Light OJ 1138

Trailing Zeroes (III)

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld &%llu

Submit Status

Description

You task was to find minimal natural number N, so that n! contains exactly Q zeroes on the trail in decimal notation. As you know n! = 1*2*...*n. For example, 5! = contains one zero on the trail.

Input

Input starts with an integer T (≤10000), denoting the number of test cases.

Each case contains an integer Q (1≤q≤108) in a line.

Output

For each case, print the case number and N. If no solution is found then print ' impossible '.

Sample Input

3

1

2

5

Sample Output

Case 1:5

Case 2:10

Case 3:impossible

Source

Problem Setter:jane Alam Jan

Arithmetical!!! Add two points ....

1#include <stdio.h>2#include <string.h>3#include <stdlib.h>4#include <math.h>5#include <iostream>6#include <algorithm>7#include <climits>8#include <queue>9 #definell Long LongTen  One  A using namespacestd; - ll ans; - ll Zero (ll num) the { -ll i,sum =0; -      for(i =5; I <= num; I *=5) -Sum + = num/i; +     returnsum; - } + voidCheckintN) A { atll low =5, Mid,high; -High =0x3f3f3f3f3f; -      while(Low <=High ) -     { -Mid = (Low+high)/2; -ll temp =Zero (mid); in          if(N <temp) -High = mid-1; to         Else if(N >temp) +Low = mid+1; -         Else if(n = =temp) the         { *             if(Mid = =Low ) $             {Panax NotoginsengAns =mid; -                 return; the             } +High =mid; A         } the     } + } -  $ intMainvoid) $ { -     intt,n,cnt =1; -Cin>>T; the      while(t--) -     {Wuyiscanf"%d",&n); theAns =-1; - check (n); Wu         if(ans = =-1) -printf"Case %d:impossible\n", cnt++); About         Else $printf"Case %d:%lld\n", cnt++, ans); -     } -     return 0; -}

Light OJ 1138