C :

#include "stdio.h"
int n,s;
int a[50],f[50001];
int main()
{
	int i,j;
	scanf("%d",&n);
	s=0;
	for(i=0;i<n;i++)
	{
		scanf("%d",&a[i]);
		s+=a[i];
	}
	if(s%2)
		printf("0\n");
	else
	{
		f[0]=1;
		s/=2;
		for(i=0;i<n;i++)
		{
			for(j=s;j>=a[i];j--)
				if(f[j-a[i]])
					f[j]+=f[j-a[i]];
		}
		printf("%d\n",f[s]/2);
	}
	return 0;
}

C++ :

#include<iostream>  
#include<cstdio>  
#include<cstring>  
using namespace std;  
int dp[51][3010], p[55];  
int main()  
{  
//  freopen("a.txt","r",stdin);  
    int n, i, j, k;  
    while(cin >> n)  
    {  
        memset(dp,0,sizeof(dp));  
        int sum(0);  
        for(i = 1; i <= n; ++ i)  
        {  
            cin >> p[i];  
            sum += p[i];  
        }  
        if(sum % 2) { cout << '0' << endl; continue; }  
        dp[0][0] = 1;  
        for(i = 1; i <= n; ++ i)  
        {  
            for(j = 0; j <= sum/2; ++ j)  
            {  
                dp[i][j] = dp[i-1][j];  
                if(j - p[i] >= 0)  
                     dp[i][j] = dp[i][j] + dp[i-1][j-p[i]];  
            }  
        }  
        cout << dp[n][sum/2]/2 << endl;  
    }  
    return 0;  
}  

Pascal :

{$N+}
const
  ipf = 'compete.in';
  opf = 'compete.out';
var
  a       : array[0..2500]of extended;
  c       : array[1..50]of byte;
  i,j,k,n : word;
  s       : longint;

begin
  {assign(input,ipf); reset(input);
  assign(output,opf); rewrite(output);}
  repeat
    readln(n);
    if n = 0 then break;
    s := 0;
    for i := 1 to n do begin
      read(c[i]); s := s+c[i];
    end;
    readln;
    if odd(s) then writeln(0)
    else begin
      fillchar(a,sizeof(a),0);
      a[0] := 1;
      for i := 1 to n do
        for j := s div 2-c[i] downto 0 do a[j+c[i]] := a[j+c[i]]+a[j];
      writeln(a[s div 2]/2:0:0);
    end;
  until false;
  {close(input); close(output);}
end.