HDU-3280 equal sum partitions

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 3280

Simple enumeration is used.

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 descriptionan Equal sum PartitionOf 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 7May be grouped: (2 5) (1 3 3) (7)To yield an equal sum 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. inputthe 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 in less than 10 values. outputfor 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 input31 62 5 1 3 72 61 2 3 4 5 63 201 1 2 1 1 1 2 1 2 2 11 2 1 1 1 1 1 2 1 1 sample output1 72 213 2

#include<iostream>#include<cstring>#include<cstdio>using namespace std;int a[10005];int main(){    int i,j,t,n,m,sum,cursum,flag ,ans;    scanf("%d",&t);    while(t--)    {          flag=0;        memset(a,0,sizeof(a));        scanf("%d%d",&n,&m);        for(i=0;i<m;i++)          scanf("%d",&a[i]);            for(i=0;i<m;i++)           {                 sum=0;            for(j=0;j<=i;j++)                  sum+=a[j];               cursum=0;             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=1;                        }                        cursum=0;                   }                 else                     j++;                 if(flag)                   break;             }            if(flag)                break;           }          if(i==m)           printf("%d %d\n",n,sum);     }    return 0;}/*31 62 5 1 3 3 72 61 2 3 4 5 63 201 1 2 1 1 2 1 1 2 11 2 1 1 2 1 1 2 1 1*/