Trailing Zeroes (III) (数学)

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】输入一个数，表示阶乘值末尾0的个数，求满足条件的最小数。

  这道题的关键在于怎么知道一个数阶乘值末尾0的个数，因为数据庞大，不可能正常计算出来，所以就要采取一定的数学方法，经过尝试可以知道，1~4的阶乘值末尾没有0,5~9的阶乘值末尾一个0,10~14的阶乘值末尾2个0,但是！！！末尾三个0的数没有，所以用而二分就可以找出这样的数，具体看代码：

#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;LL ans(LL n){    LL ss=0;    while(n)    {        ss+=n/5;        n/=5;    }    return ss;}int main(){    int t;    LL m,a;    scanf("%d",&t);    for(int i=1;i<=t;i++)    {        scanf("%lld",&m);        LL l=1,r=1e12;        a=0;        while(r>=l)        {            LL mid=(l+r)>>1;            if(ans(mid)==m)            {                a=mid;                r=mid-1;            }            else if(ans(mid)>m)                r=mid-1;            else                l=mid+1;        }        if(a)        printf("Case %d: %lld\n",i,a);        else            printf("Case %d: impossible\n",i);    }    return 0;}