hdu3280Equal Sum Partitions (区间DP)

Problem Description

An equal sum partition of a sequence of numbers is a grouping of the numbers (in the same order as the original sequence) in such a way that each group has the same sum. For example, the sequence:

2 5 1 3 3 7

may be grouped as:

(2 5) (1 3 3) (7)

to yield an equal sum of 7.

Note: The partition that puts all the numbers in a single group is an equal sum partition with the sum equal to the sum of all the numbers in the sequence.

For this problem, you will write a program that takes as input a sequence of positive integers and returns the smallest sum for an equal sum partition of the sequence.

Input

The first line of input contains a single integer
P, (1 ≤ P ≤ 1000), which is the number of data sets that follow. The first line of each data set contains the data set number, followed by a space, followed by a decimal integer
M, (1 ≤ M ≤ 10000), giving the total number of integers in the sequence. The remaining line(s) in the dataset consist of the values, 10 per line, separated by a single space. The last line in the dataset may contain less than
10 values.

Output

For each data set, generate one line of output with the following values: The data set number as a decimal integer, a space, and the smallest sum for an equal sum partition of the sequence.

Sample Input

3
1 6
2 5 1 3 3 7
2 6
1 2 3 4 5 6
3 20
1 1 2 1 1 2 1 1 2 1
1 2 1 1 2 1 1 2 1 1

Sample Output

1 7
2 21
3 2

题意：给出一个序列，假设能把该序列分成若干段，使每段和相等，求最小和，若不能分则和为这一整序列。
#include<stdio.h>
int dp[7000][7000];
int min(int a,int b)
{
    return a>b?b:a;
}
int main()
{
    int a[10005],ans[10005],t,c,m;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&c,&m); ans[0]=0;
        for(int i=1;i<=m;i++)
        {
            scanf("%d",&a[i]);
            ans[i]=ans[i-1]+a[i];
        }
        for(int r=0;r<m;r++)
        for(int i=1;i<=m-r;i++)
        {
            int j=i+r;
            dp[i][j]=ans[j]-ans[i-1];
            for(int k=i;k<j;k++)
            {
                if(ans[k]-ans[i-1]==dp[k+1][j])
                dp[i][j]=min(dp[i][j],dp[k+1][j]);
                if(dp[i][k]==ans[j]-ans[k])
                dp[i][j]=min(dp[i][j],dp[i][k]);
                if(dp[i][k]==dp[k+1][j])
                dp[i][j]=min(dp[i][j],dp[i][k]);
            }
        }
        printf("%d %d\n",c,dp[1][m]);
    }
}