Trailing Zeroes (III) LightOJ - 1138（二分+阶乘性质）

Trailing Zeroes (III) LightOJ - 1138

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

题意：给你一个数字，这个数字代表N！后面有几个0。给出这个数字，计算N的值。

解题思路：

任何质因数都可以写成素数相乘的形式。所以计算一个数的阶乘后面几个0，只需计算这个数包含多少5即可
然后用二分查找即可
code：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll solve(ll n){
    ll num = 0;
    while(n){
        num += n / 5;
        n /= 5;
    }
    return num;
}
ll search(ll n){
    ll l = 1,r = 1844674407370;
    ll mid;
    ll res = -1;
    while(l <= r){
        mid = (l + r) >> 1;
        ll ans = solve(mid);
        if(ans == n){
            res = mid;
            r = mid - 1;
        }
        else if(ans > n){
            r = mid - 1;
        }
        else if(ans < n){
            l = mid + 1;
        }
    }
    return res;
}
int main(){
    int t,cas = 0;
    scanf("%d",&t);
    while(t--){
        ll n;
        scanf("%lld",&n);
        ll ans = search(n);
        if(ans == -1)
            printf("Case %d: impossible\n",++cas);
        else
            printf("Case %d: %lld\n",++cas,ans);
    }
    return 0;
}