USACO 2.2 Subset Sums

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

太天真了，前几天dfs做多了，以为又是一个二值搜索，简单的写了一个dfs到第四个test就开始超时了。。。哦，果然这种集合里挑几个数的问题应该归为背包问题。设b[i][j] 是前i个数和为j的种数。转移方程附图。

代码如下：

/*ID: gjj50201LANG: C++TASK: subset*/#include <stdio.h>#include <iostream>#include <stdlib.h>#include <algorithm>using namespace std;int n,sum;int b[40][850] = {0};int main(){freopen("subset.in","r",stdin);freopen("subset.out","w",stdout);cin>>n;if((1+n)*n/2 % 2 == 1){cout<<"0"<<endl;return 0;}b[1][1] = 1;for(int i=2;i<=n;i++){sum = (1+i)*i/2;for(int j=1;j<=sum;j++){if(j>=i)b[i][j] = b[i-1][j-i] + b[i-1][j];elseb[i][j] = b[i-1][j];}}cout<<b[n][(1+n)*n/4]<<endl;return 0;}