zoukankan      html  css  js  c++  java
  • HDOJ 5088 Revenge of Nim II 位运算


    位运算。。

    。。

    Revenge of Nim II

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 229    Accepted Submission(s): 79


    Problem Description
    Nim is a mathematical game of strategy in which two players take turns removing objects from distinct heaps. On each turn, a player must remove at least one object, and may remove any number of objects provided they all come from the same heap.
    ---Wikipedia

    Today, Nim takes revenge on you, again. As you know, the rule of Nim game is rather unfair, only the nim-sum (⊕) of the sizes of the heaps is zero will the first player lose. To ensure the fairness of the game, the second player has a chance to move some (can be zero) heaps before the game starts, but he has to move one heap entirely, i.e. not partially. Of course, he can’t move all heaps out, at least one heap should be left for playing. Will the second player have the chance to win this time?

     

    Input
    The first line contains a single integer T, indicating the number of test cases. 

    Each test case begins with an integer N, indicating the number of heaps. Then N integer Ai follows, indicating the number of each heap.

    [Technical Specification]
    1. 1 <= T <= 100
    2. 1 <= N <= 1 000
    3. 1 <= Ai <= 1 000 000 000 000
     

    Output
    For each test case, output “Yes” if the second player can win by moving some (can be zero) heaps out, otherwise “No”.
     

    Sample Input
    3 1 2 3 2 2 2 5 1 2 3 4 5
     

    Sample Output
    No Yes Yes
    Hint
    For the third test case, the second player can move heaps with 4 and 5 objects out, so the nim-sum of the sizes of the left heaps is 1⊕2⊕3 = 0.
     

    Source
     


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    
    using namespace std;
    
    typedef long long int LL;
    
    int n;
    LL c[60];
    LL a[1100];
    
    int main()
    {
        int T_T;
        scanf("%d",&T_T);
        while(T_T--)
        {
            memset(c,0,sizeof(c));
            scanf("%d",&n);
            for(int i=1;i<=n;i++)
                cin>>a[i];
            bool flag=false;
            for(int i=1;i<=n;i++)
            {
                for(int j=60;j>=0;j--)
                {
                    if(c[j]==0&&(a[i]&(1LL<<j)))
                    {
                        c[j]=a[i];
                        break;
                    }
                    else if(a[i]&(1LL<<j))
                    {
                        a[i]^=c[j];
                        if(a[i]==0) flag=true;
                    }
                }
            }
            if(flag==true) puts("Yes");
            else puts("No");
        }
        return 0;
    }
    



  • 相关阅读:
    POJ 2750 Potted Flower (单点改动求线段树上最大子序列和)
    [异能程序猿]第四章 偶遇(第四更)
    【web开发学习笔记】Struts-Tags学习笔记1
    UVa 10100
    【算法拾遗】大数相加(不开辟额外空间)
    cocos2d-html5学习笔记(六)--alpha2中cc.Sequence.create中的bug
    大一暑假和大二专业学习的规划
    Android如何获得系统版本
    Struts2——(1)Struts2入门
    设置m_pszAppName值的做法
  • 原文地址:https://www.cnblogs.com/lcchuguo/p/5159995.html
Copyright © 2011-2022 走看看