CodeForces 248B Chilly Willy

最新推荐文章于 2019-08-12 09:55:26 发布

raptor.

最新推荐文章于 2019-08-12 09:55:26 发布

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文章标签： codeforces

本文链接：https://blog.youkuaiyun.com/C18854805113/article/details/78555482

本文探讨了一个有趣的数学问题：如何找到满足特定条件的最小n位数，该数需同时为2、3、5和7的倍数。通过分析与编程实践，文章提供了一种高效的方法来解决这一问题，并分享了实现这一解决方案的C++代码。

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Chilly Willy loves playing with numbers. He only knows prime numbers that are digits yet. These numbers are 2, 3, 5 and 7. But Willy grew rather bored of such numbers, so he came up with a few games that were connected with them.

Chilly Willy wants to find the minimum number of length n, such that it is simultaneously divisible by all numbers Willy already knows (2, 3, 5 and 7). Help him with that.

A number's length is the number of digits in its decimal representation without leading zeros.

Input

A single input line contains a single integer n (1 ≤ n ≤ 105).

Output

Print a single integer — the answer to the problem without leading zeroes, or "-1" (without the quotes), if the number that meet the problem condition does not exist.

Examples

input

output

-1

input

output

10080

题意：n位数，最小的，是2 3 5 7的倍数。。。

思路：先打表找规律

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<cmath>
using namespace std;
#define mod 1000000007
#define ll long long
int main()
{
    int n;
    /*while(cin>>n)
    {
        ll i=pow(10,n-1);
        for(;i<i*10;i++)
        {
            if(i%210==0)
                break;
        }
        cout<<i<<endl;
    }*/
    string map[]={"","050","080","170","020","200","110"};
    while(cin>>n)
    {
        if(n<3)
            cout<<-1<<endl;
        else if(n==3)
            cout<<210<<endl;
        else
        {
            cout<<1;
            for(int i=1;i<n-3;i++)
                cout<<0;
            n=(n+3)%6;
            if(n==0)
                n=6;
            cout<<map[n]<<endl;
        }
    }
    return 0;
}