TZOJ 2941 Equal Sum Partitions 模拟

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.

输入

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.

输出

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.

样例输入

 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

样例输出

1 7

2 21

3 2

题意：按顺序组合成多组（可以一组）同样的值（最小）

解析：先存一个前n个和的数组，从第一个开始，依次看后面能否成立，不行就退出进行下一个比较。

#include<bits/stdc++.h>
using namespace std;
int a[11111];
int sum[11111];
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        scanf("%d%d",&m,&n);
        memset(a,0,sizeof(a));
        scanf("%d",&a[1]);
        sum[1]=a[1];
        for(int i=2;i<=n;i++)
        {
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        int ans;
        for(int i=1;i<=n;i++)
        {
            ans=0;
            for(int j=1;j<=i;j++)
                ans+=a[j];
            int j,js=i;
            for(j=js;j<n;j++)
            {
                if(ans<sum[j]-sum[js])
                    break;
                else if(ans>sum[j]-sum[js])
                    continue;
                else
                {
                    js=j;
                }
            }
            if(ans!=sum[j]-sum[js])
                ans=sum[n];
            if(j==n)
                break;
        }
        cout<<m<<" "<<ans<<endl;
    }
}