CodeForces 584A

最新推荐文章于 2019-06-07 17:43:13 发布

原创最新推荐文章于 2019-06-07 17:43:13 发布 · 361 阅读

0 ·

CC 4.0 BY-SA版权

比赛同时被 2 个专栏收录

60 篇文章

订阅专栏

codeforces

42 篇文章

订阅专栏

本文介绍了一道编程题目，任务是找到一个由指定数量的数字组成的数，该数需能被另一个给定的数整除。文章提供了实现这一目标的C++代码示例。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

C -C

Crawling in process... Crawling failed

Time Limit: 1000MS Memory Limit: 262144KB 64bit IO Format: %I64d & %I64u

Submit Status Practice CodeForces 584A

Description

Olesya loves numbers consisting of n digits, and Rodion only likes numbers that are divisible byt. 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 andt (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.

Sample Input

Input

3 2

Output

思路：水题，一想便知；

#include<cstdio>
int main()
{
	int n,t,i;
	scanf("%d %d",&n,&t);
	if(n==1 && t==10)
		printf("-1\n");
	else if(t==10)
	{
		printf("1");
		for(i=0;i<n-1;i++)
			printf("0");
		printf("\n");
	}
	else
	{
		for(i=0;i<n;i++)
		{
			printf("%d",t);
		}
		printf("\n");
	}
	return 0;
}