light oj 1138

Trailing Zeroes (III)

Time Limit: 2000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

Submit Status

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

Source

Problem Setter: Jane Alam Jan

数论！！！加二分....

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <math.h>
#include <iostream>
#include <algorithm>
#include <climits>
#include <queue>
#define ll long long


using namespace std;
ll ans;
ll zero(ll num)
{
    ll i,sum = 0;
    for(i = 5; i <= num; i *= 5)
        sum += num/i;
    return sum;
}
void check(int n)
{
    ll low = 5,mid,high;
    high = 0x3f3f3f3f3f;
    while(low <= high)
    {
        mid = (low+high)/2;
        ll temp = zero(mid);
         if(n < temp)
            high = mid-1;
        else if(n > temp)
            low = mid+1;
        else if(n == temp)
        {
            if(mid == low)
            {
                ans = mid;
                return;
            }
            high = mid;
        }
    }
}

int main(void)
{
    int t,n,cnt = 1;
    cin>>t;
    while(t--)
    {
        scanf("%d",&n);
        ans = -1;
        check(n);
        if(ans == -1)
            printf("Case %d: impossible
",cnt++);
        else
            printf("Case %d: %lld
",cnt++,ans);
    }
    return 0;
}