Codeforces--584A--Olesya and Rodion（水题）

Olesya and Rodion
Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u
Submit Status

Description

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.

Sample Input

Input

3 2

Output

712

题意:输出一个长度为n的数，这个数可以被t整除，，很水的题，如果n是2--9的一个数，直接把这个数输出n次，如果n为1并且t=10，那么直接输出-1，如果t=0，n>1的话输出一个1，后边加上n-1个0

#include<cstdio>
#include<algorithm>
using namespace std;
int main()
{
int n,t;
while(scanf("%d%d",&n,&t)!=EOF)
{
if(t>=2&&t<10)
{
for(int i=0;i<n;i++)
printf("%d",t);
printf("\n");
}
else
{
if(n==1&&t==10)
printf("-1\n");
else
{
printf("1");
for(int i=0;i<n-1;i++)
printf("0");
printf("\n");
}
}
}
return 0;
}