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

问题等价于求两个数是否互质。考察的应该是对于数学的应用。
 #include<stdio.h>
int main()
{
int m,n,i,k,flag,t;
scanf("%d",&k);
for(i=;i<=k;i++)
{
scanf("%d%d",&n,&m);
while(m!=)
{
t=n%m;
n=m;
m=t;
}
if(n==)
printf("NO\n");
else
printf("YES\n");
}
return ;
}

hdu 1222 Wolf and Rabbit的更多相关文章

  1. HDU 1222 Wolf and Rabbit(gcd)

    HDU 1222   Wolf and Rabbit   (最大公约数)解题报告 题目链接:http://acm.hust.edu.cn/vjudge/contest/view.action?cid= ...

  2. HDU 1222 Wolf and Rabbit(欧几里得)

    Wolf and Rabbit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  3. HDU 1222 Wolf and Rabbit(数学,找规律)

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission( ...

  4. HDU 1222 - Wolf and Rabbit & HDU 1108 - [最大公约数&最小公倍数]

    水题,只是想借此记一下gcd函数的模板 #include<cstdio> int gcd(int m,int n){return n?gcd(n,m%n):m;} int main() { ...

  5. HDU 1222 Wolf and Rabbit( 简单拓欧 )

    链接:传送门 题意:狼抓兔子,狼从 0 出发沿逆时针寻找兔子,每走一步的距离为 m ,所有洞窟的编号为 0 - n-1 ,问是否存在一个洞窟使得兔子能够安全躲过无数次狼的搜捕. 思路:简单的拓展欧几里 ...

  6. 【HDOJ】1222 Wolf and Rabbit

    最大公约数,辗转相除. #include <stdio.h> long long gcd(long long a, long long b) { if (a<b) return gc ...

  7. Wolf and Rabbit

    http://acm.hdu.edu.cn/showproblem.php?pid=1222 Wolf and Rabbit Time Limit: 2000/1000 MS (Java/Others ...

  8. Wolf and Rabbit(gcd)

    Wolf and Rabbit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)T ...

  9. gcd(最大公约数)lcm(最小公倍数)E - Wolf and Rabbit

    1.gcd 递归实现 int gcd(int i,int j){ if(j==0) return i; else return gcd(j,i%j);} 2.lcm int gcd(int i,int ...

随机推荐

  1. 汽车常用的ECU芯片

    Power Train ECU的CPU用的比较多的基本来自于Infineon,ST,Freescale BOSCH的16位ECU M(E)7系列主要使用C167内核的CPU,早期的M(E)7系列使用西 ...

  2. OAB配置

    OAB管理: http://blogs.technet.com/b/exchange_chs/archive/2013/01/31/exchange-server-2013-oab-managing- ...

  3. Codeforces Round #335 (Div. 2) D. Lazy Student 构造

    D. Lazy Student Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/contest/606/probl ...

  4. 树链剖分||dfs序 各种题

    1.[bzoj4034][HAOI2015]T2 有一棵点数为 N 的树,以点 1 为根,且树点有边权.然后有 M 个 操作,分为三种: 操作 1 :把某个节点 x 的点权增加 a . 操作 2 :把 ...

  5. bzoj 1026 [SCOI2009]windy数 数位dp

    1026: [SCOI2009]windy数 Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://www.lydsy.com/JudgeOnline ...

  6. windows 32位系统中进程最大可用内存空间为3GB (转)

    http://msdn.microsoft.com/zh-cn/library/ms189334.aspx 进程地址空间 所有 32 位应用程序都有 4 GB 的进程地址空间(32 位地址最多可以映射 ...

  7. 【机器学习算法-python实现】决策树-Decision tree(1) 信息熵划分数据集

    (转载请注明出处:http://blog.csdn.net/buptgshengod) 1.背景 决策书算法是一种逼近离散数值的分类算法,思路比較简单,并且准确率较高.国际权威的学术组织,数据挖掘国际 ...

  8. 2013 French Open Semifinal Press

    http://v.youku.com/v_show/id_XNTY4MTgzOTEy.html?firsttime=0 Novak, can you take any confirt   for qu ...

  9. TP复习15

    ## ThinkPHP 3.1.2 URL#讲师:赵桐正微博:http://weibo.com/zhaotongzheng 本节课大纲:一.URL规则 1.默认是区分大小写的 2.如果我们不想区分大小 ...

  10. Visual Studio Code 1.0.1 for python

    1. 安 F1健 ext install python E:\test\.vscode下的三个文件 2.launch.json { "version": "0.1.0&q ...