# lengthy explanation give more attention
#here we are finding sums on all valid partition and storing all four
possible sums in variable a,b,c,d and and for all possible a,b,c,d we will
keep runninf max and min/
lets take an example parttion is done at row=0, coloumn=1
00 01| 02 03
--------|---------
10 11| 12 13
20 21| 21 22
a=arr[0][0]+arr[0][1]
b=arr[0][2]+arr[0][3]
c=arr[1][0]+arr[1][1]+arr[2][0][2][1]
d=arr[1][2]+arr[1][3]+arr[2][1]+arr[2][2];
this can be coded like this
int a,b,c,d,max,main;
a=b=c=d=max=min=0;
for(int row=0;row++;row<n-1)
for(int col=0;col++;col<n-1)
{
for(int i=0;i<=row;i++)
{
for(int j=0;j<=col;j++)
a+=arr[i][j];
for(int j=col;j<=n-1;j++)
b+=arr[i][j];
}
for(int i=row+1;i<=n-1;i++)
{
for(int j=0;j<=col;j++)
c+=arr[i][j];
for(int j=col;j<=n-1;j++)
d+=arr[i][j];
}
max=maximum(max,maximum(maximum(a,b),maximum(c,d))); //maximum(a,b)
is predefined function to find maximum of two nos.
min=minimum(min,minimum(minimum(a,b),minimum(c,d)));
}
cout<<"max:"<<max;
cout<<"min:"<<min;
}
please check it and let me know if any m
On Thursday, August 16, 2012 2:09:20 PM UTC+5:30, g4ur4v wrote:
>
> @sahil Can you please explain your question with an example ?
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