2.2 subset

ubset Sums
JRM

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.

PROGRAM NAME: subset

INPUT FORMAT

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

SAMPLE INPUT (file subset.in)

7

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 OUTPUT (file subset.out)

4



/*
ID: xdzhbin1
LANG: C++
TASK: subset
*/

//典型的动态规划，为什么自己就想不到呢，那就记住！
//自己首先想到的是递归枚举，明显效率低会超时
//找到子问题，目标问题到子问题

#include <fstream>
#include <iostream>
using namespace std;

int main()
{
	ifstream in("subset.in",ios::in);
	ofstream out("subset.out",ios::out);
	unsigned int array[40][781];
	int n;
	in>>n;
	
	int i,j;
	for (i=0; i<40; i++)
	{
		for (j=0; j<781; j++)
		{
			array[i][j] = 0;
		}
	}

	for (i=0; i<40; i++)
	{
		array[i][0] = 1;
		array[i][1] = 1;
	}
	for (i=2; i<=n; i++)
	{
		int temp = i*(i+1)/2;
		for (j=2; j<=temp; j++)
		{
			//数字i取与不取，两种情况
			array[i][j] = array[i-1][j] + array[i-1][j-i];
		}
	}
	if ((n*(n+1)/2)%2 == 1)
	{
		out<<"0"<<endl;
	}
	else
	{
		out<<array[n][n*(n+1)/4]/2<<endl;
	}
	in.close();
	out.close();
	return 0;
}