山东省acm省赛 I Sequence（动态规划）

Sequence

Time Limit: 1000ms Memory limit: 65536K 有疑问？点这里^_^

题目描述

Given an integer number sequence A of length N (1<=N<=1000), we define f(i,j)=(A[i]+A[i+1]+...+A[j])^2 (i<=j). Now you can split the sequence into exactly M (1<=M<= N) succesive parts, and the cost of a part from A[i] to A[j] is f(i,j). The totle cost is the sum of the cost of each part. Please split the sequence with the minimal cost.

输入

At the first of the input comes an integer t indicates the number of cases to follow. Every case starts with a line containing N ans M. The following N lines are A[1], A[2]...A[N], respectively. 0<=A[i]<=100 for every 1<=i<=N.

输出

For each testcase, output one line containing an integer number denoting the minimal cost of splitting the sequence into exactly M succesive parts.

示例输入

1
5 2
1 3 2 4 5

示例输出

提示

来源

山东省第二届ACM大学生程序设计竞赛

示例程序

（动态规划）不过对时间要求很高，考试时用几种方法都是超时，之后看到了比较有技巧的思路，

 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <cstring>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <stdlib.h>
 7 
 8 using namespace std;
 9 typedef  long long ll;
10 ll sum[1010];
11 ll ans[1010];
12 ll min (ll a,ll b)
13 {
14     return a<b?a:b;
15 }
16 int main()
17 {
18     int T;
19     scanf("%d",&T);
20     while(T--)
21     {
22         int n,m;
23         scanf("%d %d",&n,&m);
24         ll t;
25         int arr;
26         sum[0]=0;
27         for(int i=1;i<=n;i++)
28         {
29             scanf("%d",&arr);
30             sum[i]=sum[i-1]+arr;
31             ans[i]=sum[i]*sum[i];
32 
33         }
34         for(int i=2;i<=m;i++)//刷新数组次数
35         {
36             for(int j=n-m+i;j>=i;j--)//倒叙刷新
37             {
38                 for(int k=i-1;k<j;k++)
39                 {
40                     t=sum[j]-sum[k];
41                     ans[j]=min(ans[k]+t*t,ans[j]);
42                 }
43             }
44         }
45         //printf("%I64d
",ans[n]);
46         cout<<ans[n]<<endl;
47     }
48     return 0;
49 }

最大的收获就是在linux系统中%I64d表示longlong类型可能会出错，就是因为这个，提交了好几遍PE.