摘要：
=n）｛hh+=temp；pos=upper_bound（sum+1，sum+n+1，hh）-（sum+1）；if（sum[pos]！=hh）｛returnfalse；｝｝returntrue；｝intmain（）{inti，j，t，ca；sum[0]=0；scanf（“%d”，&t）；while（t--）{scanf（”%d%d“，&ca，&n）；intx；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”，ca，sum[i]）；}return0；｝
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
Total Submission(s): 551 Accepted Submission(s): 409
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.
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
Recommend
/*
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<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
typedef __int64 ll;
#define N 10005
#define INF 0x3f3f3f3f
int 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
",ca,sum[i]);
}
return 0;
}