CodeForces 260A Adding Digits 에뮬레이션

a와 b를 주고 a의 오른쪽에 숫자 n번을 추가하고 그 다음은 b의 배수를 추가해야 한다.
제목에 따라 시뮬레이션을 하면 되고 결과가 나오면 프로그램을 끝냅니다.
throw+try catch에서 dfs를 종료하는 것이 매우 편리함을 발견했습니다.

#include <cstdio>
int a, b, n;
int num[100001];
void dfs(int x, int d) {
	if (x > n) {
		printf("%d", a);
		for(int i=1;i<=n;i++) printf("%d", num[i]);
		throw 1;
	} else
		for(int i=0;i<10;i++)
			if((d*10+i)%b==0) num[x]=i,dfs(x+1,0);
}
int main() {
	scanf("%d%d%d", &a, &b, &n);
	try {
		dfs(1,a);
		printf("-1");
	} catch(int x) {
	}
	return 0;
}

A. Adding Digits
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
Vasya has got two number: a and b. However, Vasya finds number a too short. So he decided to repeat the operation of lengthening number a n times.
One operation of lengthening a number means adding exactly one digit to the number (in the decimal notation) to the right provided that the resulting number is divisible by Vasya's number b. If it is impossible to obtain the number which is divisible by b, then the lengthening operation cannot be performed.
Your task is to help Vasya and print the number he can get after applying the lengthening operation to number a n times.
Input
The first line contains three integers: a, b, n (1 ≤ a, b, n ≤ 105).
Output
In a single line print the integer without leading zeros, which Vasya can get when he applies the lengthening operations to number a n times. If no such number exists, then print number -1. If there are multiple possible answers, print any of them.
Sample test(s)
input

5 4 5

output

524848

input

12 11 1

output

121

input

260 150 10

output

-1