Problem:
Score : 100 points

Problem Statement

You are given positive integers A and B.

If A is a divisor of B, print A + B; otherwise, print B - A.

Constraints

All values in input are integers.

1 \leq A \leq B \leq 20

Input

Input is given from Standard Input in the following format:

A B

Output

If A is a divisor of B, print A + B; otherwise, print B - A.

Sample Input 1

4 12

Sample Output 1

16

As 4 is a divisor of 12, 4 + 12 = 16 should be printed.

Sample Input 2

8 20

Sample Output 2

12

Sample Input 3

1 1

Sample Output 3

2

1 is a divisor of 1.

Solution in C (GCC 5.4.1) (judge verdict: Accepted)


//#include "pch.h"
#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>

#define ll long long
#define fr(i,l,r) for(i=(l);i<(r);i++)
#define min(p,q) ((p)<(q)?(p):(q))
#define max(p,q) ((p)>(q)?(p):(q))
#define INF 1000000000000//10^12

int main(void)
{
	//変数の宣言
	int a,b;
	
	//データの読み込み
	
	scanf("%d %d",&a,&b);
	
	
//	printf("nは%dです\n", n);
//	printf("データの読み込み終了\n");
	//実際の処理
	
	
//	printf("計算部分終了\n");
	
	//出力
	if(b%a==0){
		printf("%d",a+b);
	}else{
		printf("%d",b-a);
	}
	
	//('・ω・'）
	
//	printf("結果の出力終了\n");
	
	return 0;
}