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：
#include<stdio.h>
int num(int n) //求n的阶乘末尾连续0的个数
{ //百度说是定理 记住吧... 5 int ans=0;
6 while(n)
7 {ans+=n/5;
8 n=n/5;
9
10 }
return ans; }
int main()
{
int mid,i=;
int t ,q;
long long l,r;
scanf("%d",&t);
while(t--)
{scanf("%d",&q);
l=;
r=; //r要大于1e8
long long m=;
while(l<=r)
{mid=(l+r)/;
if(num(mid)==q)
{r=mid-;
m=mid;
}
else
{if(num(mid)>q)
r=mid-;
else l=mid+;
} } if(m==) printf("Case %d: impossible\n",i);
else printf("Case %d: %lld\n",i,m);
i++;
}
return ;
}