codeforces——584A——Olesya and Rodion

最新推荐文章于 2020-10-28 07:39:15 发布

lose__way

最新推荐文章于 2020-10-28 07:39:15 发布

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分类专栏： codeforces 文章标签： ACM codeforces c语言

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这篇博客介绍了如何解决codeforces上的584A问题，即找到一个由n位组成的数字，同时该数字能被t整除。题目要求输出一个大于0且不包含前导零的数，如果不存在则输出-1。内容包括问题描述、条件限制及可能的答案。

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Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible by t. Find some number that satisfies both of them.

Your task is: given the n and t print an integer strictly larger than zero consisting of n digits that is divisible by t. If such number doesn't exist, print - 1.

Input

The single line contains two numbers, n and t (1 ≤ n ≤ 100, 2 ≤ t ≤ 10) — the length of the number and the number it should be divisible by.

Output

Print one such positive number without leading zeroes, — the answer to the problem, or - 1, if such number doesn't exist. If there are multiple possible answers, you are allowed to print any of them.

Examples

Input

3 2

Output

712

输入n,m，n为位数，m为所输出数可以整除的，没有可以输出的数，输出-1

水题

#include <iostream>
#include<cstring>
#include<string>
#include<stdio.h>
#include<algorithm>
#include<set>
using namespace std;
int main()
{
    int n,m;
    while(cin>>n>>m!=NULL)
    {
        if(n==1&&m==10)
        {
            cout<<"-1"<<endl;
            continue;
        }
        cout<<m;
        m==10?n=n-2:n--;
        while(n--)
            cout<<'0';
        cout<<endl;
    }
    return 0;
}