Trailing Zeroes (III)------二分搜索+阶乘分解

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：输入一个数字，这个数字代表N!后面0的个数。输出N最小值。

参考理解：

https://blog.csdn.net/ee8736199/article/details/48105017

注:计算N阶乘后面有几个0:

有算数基本定理可知 N！可划分为 质因数相乘的形式  N!=2^a*3^b*5^c*7^d........

因为只有2*5 才会出现 0   又因为2的数量肯定比5的多  所以计算阶乘中5的数量就可以得到该阶乘后有几个0。

50/5=10  10/5=2  所以50！后有10+2=12个0

#include 
#include 
#include 
 
using namespace std;
 
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);
		int res=0;
		while(n)
		{
			res+=n/5;
			n/=5;
		}
		printf("%d\n",res);
	}
	return 0;
} 

 

#include
#include
using namespace std;
typedef long long ll;
ll slove(ll n)
{
    ll num=0;
    while(n)
    {
        num+=n/5;
        n/=5;
    }
    return num;

}
ll er(ll n)//二分查找
{
    ll x=1;
    ll y=500000000;
    ll mid;
    ll res=-1;
    while(y>=x)
    {
        mid=(x+y)/2;
        ll ans=slove(mid);
        if(ans==n)
        {
            res=mid;
            y=mid-1;//找到 最小 的数,其阶层末尾的0为N个
        }
        else if(ans>n)
            y=mid-1;
        else if(ans