# 11318: 【原1318】subset

## 题目

### 题目描述

author: USACO 原OJ链接：https://acm.sjtu.edu.cn/OnlineJudge-old/problem/1318

## Description

For many sets of consecutive integers from 1 through N (1 <= N <= 39), one can partition the set into two sets whose sums are identical.

For example, if N=3, one can partition the set {1, 2, 3} in one way so that the sums of both subsets are identical:

{3} and {1,2}

This counts as a single partitioning (i.e., reversing the order counts as the same partitioning and thus does not increase the count of partitions).

If N=7, there are four ways to partition the set {1, 2, 3, ... 7} so that each partition has the same sum:

{1,6,7} and {2,3,4,5}

{2,5,7} and {1,3,4,6}

{3,4,7} and {1,2,5,6}

{1,2,4,7} and {3,5,6}

Given N, your program should print the number of ways a set containing the integers from 1 through N can be partitioned into two sets whose sums are identical. Print 0 if there are no such ways.

Your program must calculate the answer, not look it up from a table.

## Input Format

The input file contains a single line with a single integer representing N, as above.

## Output Format

The output file contains a single line with a single integer that tells how many same-sum partitions can be made from the set {1, 2, ..., N}. The output file should contain 0 if there are no ways to make a same-sum partition.

## Sample Input:

```
7
```

## Sample Output:

```
4
```

## yyong119's solution

```
#include<iostream>
using namespace std;
int dp[50][10000];
int main()
{
int N; cin >> N;
int sum = (1 + N) * N / 2;
if(sum % 2 != 0) {cout << 0 << endl; return 0;}
dp[1][1] = 1;
sum /= 2;
for(int i = 2; i <= N; ++i)
{
for(int j = 1; j <= i * (i+1) / 2; ++j)
dp[i][j] = dp[i-1][j] + dp[i-1][j-i];
}
cout << dp[N][sum] << endl;
return 0;
}
```