1014 - Ifter Party   PDF (English) Statistics Forum Time Limit: 2 second(s)Memory Limit: 32 MB

I have an Ifter party at the 5^th day of Ramadan for the contestants. For this reason I have invited C contestants and arranged P piaju's (some kind of food, specially made for Ifter). Each contestant ate Q piaju's and L piaju's were left (L < Q).

Now you have to find the number of piaju's each contestant ate.

Input

Input starts with an integer T (≤ 325), denoting the number of test cases.

Each case contains two non-negative integers P and L (0 ≤ L < P < 2³¹).

Output

For each case, print the case number and the number of possible integers in ascending order. If no such integer is found print 'impossible'.

Sample Input

Output for Sample Input

4

10 0

13 2

300 98

1000 997

Case 1: 1 2 5 10

Case 2: 11

Case 3: 101 202

Case 4: impossible

 题意：有p克蛋糕，每人吃q克，剩余l克，问q的值有多少种情况，从小到大输出   也就是求p-l的所有大于l的因数

         int数组会超时。。

code：

#include<cstdio> #include<algorithm> using namespace std ; #include<cmath> #include<vector> vector < int >v ; int main ( ) {     int t ;     long long p,l ;     scanf ( "%d", &t ) ;     int k = 1 ;     while (t -- )     {         v. clear ( ) ;         scanf ( "%lld%lld", &p, &l ) ;         long long q ;         q =p -l ;         int m = ( int ) sqrt (q ) ;         for ( long long i = 1 ;i <=m ;i ++ )         {             if (q %i == 0 ) {                 if (i >l ) v. push_back (i ) ;                 if (q /i ! =i &&q /i >l ) v. push_back (q /i ) ;             }                     }         if (v. empty ( ) ) printf ( "Case %d: impossible\n",k ++ ) ;         else {             printf ( "Case %d:",k ++ ) ;         sort (v. begin ( ),v. end ( ) ) ; int s =v. size ( ) ;         for ( int i = 0 ;i <s ;i ++ )         printf ( " %d",v [i ] ) ; printf ( "\n" ) ;         }             }     return 0 ;   }