Trailing Zeroes (III)

A - Trailing Zeroes (III)

Time Limit:2000MS     Memory Limit:32768KB     64bit IO Format:%lld & %llu

Submit Status

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

#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<stack>#include<math.h>#include<iostream>#include<algorithm>using namespace std;const int MAXN=0x3f3f3f3f;int get(int x){int q = 0;while(x){q += x/5;x /= 5;}return q;}void fun(int n){int l = 0,r = MAXN;while(l <= r){int mid = (l+r)>>1;if(get(mid) >= n){r = mid - 1;}else{l = mid+1;}}if(get(l) == n){printf("%d\n",l);}else{printf("impossible\n");}}int main(){int t,q,text = 0;cin>>t;while(t--){cin>>q;printf("Case %d: ",++text);fun(q);} return 0;}