Equal Sum Partitions
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 720 Accepted Submission(s): 512
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
Source
2009 Greater New York Regional
题意：给一串数字，问能否把这些数字分成连续的几个区间，并且每部分和相等，输出最小的区间和；
思路：直接暴力，算出所有的数字的和sum，枚举1~sum，第一个符合条件的值便是答案。
AC代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int t,cas,n,a[10005];
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&cas,&n);
int sum=0,ans;
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
sum+=a[i];
}
int s,flag;
for(int i=1;i<=sum;i++)
{
if(sum%i==0)
{
s=0,flag=1;
for(int j=1;j<=n;j++)
{
s+=a[j];
if(s==i)s=0;
if(s>i){flag=0;break;}
}
}
if(flag){ans=i;break;}
}
printf("%d %d\n",cas,ans);
}
return 0;
}
