trailing zeroes

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

注意以下是超時的代碼

#include<iostream>
#include<algorithm>
#include<cstdio>

#include<cstring>
#include<cmath>
using namespace std;
#define ll long long


int main(){
	int t;
	scanf("%d",&t);int ccase=0;
	while(t--){
		int n;
		scanf("%d",&n);int ans=0;
		int res=n*5;int tem=5; bool flag=1;int ress;
		while(1){
			ress=res;
			ans=0;
			while(res){
				ans+=res/tem;
				res/=tem;
				
			}
		//	cout<<ans<<endl;
			if(ans>n){
				res=ress-5;
				
			}
			else if(ans==n){
				break;
			}
			else{
				flag=0;
				break;
			}
		}
		printf("Case %d: ",++ccase);
		if(!flag) cout<<"impossible"<<endl;
		else printf("%d\n",ress);
	}
	return 0;
}

首先偶數的個數顯然大於5的個數，所以看5的個數就可以了。超時的話，也就是1e8還是挺大的（我準備豐富一下自己的博客，寫的太醜會很丟人的，儘量寫的清晰易懂，不過不知道什麼樣的該寫博客，由於我自己的水平比較低，基本上都會寫的

所以二分，二分還是需要多寫寫的，邊界的問題總是出錯

我的思維正常是解方程組的思想，但是計算機是可以一個個試的，所以二分我不是很習慣用。需要注意一下範圍，保證1e8個0的話，二分的範圍要大一些

#include<iostream>
#include<algorithm>
#include<cstdio>

#include<cstring>
#include<cmath>
using namespace std;
#define ll long long
const int maxn=5e8;//要適當的大點保證範圍 
int judge(int n){
	int res=0;
	while(n){
		res+=n/5;
		n/=5;
	}
	return res;
}
int erfen(int n){
	int l=0; int r=maxn;
	while(l<=r){
		int mid=(l+r)>>1;
		if(judge(mid)<n){
			l=mid+1;
		}
		else r=mid-1;
	}
	if(judge(l)==n) return l;
	return 0;
}
int main(){
	int t;
	scanf("%d",&t);int ccase=0;
	while(t--){
		int n;
		scanf("%d",&n);int ans=0;
		//if(erfen(n)==0)
		
	
		printf("Case %d: ",++ccase);
		if(erfen(n)==0) cout<<"impossible"<<endl;
		else printf("%d\n",erfen(n));
	}
	return 0;
}