LightOJ 1138 Trailing Zeroes (III)

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=70017#overview

Trailing Zeroes (III)

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

题目描述：

　　假设有一个数n，它的阶乘末尾有Q个零，现在给出Q，问n最小为多少？

解题思路:

　　由于数字末尾的零等于min（因子2的个数，因子5的个数），又因为2<5,那么假设有一无限大的数n，n=2^x=5^y，可知x<<y。

所以我们可以直接根据因子5的个数，算阶乘末尾的零的个数。1<=Q<=10^8,所以可以用二分快速求解。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>

using namespace std;

#define N 300
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))

typedef long long LL;

LL find_zero (LL n);

int main ()
{
int t;
LL n, nCase = 1, l, r, mid;
scanf ("%d", &t);

while (t--)
{
scanf ("%lld", &n);

l = 0, r = 1000000000000;

while (l <= r)
{
mid = (l + r) / 2;
LL temp = find_zero (mid);

if (temp < n)
l = mid + 1;
else r = mid - 1;
}

if (find_zero (l) == n) printf ("Case %lld: %lld\n", nCase++, l);
else printf ("Case %lld: impossible\n", nCase++);
}
return 0;
}

LL find_zero (LL n)
{
LL ans = 0;

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