minn);
}
if(ans>=0) l=mid;
else r=mid;
}
cout<<int(r*1000)<<endl;
return 0;
}