USACO: Subset Sums



/*
ID: tausiq11
PROG: subset
LANG: C++
 */

// @BEGIN_OF_SOURCE_CODE

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cctype>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <sstream>
#include <cmath>
#include <bitset>
#include <utility>
#include <set>
#include <numeric>
#include <ctime>

#define Inf 2147483647
#define Pi acos(-1.0)
#define N 1000000
#define LL long long

inline LL Power(int b, int p) { LL ret = 1; for ( int i = 1; i <= p; i++ ) ret *= b; return ret; }
const int dr [] = {-1, -1, 0, 1, 1, 1, 0, -1};
const int dc [] = {0, 1, 1, 1, 0, -1, -1, -1};

#define F(i, a, b) for( int i = (a); i < (b); i++ )
#define Fs(i, sz) for( size_t i = 0; i < sz.size (); i++ )
#define Fe(i, x) for(typeof (x.begin()) i = x.begin(); i != x.end (); i++)
#define Set(a, s) memset(a, s, sizeof (a))
#define max(a, b)  (a < b ? b : a)
#define min(a, b)  (a > b ? b : a)

using namespace std;

long long dp [1600];

int main () {

	freopen ("subset.in", "r", stdin);
	freopen ("subset.out", "w", stdout);

	int n; scanf ("%d", &n);

	if ( ((n * (n + 1)) / 2) % 2 ) {
        printf ("0\n"); return 0;
	}

	Set (dp, 0);

	dp [0] = 1;		// we can make zero in 1 ways

	for ( int i = 1; i <= n; i++ ) {
		for ( int j = 1599; j >= 0; j-- ) {
			if ( dp [j] && j + i < 1600 )
				dp [j + i] += dp [j];
		}
	}

	//printf ("%d\n", ((n * (n + 1)) / 2));


	cout << dp [(n * (n + 1) / 2) / 2] / 2 << endl;

	return 0;
}

// @END_OF_SOURCE_CODE

Response


USER: Shadow King [tausiq11]
TASK: subset
LANG: C++

Compiling...
Compile: OK

Executing...
   Test 1: TEST OK [0.000 secs, 3356 KB]
   Test 2: TEST OK [0.000 secs, 3356 KB]
   Test 3: TEST OK [0.000 secs, 3356 KB]
   Test 4: TEST OK [0.000 secs, 3356 KB]
   Test 5: TEST OK [0.000 secs, 3356 KB]
   Test 6: TEST OK [0.000 secs, 3356 KB]
   Test 7: TEST OK [0.000 secs, 3356 KB]

All tests OK.
Your program ('subset') produced all correct answers!  This is your
submission #3 for this problem.  Congratulations!

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