This is a explanation of this problem from USACO's training website. I have converted it to markdown. Please do not just copy code; you will not learn anything; at least type it out and understand so you can do it yourself in the future!

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 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
``````

Java

C++

Pascal

## ANALYSIS

This is a classic dynamic programming problem.

The basis of a handful of DP algorithms is the “take-an-old-count, add some to it, and carry it forward again”. Subset sums is a classic example of this.

First, let’s rephrase the task as, “Given N, calculate the total number a partition must sum to {n*(n+1)/2 /2}, and find the number of ways to form that sum by adding 1, 2, 3, … N.”

Thus, for N=7, the entire set of numbers 1..7 sums to 7*8/2 which is 56/2=28. And, if we partitioned that into two equal-sum sets, they’d pretty much have to sum to ** 14 **.

Now comes the ‘DP trick’ for this class of DP tasks. We’ll set up a little one-dimensional array called… ‘buckets’. In this array reside the counts of ways you can build various numbers given a set of numbers 1.. m. Thus, over time, buckets looks like this:

``````Lev::   0  1  2  3  4  5  6  7  8  9 10 11  ...
1::   1  1  0  0  0  0  0  0  0  0  0  0  ...
2::   1  1  1  1  0  0  0  0  0  0  0  0  ...
3::   1  1  1  2  1  1  1  0  0  0  0  0  ...
4::   1  1  1  2  2  2  2  2  1  1  1  0  ...
``````

If we look at row 1::, we see “There is 1 way to make a 0 and one way to build a 1 using the set {1}”

If we look at row 2::, we see “There is 1 way to make a 0 (this is always true), one way to make a 1, one way to make a 2, and one way to make a 3” – the 3 is from 1 + 2

If we look at row 3::, we see: “… two ways to make a 3 …”. Since you can make a 3 with: 3 or 1+2, now we’re seeing some growth.

Row 4:: says “… two ways for a 3, two ways for a 4, two ways for a 5, two ways for a 6, two ways for a 7, …” and one way for many others.

We calculate row N+1 from row N.

For each bucket j in our set:

``````newbucket[j + added_number] += olderbucket[j]
``````

So if the added_number is 3, and you’re processing column 4,

``````newbucket [4 + 3] += olderbucket[4]
``````

which is to say, “With a 3 in our hand, realize we can create a 7 by olderbucket[4] ways” and mark that down. The above sentence is the key. Once you see that, everything becomes clear.

The answer is found for N=7 by looking in bucket 14 and dividing the answer by 2, since we count each partitioning twice (once in each of two orderings).

Hal Burch

``````/* Calculate how many two-way partitions of {1, 2, ..., N} are
even splits (the sums of the elements of both partition are equal) */

#include <stdio.h>
#include <string.h>

#define MAXSUM 637

unsigned int numsets[637][51];

int max;
unsigned int sum;

main(int argc, char **argv)
{
int lv, lv2, lv3;
int cnt;
FILE *fin, *fout;

fin = fopen ("subset.in", "r");
fscanf(fin, "%d", &max);
fclose (fin);
fout = fopen("subset.out", "w");

if ((max % 4) == 1 || (max % 4) == 2) {
fprintf (stderr, "0\n");
exit(1);
}

sum = max * (max+1) / 4;

memset(numsets, 0, sizeof(numsets[0]));
numsets[0][0] = 1;
for (lv = 1; lv < max; lv++) {
for (lv2 = 0; lv2 <= sum; lv2++)
numsets[lv2][lv] = numsets[lv2][lv-1];
for (lv2 = 0; lv2 <= sum-lv; lv2++)
numsets[lv2+lv][lv] += numsets[lv2][lv-1];
}

fprintf (fout, "%u\n", numsets[sum][max-1]);
fclose (fout);
exit (0);
}
``````

Nick Tomitov

Here’s an even more concise solution from Nick Tomitov of Bulgaria:

``````#include <fstream>
using namespace std;
const unsigned int MAX_SUM = 1024;
int n;
unsigned long long int dyn[MAX_SUM];
ifstream fin ("subset.in");
ofstream fout ("subset.out");

int main() {
fin >> n;
fin.close();
int s = n*(n+1);
if (s % 4) {
fout << 0 << endl;
fout.close ();
return ;
}
s /= 4;
int i, j;
dyn [0] = 1;
for (i = 1; i <= n; i++)
for (j = s; j >= i; j--)
dyn[j] += dyn[j-i];
fout << (dyn[s]/2) << endl;
fout.close();
return 0;
}
``````