HDU4951: Multiplication table

HDU4951: Multiplication table
Problem DescriptionTeacher Mai has a multiplication table in base p.

For example, the following is a multiplication table in base 4:

* 0 1 2 3
0 00 00 00 00
1 00 01 02 03
2 00 02 10 12
3 00 03 12 21

But a naughty kid maps numbers 0 .. P-1 into another permutation and shuffle the multiplication table.

For example Teacher Mai only can see:

1*1 = 11 1*3 = 11 1*2 = 11 1*0 = 11
3*1 = 11 3*3 = 13 3*2 = 12 3*0 = 10
2*1 = 11 2*3 = 12 2*2 = 31 2*0 = 32
0*1 = 11 0*3 = 10 0*2 = 32 0*0 = 23

Teacher Mai wants you to recover the multiplication table. Output the permutation number 0... P-1 mapped.

It's guaranteed the solution is unique.
InputThere are multiple test cases, terminated by a line "0 ".

For each test case, the first line contains one integer p (2 <= p <= 500 ).

In following p lines, each line contains 2 * p integers. The (2 * j + 1)-th number x and (2 * j + 2) -th number y in the I-th line indicates equation I * j = xy in the shuffled multiplication table.

Warning: Large IO!
OutputFor each case, output one line.

First output "Case # k:", where k is the case number counting from 1. The following are p integers, indicating the permutation number 0... P-1 mapped.
Sample Input

42 3 1 1 3 2 1 01 1 1 1 1 1 1 13 2 1 1 3 1 1 21 0 1 1 1 2 1 30

Sample Output

Case #1: 1 3 2 0 it can be found that, in the Case of 0, all numbers in this row are the same as other numbers, then the first digit is taken. Suppose there are n different situations, the corresponding number is that in the case of n1, the corresponding 0 must be used for special processing. G ++ times out and C ++ is used. Because I did not pay attention to the details, WA performed several times, it took a long time to correct errors...
#include 
  
   #include 
   
    #include using namespace std;#define up(i,x,y,z) for(i=x;i<=y;i+=z)#define mem(a,b) memset(a,b,sizeof(a))#define w(x) while(x)int n,a[505][1005],i,j,vis[505],ans[505],cas=1;int main(){    w((scanf("%d",&n),n))    {        up(i,0,n-1,1)        up(j,0,2*n-1,1)        scanf("%d",&a[i][j]);        up(i,0,n-1,1)        {            up(j,1,2*n-1,1)            {                if(a[i][j]!=a[i][j-1])                    break;            }            if(j==2*n)            {                ans[0]=i;                break;            }        }        up(i,0,n-1,1)        {            int cnt=0;            mem(vis,0);            up(j,0,2*n-1,2)            {                if(!vis[a[i][j]])                {                    vis[a[i][j]]=1;                    cnt++;                }            }            if(cnt==1&&ans[0]!=i) ans[1]=i;            else if(cnt!=1)ans[cnt]=i;        }        printf("Case #%d:",cas++);        up(i,0,n-1,1)        printf(" %d",ans[i]);        printf("\n");    }    return 0;}