Codeforces Round #324 (Div. 2) A. Olesya and Rodion

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 test(s)

input

3 2

output

712

这道题的意思是给出n和t，n是指输出数字的位数，t是指输出的数字要可以整除t，如果直接暴力，肯定会爆，所以来个选择性暴力，在前面的200以内直接暴力，也就是说，两位数的直接暴力，其余的，直接就是输出 t ，然后在 t 的后面补n-1个0，就是要输出的数字，当然，如果t是10的话，就是补n-2个0，好吧，脑洞要开。

#include<stdio.h>
#include<math.h>
int main()
{
int n,t;
scanf("%d %d",&n,&t);
if(n==1&&t<10) printf("%d\n",t);
else if(n==1&&t==10) printf("-1\n");
else if(n==2)
{
long long ans=-1;
for(long long i=pow(10,n-1);i<pow(10,n);i++)
{
if(i%t==0)
{
ans=i;
break;
}
}
printf("%I64d\n",ans);
}
else
{
printf("%d",t);
if(t==10)
{
for(int i=0;i<n-2;i++) printf("0");
}
else
{
for(int i=0;i<n-1;i++) printf("0");
}
printf("\n");
}
return 0;
}