HDU-3280 Equal Sum Partitions

http://acm.hdu.edu.cn/showproblem.php?pid=3280

用了简单的枚举。

Equal Sum Partitions

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 453    Accepted Submission(s): 337

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

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int a[];
int main()
{
    int i,j,t,n,m,sum,cursum,flag ,ans;
    scanf("%d",&t);
    while(t--)
    {
          flag=;
        memset(a,,sizeof(a));
        scanf("%d%d",&n,&m);
        for(i=;i<m;i++)
          scanf("%d",&a[i]);
            for(i=;i<m;i++)
           {
                 sum=;
            for(j=;j<=i;j++)
                  sum+=a[j];
               cursum=;
             while(j<m)
              {
                cursum+=a[j];
               if(cursum>sum)
                   break;
               else if(cursum==sum)
                   {
                        j++;
                        if(j==m)
                        {
                           printf("%d %d\n",n,sum);
                           flag=;
                        }
                        cursum=;
                   }
                 else
                     j++;
                 if(flag)
                   break;
             }

            if(flag)
                break;
           }
          if(i==m)
           printf("%d %d\n",n,sum);
     }
    return ;
}
/*
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
*/

区间dp

#include<iostream>
#include<cstdio>
using namespace std;
int dp[][],ans[];
int main()
{
    int t,n,m,i,j,k,g,a[];
    cin>>t;
    while(t--)
    {
           cin>>n>>m;
            ans[]=;
        for(i=;i<=m;i++)
          {
              cin>>a[i];
              ans[i]=ans[i-]+a[i];
          }
         for(k=;k<m;k++)//k不能从1-m，虽然同样个数相同，但是j=2开始，就会使区间减少了一层，
         {               //比如i=1，j=2就没有这个区间。
             for(i=;i<=m-k;i++)
                 {
                     j=i+k;
                     dp[i][j]=ans[j]-ans[i-];//初始化dp，求出每个区间的和。
                      for(g=i;g<j;g++)
                       {//三者的顺序可以随便调换。
                           if((ans[g]-ans[i-])==dp[g+][j])
                                   dp[i][j]=min(dp[i][j],dp[g+][j]);
                           if(dp[i][g]==ans[j]-ans[g])
                                dp[i][j]=min(dp[i][j],dp[i][g]);
                                if(dp[i][g]==dp[g+][j])
                                 dp[i][j]=min(dp[i][j],dp[i][g]);

                       }

                 }
         }
         printf("%d %d\n",n,dp[][m]);
    }

}
/*
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
*/