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HDU-3280 Equal Sum Partitions

2024-07-13 08:28:29   214人阅读

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 11 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[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*/

 


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