1020 : The Funny Number

时间限制Time Limit 1 秒Sec 内存限制Memory Limit 128 兆MB 提交次数Submitted 1 次Times 通过次数Solved 1 次Times

标准评测 Standard

从标准输入读入，结果写到标准输出。评测将输出拆成 token，与标准答案逐项比较，不按整段逐字节比对。 Read from standard input and write to standard output. The judge splits the output into tokens and compares them with the official answer; it does not compare raw bytes.

比对Compare	token 相同即通过。中间的空格、制表符、换行可多可少。Matching tokens pass. Extra spaces, tabs, or newlines between them are ignored.
不同则错Differs	token 个数或内容不同即错误。23 与 2 3、02 与 2、2.0 与 2 均视为不同。A different token count or value is wrong. 23 vs 2 3, 02 vs 2, and 2.0 vs 2 all differ.
输入Input	标准输入，格式见题面。Standard input; format as in the statement.
输出Output	标准输出。Standard output.

模板Template

C++ C Java Python

#include <iostream>
using namespace std;
int main() {
    int a, b;
    cin >> a >> b;
    cout << (a + b) << endl;
    return 0;
}

#include <stdio.h>
int main() {
    int a, b;
    scanf("%d%d", &a, &b);
    printf("%d\n", a + b);
    return 0;
}

import java.util.Scanner;
public class Main {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int a = in.nextInt();
        int b = in.nextInt();
        System.out.println(a + b);
    }
}

a, b = map(int, input().split())
print(a + b)

题目描述Description

Add a specific digit p before and after a decimal positive integer k to get a new number. If the new number obtained is divisible by w, we call k a funny number.

Given p and w, could you get the smallest k?

For example: we set p=1 and w=9. The samllest k is 7, since add 1 before and after 7 we could get 171, which is divisible by 9. But numbers 1 ~ 6 (i.e. 111, 121, ... 161) could not.

输入格式Input

Each line contains two integer numbers p and w.

1 <= p <= 9 and 1 <= w <= 1000.

输出格式Output

For each case output oneline, The smallest funny number k in range [1, 10000].

If there’s no qualified k in range [1, 10000], output -1 instead.

样例Sample

出题Author

CSGrandeur

来源Source

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