最大公约数问题（Wolf and Rabbit ）
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.
InputThe 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). 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.
OutputFor 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
刚开始是用数组写的，但是提交提示超出内存限制，然后改用最大公约数来写。
代码：
- #include <iostream>
- #include <algorithm>
- using namespace std;
- int gcd(int a,int b)
- {
- return b%a==0?a:gcd(b%a,a);
- }
- int main()
- {
- int m,n,p;
- cin>>p;
- while(p--)
- {
- cin>>m>>n;
- //cout<<gcd(m,n)<<endl;
- if(gcd(m,n)==1) cout<<"NO"<<endl;
- else cout<<"YES"<<endl;
- }
- return 0;
- }
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