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

117

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，

#include <iostream>
#include <stdio.h>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stdlib.h>

using namespace std;
typedef  long long ll;
ll sum[1010];
ll ans[1010];
ll min (ll a,ll b)
{
    return a<b?a:b;
}
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d %d",&n,&m);
        ll t;
        int arr;
        sum[0]=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&arr);
            sum[i]=sum[i-1]+arr;
            ans[i]=sum[i]*sum[i];

        }
        for(int i=2;i<=m;i++)//刷新数组次数
        {
            for(int j=n-m+i;j>=i;j--)//倒叙刷新
            {
                for(int k=i-1;k<j;k++)
                {
                    t=sum[j]-sum[k];
                    ans[j]=min(ans[k]+t*t,ans[j]);
                }
            }
        }
        //printf("%I64d
",ans[n]);
        cout<<ans[n]<<endl;
    }
    return 0;
}

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.