HDU-1222 Wolf and Rabbit (欧几里得定理)

Posted 2020-08-19 可惜没如果=_=

Wolf and Rabbit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7684    Accepted Submission(s): 3870

Problem Description

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

 

 

Author

weigang Lee

 

 

Source

杭州电子科技大学第三届程序设计大赛

 

 

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其实此题稍加思索就可以知道，若n,m不互质，则一定有安全洞口的存在

 1 #include "bits/stdc++.h"
 2 using namespace std;
 3 typedef long long LL;
 4 int cas;
 5 int n,m;
 6 int gcd(int a,int b){
 7     if (b==0) return a;
 8     else return gcd(b,a%b);
 9 }
10 int main(){
11     freopen ("wolf.in","r",stdin);
12     freopen ("wolf.out","w",stdout);
13     int i,j;
14     scanf("%d",&cas);
15     while (cas--){
16         scanf("%d%d",&n,&m);
17         if (gcd(n,m)==1)
18          puts("NO");
19         else
20          puts("YES");
21     }
22     return 0;
23 }

水题还想看代码……