Aizu1072 Rearranging Seats(水题)

题目：

Rearranging Seats

Time Limit : 1 sec, Memory Limit : 65536 KB 
Japanese version is here

Problem K: Rearranging Seats

Haruna is a high school student. She must remember the seating arrangements in her class because she is a class president. It is too difficult task to remember if there are so many students.

That is the reason why seating rearrangement is depress task for her. But students have a complaint if seating is fixed.

One day, she made a rule that all students must move but they don't move so far as the result of seating rearrangement.

The following is the rule. The class room consists of r*c seats. Each r row has c seats. The coordinate of the front row and most left is (1,1). The last row and right most is (r,c). After seating rearrangement, all students must move next to their seat. If a student sit (y,x) before seating arrangement, his/her seat must be (y,x+1) , (y,x-1), (y+1,x) or (y-1,x). The new seat must be inside of the class room. For example (0,1) or (r+1,c) is not allowed.

Your task is to check whether it is possible to rearrange seats based on the above rule.

Input

Input consists of multiple datasets. Each dataset consists of 2 integers. The last input contains two 0. A dataset is given by the following format.

r c

Input satisfies the following constraint.
1 ≤ r ≤ 19, 1 ≤ c ≤ 19

Output

Print "yes" without quates in one line if it is possible to rearrange the seats, otherwise print "no" without quates in one line.

Sample Input

1 12 20 0

Sample Output

noyes

Hint

For the second case, before seat rearrangement, the state is shown as follows.

1 23 4

There are some possible arrangements. For example

2 41 3

or

2 14 3

is valid arrangement.

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Source: University of Aizu Programming Contest , Aizu-Wakamatsu, Japan, 2011-06-05 
Problem Setter:  Tomoya Sakai

思路：

这题主要是题意不好懂。。。题意懂了以后就知道如果两个数的乘积为奇数就输出no,为偶数就输出yes。。。

代码：

#include <cstdio>#include <cstring>#include <cctype>#include <string>#include <set>#include <iostream>#include <stack>#include <cmath>#include <queue>#include <vector>#include <algorithm>#define mem(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3f#define mod 10000007#define N 3#define M 1000000+10#define ll long longusing namespace std;int main(){    int a,b;    while(1)    {        cin>>a>>b;        if(a==0&&b==0)return 0;        if((a*b)&1)            puts("no");        else            puts("yes");    }    return 0;}