Trailing Zeroes (III)

You task is 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! = 120, 120 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

题意：给出数字，代表某个数的阶乘末尾连续0的个数，求出这个数是多少

代码：

 1 #include<stdio.h>
 2 int num(int n) //求n的阶乘末尾连续0的个数
 3 {              //百度说是定理 记住吧...
 4
 5     int ans=0;
 6     while(n)
 7     {ans+=n/5;
 8      n=n/5;
 9
10     }
11     return ans;
12
13 }
14 int main()
15 {
16     int mid,i=1;
17     int t ,q;
18     long long l,r;
19     scanf("%d",&t);
20     while(t--)
21     {scanf("%d",&q);
22      l=0;
23      r=100000000000000;  //r要大于1e8
24      long long m=0;
25      while(l<=r)
26        {mid=(l+r)/2;
27         if(num(mid)==q)
28           {r=mid-1;
29            m=mid;
30           }
31         else
32           {if(num(mid)>q)
33                r=mid-1;
34            else l=mid+1;
35           }
36
37        }
38
39       if(m==0) printf("Case %d: impossible\n",i);
40       else printf("Case %d: %lld\n",i,m);
41       i++;
42     }
43     return 0;
44 }