C++ :

#include <cstdio>

#define  maxn  101

int n , S , ans = 0 ; 
int sum[maxn][maxn] = {0} ;

int main()
{
	scanf("%d", &n );
	int t ;
	for (int i = 1 ; i <= n ; i ++ )
	{
		for (int j = 1 ; j <= n ; j ++ )
		{
			scanf("%d", &t ) ;
			for (int p = i ; p <= n ; p ++ )
			{
				for (int q = j ; q <= n ; q ++ )
				{
					sum[p][q] += t ;
					//printf("s[%d][%d] = %d\n",p,q,sum[p][q]);
				}
			}
		}
	}
	scanf("%d", &S ) ;
	if (S > n*n) S=n*n;
	for (int s = 1 ; s <= S ; s ++ )
	{
		for (int a = 1 ; a <= s ; a ++ )
		{
			int b = s/a ;
			if (a*b != s) continue ;
			for (int i = 1 ; i <= n-a+1 ; i ++ )
			{
				for (int j = 1 ; j <= n-b+1 ; j ++ )
				{
					int i2 = i+a-1 , j2 = j+b-1 ;
					int t = sum[i2][j2]-sum[i-1][j2]-sum[i2][j-1]+sum[i-1][j-1];
					if (t > ans) ans = t ;
				}
			}
		}
	}
	printf("%d\n",ans);
	return 0 ;
}

Pascal :

var
  n,i,j,k,k1,max,s,s1,i1:longint;
  g,a:array[0..100,0..100] of longint;
begin
  
  readln(n);    
  for i:=1 to n do    
    begin
      for j:=1 to n do         
	    begin         
		  read(g[i,j]);         
		  a[i,j]:=a[i-1,j-1];         
		  for i1:=1 to i-1 do a[i,j]:=g[i1,j]+a[i,j];         
		  for i1:=1 to j-1 do a[i,j]:=g[i,i1]+a[i,j];         
		  a[i,j]:=a[i,j]+g[i,j];
         //统计从点(1,1)到点(i,j)的矩阵之和         
		end;      
	readln;    
   end;    
   readln(s1);    
   max:=0;    
   for i:=1 to n do    
   for j:=1 to n do      
     begin        
	   for k:=0 to n-i do    //枚举长和宽        
	   for k1:=0 to n-j do        
	   if (k+1)*(k1+1)<=s1 then    //如果这样的矩形符合要求        
	     begin          
		   s:=a[i+k,j+k1]-a[i+k,j-1]-a[i-1,j+k1]+a[i-1,j-1];    //请参见注释          
		   if s>max then max:=s;        
		 end;     
	 end;    
  writeln(max);    
  
end.