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

动态转移方程：当j<i时dp[i][j] = dp[i-1][j]；当j>=i时，dp[i][j] = dp[i-1][j]+dp[i-1][j-i]；dp[i][j]表示在前i--1个数的前提下，前i个数能够凑成和为j的个数。dp[i-1][j]表示前i-1个数能凑成和为j的个数，dp[i-1][j-i]表示在前i-1个数的和的基础上加上第i个数能凑成和为j的个数。两个加起来就是前i个数能凑成和为j的总的个数。

/*
ID:hanzhic1
PROG:subset
LANG:C++
 */


#include<iostream>
#include<fstream>
#include<cstring>


using namespace std;


int dp[40][400];


void subdp(int n,int sum)
{
    int i,j;
    dp[1][1] = 1;
    for(i = 2;i<=n;i++)
    {
for(j = 1;j<=sum;j++)
{
   dp[i][j] = dp[i-1][j];
   if((j-i)>0)
   {
dp[i][j]+=dp[i-1][j-i];
   }
}
    }
}


int main()
{
    ifstream fin("subset.in");
    ofstream fout("subset.out");
    int n;
    int ans;
    int sub;
    fin>>n;
    memset(dp,0,sizeof(dp));
    sub = n*(1+n)/2;
    ans = 0;
    if(sub%2==0)
    {
subdp(n,sub/2);
ans = dp[n][sub/2];
    }
    fout<<ans<<endl;
    return 0;
}

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3232 KB]
   Test 2: TEST OK [0.000 secs, 3232 KB]
   Test 3: TEST OK [0.000 secs, 3232 KB]
   Test 4: TEST OK [0.000 secs, 3232 KB]
   Test 5: TEST OK [0.000 secs, 3232 KB]
   Test 6: TEST OK [0.000 secs, 3232 KB]
   Test 7: TEST OK [0.000 secs, 3232 KB]

All tests OK.
Your program ('subset') produced all correct answers!  This is your
submission #6 for this problem.  Congratulations!

Here are the test data inputs:

------- test 1 ----
7
------- test 2 ----
15
------- test 3 ----
24
------- test 4 ----
31
------- test 5 ----
36
------- test 6 ----
39
------- test 7 ----
37