N - Trailing Zeroes (III)

N - 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

思路：

一开始自己想找阶乘末尾0的个数的规律，打表。但是发现数据有1e8大。太大。
参考了别人的，其实是一个二分查找求下界的题。
l=0,r=100000000*5+10;
每次取中间值求其末尾0的个数。
当然最后得判断这个下界的末尾0的个数是不是等于输入的值。

代码：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;

/*二分查找求下界+求阶乘末尾0*/

int solve(int n)
{
    int num=0;
    while(n)
    {
        num+=n/5;
        n/=5;
    }
    return num;
}


int main()
{
    int T;
    scanf("%d",&T);
    int tmp;
    int ks=1;
    while(T--)
    {
       scanf("%d",&tmp);
       int l=0,r=100000000*5+10;

       int mid;
       int ans;

       while(r>l)
       {
          mid =(r+l)>>1;
          if(solve(mid)>=tmp)
          {
             ans=mid;
             r=mid;
          }
          else
          {
              l=mid+1;
          }
       }

       printf("Case %d: ",ks++);

       if(solve(ans)==tmp)
       {
           cout<<ans<<endl;
       }
       else
       {
           printf("impossible\n");
       }

    }
    return 0;
}