POJ2018 Best Cow Fences

Best Cow Fences

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 11175		Accepted: 3666

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

Source

USACO 2003 March Green

 

【题解】

　　这是一道极其恶心的卡精度题，据说有O(n)做法，但太高端了。

　　nlogn做法很显然是二分答案，关键就在如何检查了。

　　直接去求平均值是否大于某个数跟暴力无异，因此我们转换一下，把每个数都减去答案，求一个满足条件的区间且区间和大于0。

　　预处理dp[i]表示i及i向右的区间最大是多少，可以不选，为0。

　　线性做就可以了

　　放大处理，被卡了老半天精度，最终弃疗，老老实实乘了1ex

　　

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>

inline void read(long long &x)
{
    x = 0;char ch = getchar(),c = ch;
    while(ch < '0' || ch > '9')c = ch, ch = getchar();
    while(ch <= '9' && ch >= '0')x = x * 10 + ch - '0', ch = getchar();
    if(c == '-')x = -x;
}

const int MAXN = 100000 + 10;

long long n,num[MAXN],len,sum,tmp[MAXN],dp[MAXN];

int check(long long m)
{
    for(register int i = 1;i <= n;++ i)
        tmp[i] = num[i] - m;
    //dp[i]表示从i开始向右的最大和 
    for(register int i = n;i >= 1;-- i)
        if(dp[i + 1] + tmp[i] > 0)dp[i] = dp[i + 1] + tmp[i];
        else dp[i] = 0;
    for(register int i = 1;i <= n;++ i)
        tmp[i] += tmp[i - 1];
    //最短长度len + 向右最大和 
    for(register int i = len;i <= n;++ i)
        if(tmp[i] - tmp[i - len] + dp[i + 1] >= 0)
            return 1;
    return 0;
} 

int main()
{
    read(n), read(len);
    for(register int i = 1;i <= n;++ i)
        read(num[i]), num[i] *= 10000000, sum += num[i];
    register long long l = 1, r = sum, mid, ans;
    while(l <= r)
    {
        mid = (l + r) >> 1;
        if(check(mid)) l = mid + 1;
        else r = mid - 1;
    }
    printf("%lld", l/10000);
    return 0;
}