USACO 2.2：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.

Input

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

Output

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 Input

7

Sample Output

4

Sample Code

//版本1#include<cstdio>const int maxn=1000;int dp[maxn][maxn];int n;int main(){scanf("%d",&n);int sum=n*(1+n)/2;if(sum%2==1) printf("0\n");else{sum/=2;dp[0][0]=1;int i,j;for(i=1;i<=n;i++)for(j=1;j<=sum;j++)dp[i][j]=dp[i-1][j]+dp[i-1][j-i];printf("%d\n",dp[n][sum]);}return 0;}

//版本2 #include<cstdio>const int maxn=1000;long long dp[maxn];int n;int main(){scanf("%d",&n);int sum=n*(1+n)/2;if(sum%2==1) printf("0\n");else{sum/=2;dp[0]=1;int i,j;for(i=1;i<=n;i++)for(j=sum;j>=i;j--)dp[j]=dp[j]+dp[j-i];printf("%lld\n",dp[sum]/2);}return 0;}