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

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本文探讨了如何找到一个特定位数且能被指定数整除的正整数，若不存在满足条件的整数，则输出-1。通过实例演示了解决此类问题的方法，并详细阐述了实现过程。

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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.

          Sample test(s)
        
            input
          
3 2

            output
          
712

题意：求是n位且是t的倍数的数，如果没有输出-1

简单题，还是错了一次，忘了10是两位

详见代码

#include
     
      
#include
      
       
#include
       
        
using namespace std;

int main(void)
{
    int n, t;
    while(~scanf("%d%d", &n, &t))
    {
        int i;
        if(n == 1 && t == 10)
        {
            printf("-1\n");
            continue;
        }
        printf("%d", t%10==0?1:t);
        for(i = 0; i < n-1; i++)
            printf("0");
        printf("\n");
    }
    return 0;
}