Equal Sum Partitions

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 551 Accepted Submission(s): 409

Problem Description

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

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 

Source

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/* 题意:n个数，分成若干个集合，要求每一个集合的数和同样，求集合最小值 思路：枚举当前集合推断是否满足条件*/#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;typedef __int64 ll;#define N 10005#define INF 0x3f3f3f3fint sum[N];int n;bool fdd(ll temp){     int hh=0;     int pos=0;     while(pos!=n)     {         hh+=temp;         pos=upper_bound(sum+1,sum+n+1,hh)-(sum+1);         if(sum[pos]!=hh)         {            return false;         }     }     return true;}int main(){    int i,j,t,ca;    sum[0]=0;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&ca,&n);        int x;        for(i=1;i<=n;i++)        {            scanf("%d",&x);            sum[i]=sum[i-1]+x;        }        for(i=1;i<=n;i++)            if(fdd(sum[i])) break;        printf("%d %d\n",ca,sum[i]);    }    return 0;}

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