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Dr.Kong’s laboratory monitor some interference signals. The interference signals can be digitized into a series of positive integer. May be, there are N integers a1,a2,…,an.
Dr.Kong wants to know the average strength of a contiguous interference signal block. the block must contain at least M integers.
Please help Dr.Kong to calculate the maximum average strength, given the constraint.
The input contains K test cases. Each test case specifies:
* Line 1: Two space-separated integers, N and M.
* Lines2~line N+1: ai (i=1,2,…,N)
1 ≤ K≤ 8, 5 ≤ N≤ 2000, 1 ≤ M ≤ N, 0 ≤ ai ≤9999
the maximum average strength
2 10 6 6 4 2 10 3 8 5 9 4 1 5 2 10 3 8 5 9
6500 7333
题目大意： 求连续序列大于等于k个的最大平均 值
dp[i][j] 表示在j的位置的i个连续序列的最大和 dp[i][j] = max(dp[i-1][j-1] + a[j], dp[i][j])
只要公式推出来就很容易了
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<math.h> #include<algorithm> #include<iostream> #include<vector> #include <queue> using namespace std; #define N 2500 #define ESP 1e-8 #define INF 0x3f3f3f3f #define memset(a,b) memset(a,b,sizeof(a)) int a[N]; int dp[N][N]; int main() { int T,n,k; scanf ("%d", &T); while(T --) { scanf("%d %d", &n, &k); for(int i=1; i<=n; i++) scanf("%d", &a[i]); memset(dp, 0); dp[1][1]=a[0]; for(int i=1; i<=n; i++) { for(int j=1; j<=n; j++) { dp[i][j] = max(dp[i][j], dp[i-1][j-1]+a[j]); } } double Max=0; for(int i=k; i<=n; i++) { for(int j=i; j<=n; j++) { Max = max(Max, dp[i][j]*1.0/i*1.0); } } printf("%d\n", (int)(Max*1000.0)); } return 0; }
G.Interference Signal---河南省第八届程序设计大赛（dp）
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原文地址：http://www.cnblogs.com/linliu/p/5526878.html
