Trailing Zeroes (III) （二分）题解

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

思路：

有几个零就是看能分解出几个5，所以用二分查找是否存在能分解出n个5的数

代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<queue>
#include<cmath>
#include<string>
#include<map>
#include<stack> 
#include<set>
#include<vector>
#include<iostream>
#include<algorithm>
#define INF 0x3f3f3f3f
#define ll long long
const int N=1e5+5;
const int MOD=1000; 
const double C=0.57721566490153286060651209;
using namespace std;

ll judge(int x){	//判断数x能分解出5的个数 
	ll res=0;
	while(x){
		res+=x/5;
		x/=5;
	}
	return res;	
}

ll panduan(ll n){
	ll l=0,r=5e8+1;
	ll mid,res;
	while(r>=l){
		mid=(l+r)/2;
		if(judge(mid)>=n){
			res=mid;
			r=mid-1;
		}
		else{
			l=mid+1;
		}
	}
	return res;
}
int main(){
	int T,num=1;
	ll ans,n;
	ll mid,l,r;
	scanf("%d",&T);
	while(T--){
		scanf("%lld",&n);
		int ans=panduan(n);
		if(judge(ans)==n){
			printf("Case %d: %lld\n",num++,ans);
		}
		else{
			printf("Case %d: impossible\n",num++);
		}
	}
	return 0;
}