zoukankan      html  css  js  c++  java
  • poj 2348

    Euclid's Game
    Time Limit: 1000MS   Memory Limit: 65536K
    Total Submissions: 7418   Accepted: 3022

    Description

    Two players, Stan and Ollie, play, starting with two natural numbers. Stan, the first player, subtracts any positive multiple of the lesser of the two numbers from the greater of the two numbers, provided that the resulting number must be nonnegative. Then Ollie, the second player, does the same with the two resulting numbers, then Stan, etc., alternately, until one player is able to subtract a multiple of the lesser number from the greater to reach 0, and thereby wins. For example, the players may start with (25,7): 
             25 7
    
    11 7
    4 7
    4 3
    1 3
    1 0

    an Stan wins.

    Input

    The input consists of a number of lines. Each line contains two positive integers giving the starting two numbers of the game. Stan always starts.

    Output

    For each line of input, output one line saying either Stan wins or Ollie wins assuming that both of them play perfectly. The last line of input contains two zeroes and should not be processed.

    Sample Input

    34 12
    15 24
    0 0
    

    Sample Output

    Stan wins
    Ollie wins
    

    Source

     
    分两种情况 b - a < a b - a > a
    第一种情况别无选择,第二种情况必胜,由此可得答案
     
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 int a, b;
     9 
    10 bool check() {
    11         bool flag = 1;
    12         for(;;) {
    13                 if(a > b) swap(a,b);
    14                 if(b % a == 0) break;
    15                 if(b - a > a) break;
    16                 b -= a;
    17                 flag = !flag;
    18         }
    19 
    20         return flag;
    21 }
    22 int main()
    23 {
    24     while(~scanf("%d%d",&a, &b) && (a + b)) {
    25             if(check()) {
    26                     printf("Stan wins
    ");
    27             } else {
    28                     printf("Ollie wins
    ");
    29             }
    30     }
    31     //cout << "Hello world!" << endl;
    32     return 0;
    33 }
    View Code
  • 相关阅读:
    洛谷P5281 [ZJOI2019] Minimax搜索
    势函数
    Comet OJ [Contest #5] 迫真大游戏
    洛谷P3307 [SDOI2013] 项链
    洛谷P5985 [PA2019] Muzyka pop
    CF1205E Expected Value Again
    review
    CF891E Lust
    线性代数
    洛谷P4607 [SDOI2018] 反回文串
  • 原文地址:https://www.cnblogs.com/hyxsolitude/p/3718570.html
Copyright © 2011-2022 走看看