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

 

分析：1到N的阶乘结尾所含的0的数目依次增大(不减)，可以用二分法。

#include<cstdio>

int f(int n)
{
    int ans=0;
    while(n)
    {
        ans+=n/5;
        n/=5;
    }
    return ans;
}

int main()
{
    int T,cas=0,Q;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&Q);
        int l=5,r=400000015;
        while(l<r)
        {
            int mid=(l+r)>>1;
            if(f(mid)>=Q) r=mid;
            else l=mid+1;
        }
        if(f(l)==Q) printf("Case %d: %d\n",++cas,l);
        else printf("Case %d: impossible\n",++cas);
    }
    return 0;
}