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; } }