USACO Section 2.2 Subset Sums（整数划分01背包思想）

Subset 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
题意：输入一个 n ，从 1 到 n 这 n 个数分成两部分，这两部分的和相等，问这种方法有多少种。
分析：d[i][j] 表示前 i 个数组合起来的和是 j 的种数，对 j 考虑选择与不选择两种情况，不选：d[i-1][j],选择：d[i-1][j-i]；所以d[i][j]=d[i-1][j]+d[i-1][j-i];

/*
  ID: dizzy_l1
  LANG: C++
  TASK: subset
*/
#include<iostream>
#include<cstring>
#include<cstdio>

using namespace std;

long long d[1000];

int main()
{
    freopen("subset.in","r",stdin);
    freopen("subset.out","w",stdout);
    int n,sumv,i,j;
    while(cin>>n)
    {
        sumv=(n+1)*n/2;
        if(sumv%2==1)
        {
            printf("0\n");
            continue;
        }
        sumv=sumv/2;
        memset(d,0,sizeof(d));
        d[0]=1;
        for(i=1;i<=n;i++)
        {
            for(j=sumv;j>=i;j--)
            {
                d[j]+=d[j-i];
            }
        }
        cout<<d[sumv]/2<<endl;
    }
    return 0;
}