LightOJ 1138 Trailing Zeroes (III)（二分 + 思维）

http://lightoj.com/volume_showproblem.php?problem=1138

Trailing Zeroes (III)

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

Submit Status Practice LightOJ 1138

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

求最小的N使N！中0的个数等于q

0是有5乘4、2、8等等构成的，N中只要有因子5，那么N!中一定能构成0，所以我们只需要找N中因子5的个数，然后用二分来快速查找N

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
 
using namespace std;
typedef long long ll;
 
ll solve(ll n)
{
    ll ans = ;
    while(n)
    {
        ans += n / ;
        n /= ;
    }
    return ans;
}//求n中因子5的个数
 
int main()
{
    int t, p = ;
    ll q;
    scanf("%d", &t);
    while(t--)
    {
        p++;
        scanf("%lld", &q);
        ll low = , high = ;
        while(low <= high)
        {
            ll mid = (low + high) / ;
            ll m = solve(mid);
            if(m < q)
                low = mid + ;
            else
                high = mid - ;
        }//二分
        if(solve(low) == q)
            printf("Case %d: %lld\n", p, low);
        else
            printf("Case %d: impossible\n", p);
    }
    return ;
}