C++ :

#include<stdio.h>
#include<cstring>
using namespace std;
const int maxn=256*256;
const int maxx=256*256+256;
bool f[maxx+1],flag; 
int nn,i,j,n,c,a[11],now,k;
int main()
{

  scanf("%ld",&nn);n=0;
  for (i=1;i<=nn;i++) 
  {
    scanf("%ld",&c);
    if (c==1) 
    {
      printf("0\n");
      return 0;
    }
    flag=true;
    for (j=1;j<=n;j++)
      if (c%a[j]==0) {flag=false;break;}
    if (flag) a[++n]=c;
  }
  memset(f,0,sizeof(f));
  f[0]=true;
  for (i=1;i<=n;i++)
    for (j=a[i];j<=maxx;j++)
      if (f[j-a[i]]) f[j]=true;
  for (i=maxx;i>maxn;i--) if (!f[i]) {printf("0\n");return 0;}
  for (i=maxn;i>0;i--)
    if (!f[i])
    {
      printf("%ld\n",i);
      return 0;
    }
  printf("0\n");
  return 0;
  }

Pascal :

program nuggets;
var
 f:array[0..666666]of boolean; 
 a:array[0..10]of longint;
 i,j,m,max,n:longint;
begin
 readln(n);
 for i:=1 to n do
  begin
   readln(a[i]);
   if a[i]>max then max:=a[i];
  end;
 f[0]:=true;
 for i:=0 to sqr(max)+max do  //数学可证明到最大到这个范围就都可以包装；
  if f[i] then   // 这个位置可以得到
   for j:=1 to n do //把之后能达到的都赋值成true；      
    f[i+a[j]]:=true
   else
    if i>m then m:=i;
 if m>sqr(max) then writeln(0)  
               else writeln(m);
end.