time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
Olesya loves numbers consisting ofndigits, and Rodion only likes numbers that are divisible byt. Find some number that satisfies both of them.
Your task is: given thenandtprint an integer strictly larger than zero consisting ofndigits that is divisible byt. If such number doesn’t exist, print-1.
Input
The single line contains two numbers,nandt(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
解题说明：此题要求一个能被t整除的n位数，可以采用构造的方法，最容易想到的数字是以t为开始，后面全部为0. 当然，需要排除位数为1但t=10的情况。
#include<stdio.h>#include <string.h>#include<iostream>#include<algorithm>using namespace std;int main(){int n, t, i;scanf("%d %d", &n, &t);if(n==1 && t==10) {printf("-1\n");}else{if(t==10) {t=1;}printf("%d", t);for(i=0; i<n-1; i++){printf("0");}printf("\n");}return 0;}
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