LightOJ - 1138 Trailing Zeroes (III)

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

题目：给出N的阶乘最后0的个数求N。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<cmath>
#define ll long long
#define mem(a,b) memset(a,b,sizeof(a))
#define inf 0x3f3f3f3f3f3f
#define maxn 200010
#define root  1,n,1
#define lson  l,mid,rt<<1
#define rson  mid+1,r,rt<<1|1
#define mod  10007
using namespace std;
ll sum(ll x){
     ll k=5,ans=0;
    while(x>=k){
        ans+=x/k;
        k*=5;
    }
    return ans;
}
int main(){
    int t;scanf("%d",&t);
   int test=0;
    while(t--){
        ll n;scanf("%lld",&n);
        ll l=0,r=inf;
        while(l<r){
            ll mid=(l+r)/2;
            if(sum(mid)<n) l=mid+1;
            else r=mid;
        }
       if(sum(l)!=n)printf("Case %d: impossible\n",++test);
       else
        printf("Case %d: %lld\n",++test,l);
    }
}