C - Trailing zeroes(二分+阶乘） LightOJ

该博客探讨了如何找到最小的自然数N，使得N的阶乘在十进制表示中尾部恰好有Q个零。通过介绍二分搜索和数论的方法来解决这个问题。

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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

Sample Output

Case 1: 5

Case 2: 10

Case 3: impossible

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
ll find_num(ll n)
{
     ll cnt =0;
     while (n)
     {
         cnt+=n/5;
            n/=5;
     }
        return cnt ;
}
ll q;
ll  binary_search_()
{
   // cout<<q<<endl;
    ll l=1,r=1000000000000000000;
     ll ans =0;
     while (l<=r)
     {
        ll mid =(l+r)>>1;
       // cout<<mid<<endl;
        ans=find_num(mid);
     //   cout<<"ans :: "<<ans<<endl;
         if (ans>=q)
            r=mid-1;
        else
            l=mid+1;
       // cout<<l<<"\t"<<r<<endl;

     }
                ans=find_num(l);
            if(ans==q)
            return l;
            else
            return -1;
}
int main ()
{   int t,cnt =1;
    cin>>t;
    ll ans;
    while (t--)
    {
        scanf("%lld",&q);
        ans =binary_search_();
        if(ans!=-1)
        printf("Case %d: %lld\n",cnt++,ans);
        else
        printf("Case %d: impossible\n",cnt++);
    }
    return 0;
}