Trailing Zeroes

最新推荐文章于 2022-03-27 22:09:16 发布

原创最新推荐文章于 2022-03-27 22:09:16 发布 · 218 阅读

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本文探讨如何通过编程解决一个数学问题：寻找最小的自然数N，使得N的阶乘在十进制表示中恰好有Q个尾随零。采用二分查找法结合数学分析，实现快速准确地定位目标数值。

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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

Sample Output

Case 1: 5

Case 2: 10

Case 3: impossible

题意：求满足情况的阶乘末尾0的个数。

解题思路：5的个数即使0的个数；统级一个数有多少个5构成。///利用二分查找到最小的值

#include<iostream>
#include<algorithm>
using namespace std;

int solve(int n)  //统计0的个数
{
    int sum=0;
    while(n)
    {
        sum+=n/5;
        n=n/5;
    }
    return sum;
}
int er(int n)
{
    int l=0,r=500000000;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(solve(mid)<n)
            l=mid+1;
        else
            r=mid-1;
    }
    if(solve(l)==n)
        return l;
    return 0;
}
int main()
{
    int t,k=0;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        if(!er(n))
            cout<<"Case "<<++k<<": impossible"<<endl;
        else
            cout<<"Case "<<++k<<": "<<er(n)<<endl;
    }
    return 0;
}