zoukankan      html  css  js  c++  java
  • 杭电_ACM_Wolf and Rabbit

    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
    View Code
     1 #include <stdio.h>
     2 //get the greatest common divisor
     3 int gcd(int a, int b)
     4 {
     5     int r;
     6     while (b)
     7     {
     8         r = a % b;
     9         a = b;
    10         b = r;
    11     }
    12     return a;
    13 }
    14 int main()
    15 {
    16     int N, m, n;
    17     scanf("%d", &N);
    18     while(N--)
    19     {
    20         scanf("%d %d", &n, &m);
    21         if(gcd(n, m) == 1)
    22             printf("NO\n");
    23         else
    24             printf("YES\n");
    25     }
    26     return 0;
    27 }

    The reason why we can judge by greatest common divisor(gcd):

    Condition:

    firstly, r is the gcd of a, b and a is bigger than b, t is the lowest common multiple(lcm) of a, b

    then you can get 

    r = gcd(a, b)

    t = a * b / r

    secondly, the wolf find the t hold, the wolf will stop, because it's will be circulatory.

    -------------------------

    Question

    the question is to find 0 < x < t / a and 0 < y < t / b to satisfy the equality

    ax + r = by

    for the equality, you will find (by - ax) / r is a integer, so you can find x, y making (by - ax) / r = 1

    ------------------------

    Conclusion

    "ax + r = by" means the r hole will be found, in other words, the gcd will be found.

    so, if gcd is equal to 1, it means all holes will be found. if gcd is equal to other number, it means there are safe holes.

  • 相关阅读:
    day10
    day 09
    day08
    day07
    day6
    day5
    成员变量和局部变量
    (第五章)java面向对象之this的作用总结
    简单的音乐播放
    异步消息处理机制 简析
  • 原文地址:https://www.cnblogs.com/chuanlong/p/2745787.html
Copyright © 2011-2022 走看看