USACO－Subset Sums

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：给你从1－n的连续数字，求有多少种将它们平分成两部分的方法。

等差数列求和公式：S＝(n+1)*n/2。
要平分，每一部分为S1=(n+1)*n/4,于是在开头判断S1是否为整数，不能则直接不能。
接着dp，最后要除以2，因为会算重复一次。

f[i,j]=f[i-1,j]+f[i-1,j-i] ,j-i>=0
f[i,j]=f[i-1,j] ,j-i<0

代码：

/*ID:iam666PROG:subsetLANG:C++*/#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>using namespace std;long long dp[800];int n;int main (void){    //freopen("subset.in","r",stdin);    //freopen("subset.out","w",stdout);    cin>>n;    if(n*(n+1)%4)        {            printf("0\n");            return 0;        }    int re=n*(n+1)/4;    memset(dp,0,sizeof(dp));    for(int i=0;i<=n;i++)    dp[0]=1;    for(int i=1;i<=n;i++)    {        for(int j=re;j>=i;j--)        {            dp[j]=dp[j]+dp[j-i];//j比i多出来的数的分法加上原来的分法        //  printf("%lld___\n",dp[j]);        }    //  printf("*******\n");    }    printf("%lld\n",dp[re]/2);    return 0;}