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

Subset SumsJRMFor 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];

View Code

/*
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;
}