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=88159#problem/G 题目: 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…

题意: 一头狼和一头兔子在一座山中,给你一个数n表示洞的个数,编号从0~n-1.兔子可以随意躲在其中一个洞中,狼每次都从编号为0的洞出发,接下来走到第m个洞中,问兔子能不能活下来,即不被狼吃掉.例如:当n = 8,m = 6时,狼走过的洞途径:0->6->4->0->6......,它没走过的洞还有1,2,3,5,7,所以兔子可以存活下来. 分析: 狼会一直在这n个洞中来回,它的模数就是n:它走过的圈数会增加且只有这一个变量,所以可以想到模线性方程组. mx ≡ k(mod n),…

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…

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…

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit Status 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 anti…

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() { int n,m,t; scanf("%d",&t); while(t--){ scanf("%d%d",&m,&n); ) printf("NO\n"); else printf("YES\n"); }…

Wolf and Rabbit Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9485    Accepted Submission(s): 4857 Problem Description There is a hill with n holes around. The holes are signed from 0 to n-1.…

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9848    Accepted Submission(s): 5011 Problem Description There is a hill with n holes around. The holes are signed from 0 to n-1. A rabbit must hi…

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:简单的拓展欧几里德,设 st 为兔子洞窟编号( 0 <= st < n ),很简单就可以得到一个方程 0 + m * x = n * y + st,化简一下得到 m * x - n * y = st,如果这个方程有解,那么兔子一定能被狼抓到.方程有解的条件是 st % d == 0 ,当 d == 1 时,显然是…

HDU 1000 A + B Problem  I/O HDU 1001 Sum Problem  数学 HDU 1002 A + B Problem II  高精度加法 HDU 1003 Maxsum  贪心 HDU 1004 Let the Balloon Rise  字典树,map HDU 1005 Number Sequence  求数列循环节 HDU 1007 Quoit Design  最近点对 HDU 1008 Elevator  模拟 HDU 1010 Tempter of th…