zoukankan      html  css  js  c++  java
  • poj 1704

    Georgia and Bob
    Time Limit: 1000MS   Memory Limit: 10000K
    Total Submissions: 7233   Accepted: 2173

    Description

    Georgia and Bob decide to play a self-invented game. They draw a row of grids on paper, number the grids from left to right by 1, 2, 3, ..., and place N chessmen on different grids, as shown in the following figure for example: 

    Georgia and Bob move the chessmen in turn. Every time a player will choose a chessman, and move it to the left without going over any other chessmen or across the left edge. The player can freely choose number of steps the chessman moves, with the constraint that the chessman must be moved at least ONE step and one grid can at most contains ONE single chessman. The player who cannot make a move loses the game. 

    Georgia always plays first since "Lady first". Suppose that Georgia and Bob both do their best in the game, i.e., if one of them knows a way to win the game, he or she will be able to carry it out. 

    Given the initial positions of the n chessmen, can you predict who will finally win the game? 

    Input

    The first line of the input contains a single integer T (1 <= T <= 20), the number of test cases. Then T cases follow. Each test case contains two lines. The first line consists of one integer N (1 <= N <= 1000), indicating the number of chessmen. The second line contains N different integers P1, P2 ... Pn (1 <= Pi <= 10000), which are the initial positions of the n chessmen.

    Output

    For each test case, prints a single line, "Georgia will win", if Georgia will win the game; "Bob will win", if Bob will win the game; otherwise 'Not sure'.

    Sample Input

    2
    3
    1 2 3
    8
    1 5 6 7 9 12 14 17
    

    Sample Output

    Bob will win
    Georgia will win
    

    Source

     
    转化成nim游戏,两个石子之间的间隔就是石子个数
     1 #include <iostream>
     2 #include <cstdio>
     3 #include <cstring>
     4 #include <algorithm>
     5 
     6 using namespace std;
     7 
     8 const int MAX_N = 1005;
     9 int p[MAX_N];
    10 
    11 int main()
    12 {
    13     int t;
    14     scanf("%d",&t);
    15     while(t--) {
    16             int N;
    17             scanf("%d",&N);
    18             for(int i = 0; i < N; ++i) {
    19                     scanf("%d",&p[i]);
    20             }
    21             if(N % 2 == 1) p[N++] = 0;
    22             sort(p, p + N);
    23 
    24             int x = 0;
    25             for(int i = 0; i + 1 < N; i += 2) {
    26                     x ^= (p[i + 1] - p[i] - 1);
    27             }
    28 
    29             if(x == 0) printf("Bob will win
    ");
    30             else printf("Georgia will win
    ");
    31     }
    32     //cout << "Hello world!" << endl;
    33     return 0;
    34 }
    View Code
  • 相关阅读:
    《图解HTTP》读书笔记
    【译】关于vertical-align你应知道的一切
    【移动端debug-5】可恶的1px万能实现方案
    《编写高质量代码改善JavaScript程序的188个建议》读书笔记
    【移动端debug-4】iOS下setTimeout无法触发focus事件的解决方案
    一张图看懂Function和Object的关系及简述instanceof运算符
    三张图搞懂JavaScript的原型对象与原型链
    一张图看懂encodeURI、encodeURIComponent、decodeURI、decodeURIComponent的区别
    图解call、apply、bind的异同及各种实战应用演示
    centos vm 桥接 --网络配置
  • 原文地址:https://www.cnblogs.com/hyxsolitude/p/3718811.html
Copyright © 2011-2022 走看看