One

Description

154. Factorial

time limit per test: 0.25 sec. 
memory limit per test: 4096 KB

input: standard input 
output: standard output

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

One number Q written in the input (0<=Q<=10^8).

Output

Write "No solution", if there is no such number N, and N otherwise.

Sample test(s)

Input

2

Output

10

题意为输入一个数表示某阶乘的数字末尾0的个数，让你找出最小的那个n的阶乘。由末尾有0想到必然有因子2和5！然而如果有了因子5那么因子2一定存在，所以我们要做的就是寻找因子5！仔细找找规律会发现只有乘到5的倍数的阶乘时才会出现一段阶乘里面又新增了因子5。（这个靠自己仔细琢磨，想清楚了自然有思路，并没有什么难点）。

#include <iostream>
#include<memory.h>
using namespace std;

int main()
{
    int n;
   cin>>n;
    if(n==0)cout<<1<<endl;
    else
    {
        int t=0;
    for(int i=5;t<n;i+=5)
    {
        int x=i,num=0;
        while(x%5==0)
            x/=5,
            num++;
        t+=num;
        if(t==n)
        {
            cout<<i<<endl;break;
        }
    }
    if(t!=n)cout<<"No solution"<<endl;
    }
    return 0;
}