Description
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：寻找最小的自然数N，使其阶乘的末尾有Q个0.
思路：只有2和5相乘时末尾才会出现0，在阶乘过程中，因子2的个数足够多，所以每遇到一个5，末尾就会出现一个0，因此问题转化为求1到N这N个整数中包含了多少个因子5。
#include<stdio.h>
#include<string.h>
#define MAX 0x3f3f3f
#define LL long long
LL fun(LL x)
{
LL ans=0;
while(x)
{
ans+=x/5;
x=x/5;
}
return ans;
}
int main()
{
int n,k=1;
LL m;
scanf("%d",&n);
while(n--)
{
scanf("%lld",&m);
LL left=0;
LL right=400000010;
LL mid;
while(left<=right)
{
mid=(left+right)>>1;
if(fun(mid)>=m)
right=mid-1;
else
left=mid+1;
}
printf("Case %d: ",k++);
if(fun(left)!=m)
printf("impossible
");
else
printf("%lld
",left);
}
return 0;
}