题目
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
大意
给你 n 个田地的牛数,找出连续的几块田地且田地数量 >= F,使这几块田地内的每个田地的平均牛数最大。
输出最大的平均牛数∗1000 下取整 的值。
思路
二分平均值,再去检查是否存在长度 >= F 且平均值 > = mid 的子串.
平均值小技巧:
判断( a1+a2+...+an)/n >= mid
等价于 ( a1+a2+...+an) >= mid*n
等价于 (a1-mid)+(a2-mid)+(a3-mid)+...+(an-mid)>=0
-
每个数都减去 mid,问题就转化成了 看是否存在长度 >= F 且这些数的和 >= 0 的子段
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对这个序列进行前缀和操作,sum [ i ]
某个点编号为 j ,以这个点结尾的子序列,左端点编号为i,那么子序列的和就是 sum [ j ] - sum [ i ]
判断 sum [ j ] - sum [ i ] >=0
即判断是否存在序列满足sum [ j ] >= sum [ i ]
i 的范围是 1 ~( j - F )扫描二维码关注公众号,回复: 9483942 查看本文章 -
我们是要判断在 i 的范围里,是否存在 sum [ i ] ,使得sum [ j ] >= sum [ i ] 成立
既然是存在,那么只要看看在这个范围里,sum[ i ] 的最小值 是不是满足这个等式
4.二分平均值
l=0,r=2000 为平均值的初始上下限
用二分来逐渐缩小最大平均值的范围
每次判断时
假若存在区间 > mid,那么mid的最大值就在 mid ~ r 中
所以 l = mid
假若不存在区间 > mid ,那么mid的最大值就在 l ~ mid 中
所以 r = mid
AC代码
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<vector>
#include<cmath>
#include <queue>
using namespace std;
typedef long long ll;
const int N= (int)1e5;
const int inf = 0x3f3f3f3f;
int n,f;
int cows[N+10];
double sum[N+10];
bool check(double ave)
{
for(int i=1;i<=n;i++)sum[i]=sum[i-1]+cows[i]-ave;
///前缀和
double minv=inf*1.0;///sum[i]最小
for(int j=f;j<=n;j++)
{
minv=min(minv,sum[j-f]);
if(sum[j]>minv)
return 1;///存在
}
return 0;
}
int main()
{
cin>>n>>f;
for(int i=1;i<=n;i++)
{
scanf("%d",&cows[i]);
}
double l=0,r=2000;///平均值上下限
while(r-l>1e-5)
{
double mid=(l+r)/2.0;
if(check(mid))l=mid;
else r=mid;
}
printf("%d\n",int(r*1000));
return 0;
}