[LightOJ](1138)Trailing Zeroes (III) ---- 二分

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

AC代码：

#include<bits/stdc++.h>
using namespace std;
#define rep(i,s,e)      for(int i=s;i<=e;i++)
#define rev(i,s,e)      for(int i=e;i>=s;i--)
#define all(x)          x.begin(),x.end()
#define sz(x)           x.size()
#define szz(x)          int(x.size()-1)
const int INF = 0x3f3f3f3f;
const int MOD =  1e9+7;
const int MAXN = 2e5+10;
typedef long long LL;
typedef pair<LL,LL> PII;
LL judge(LL n)
{
    LL res = 0;
    while(n)
    {
        res+=n/5;
        n/=5;
    }
    return res;
}
LL bs(LL key)
{
    LL l = 0;
    LL r = 1LL<<60;
    LL res = -1;
    while(r-l>1)
    {
        LL mid = (l+r)/2;
        LL ans = judge(mid);
        if(ans>key) r = mid;
        else if(ans == key)
        {
            r = mid;
            res = mid;
        }
        else l = mid;
    }
    return res;
}
int main()
{
    #ifdef LOCAL
    freopen("in.txt","r",stdin);
    #endif // LOCAL
    ios_base::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
    int t;
    cin>>t;
    int q;
    int id = 1;
    while(t--)
    {
        cin>>q;
        LL ans = bs(q);
        if(ans == -1) cout<<"Case "<<id++<<": "<<"impossible"<<endl;
        else cout<<"Case "<<id++<<": "<<ans<<endl;
    }
    return 0;
}