狐狸找兔纸

题意：

There is a hill with n holes around. The holes are signed from 0 to n-1.   A rabbit must hide in one of the holes. A wolf searches the rabbit in anticlockwise order. The first hole he get into is the one signed with 0. Then he will get into the hole every m holes. For example, m=2 and n=6, the wolf will get into the holes which are signed 0,2,4,0. If the rabbit hides in the hole which signed 1,3 or 5, she will survive. So we call these holes the safe holes.

Input

The input starts with a positive integer P which indicates the number of test cases. Then on the following P lines,each line consists 2 positive integer m and n(0<m,n<2147483648).

Output

For each input m n, if safe holes exist, you should output "YES", else output "NO" in a single line.

Sample Input

2

1 2

2 2

Sample Output

NO

YES

思路：

辗转相处，看最后的公约数是不是1，是就代表兔纸逃不了。嘿嘿~~~~

 1 #include<iostream>
 2 using namespace std;
 3 int main()
 4 {
 5
 6     int T;
 7     cin >> T;
 8     while (T--)
 9     {
10       int a, n, m;
11       cin >> m >> n;
12       while ( n >m ? (n %= m) :(m %= n) );   //找公约数是不是1
13       if (n+m==1)
14           cout << "NO" << endl;
15       else
16           cout << "YES" << endl;
17
18     }
19     return 0;
20 }

心得：

又学到一招~~~~不怕简单，就怕不懂！