poj 2018

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题意：在所有长度大于等于m的区间中找一个平均值最大的，输出平均数。

solution：DP

f(i)，num(i) 分别表示最后取第 i 个元素的最优决策 元素值之和 与 元素个数

对于 f(i)，只有两种情况：
1. f(i)=∑nj=n−m+1cow(j),num(i)=m
2. f(i)=f(i−1)+cow(i),num(i)=num(i−1)+1

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#include<cstdio>#include<cstdlib>#include<iostream>#include<algorithm>const int MAXN = 1e5+5;int n, m;int c[MAXN] = {0}, sum[MAXN] = {0};int f[MAXN] = {0}, num[MAXN] = {0};int ans = 0;int main(){#ifndef ONLINE_JUDGE    freopen("poj2018.in","r",stdin);    freopen("poj2018.out","w",stdout);#endif    std::cin >> n >> m;    for(int i = 1; i <= n; i++)       {scanf("%d",c+i); sum[i] = sum[i-1]+c[i];}    f[m] = sum[m], num[m] = m;    for(int i = m+1; i <= n; i++)    {        f[i] = sum[i] - sum[i-m] ,num[i] = m;        if((f[i-1]+c[i])*num[i] > f[i]*(num[i-1]+1) )            f[i] = f[i-1]+c[i], num[i] = num[i-1]+1;        }    for(int i = m; i <= n; i ++)        ans = std::max(ans, f[i]*1000/num[i]);    std::cout << ans << std::endl;  #ifndef ONLINE_JUDGE    fclose(stdin);    fclose(stdout);#endif      return 0;}