zoukankan      html  css  js  c++  java
  • HDU

    Nim is a two-player mathematic game of strategy in which 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.

    Nim is usually played as a misere game, in which the player to take the last object loses. Nim can also be played as a normal play game, which means that the person who makes the last move (i.e., who takes the last object) wins. This is called normal play because most games follow this convention, even though Nim usually does not.

    Alice and Bob is tired of playing Nim under the standard rule, so they make a difference by also allowing the player to separate one of the heaps into two smaller ones. That is, each turn the player may either remove any number of objects from a heap or separate a heap into two smaller ones, and the one who takes the last object wins.

    Input

    Input contains multiple test cases. The first line is an integer 1 ≤ T ≤ 100, the number of test cases. Each case begins with an integer N, indicating the number of the heaps, the next line contains N integers s[0], s[1], ...., s[N-1], representing heaps with s[0], s[1], ..., s[N-1] objects respectively.(1 ≤ N ≤ 10^6, 1 ≤ S[i] ≤ 2^31 - 1)

    Output

    For each test case, output a line which contains either "Alice" or "Bob", which is the winner of this game. Alice will play first. You may asume they never make mistakes.

    Sample Input

    2
    3
    2 2 3
    2
    3 3

    Sample Output

    Alice
    Bob

    题解:

    打SG函数表代码:

    #include <bits/stdc++.h>
    
    using namespace std;
    
    const int MAXN = 10005;
    
    int SG[MAXN];
    bool SGhash[MAXN];
     
    inline int getSG(int x){
    	memset(SGhash,false,sizeof SGhash);
    	for(int i=1 ; i<=x/2 ; ++i)SGhash[SG[i]^SG[x-i]] = true;
    	for(int i=0 ; i<=x ; ++i)SGhash[SG[i]] = true;
    	for(int i=0 ; ; ++i)if(!SGhash[i])return i;
    }
     
    inline void BuildSG(int x){
    	memset(SG,0,sizeof SG);
    	for(int i=1 ; i<=x ; ++i){
    		SG[i] = getSG(i);
    	}
    }
    
    int main(){
    	
    	BuildSG(100);
    	for(int i=0 ; i<11 ; ++i)printf("%d:%d ",i,SG[i]);
    	printf("
    ");
    	for(int i=11 ; i<20 ; ++i)printf("%d:%d ",i,SG[i]);
    	
    	return 0;
    }

    结果:

    发现当 x%4==0 时SG[x]=x-1,当 x%4==3 时SG[x]=x+1,其余 SG[x]=x。

    AC代码:

    #include <cstdio>
    
    using namespace std;
    
    int getSG(int x){
    	int t = x%4;
    	if(t == 0)return x-1;
    	else if(t == 3)return x+1;
    	else return x;
    }
    
    int main(){
    	
    	int T;
    	scanf("%d",&T);
    	while(T--){
    		int N;
    		scanf("%d",&N);
    		int t;
    		int re = 0;
    		while(N--){
    			scanf("%d",&t);
    			re ^= getSG(t);
    		}
    		if(re)printf("Alice
    ");
    		else printf("Bob
    ");
    	}
    	
    	return 0;
    }
  • 相关阅读:
    1JavaScript的历史
    扯扯 require .js
    查看js 执行效率工具
    php 输入输出流
    词法分析096
    Python基础综合练习
    096
    大数据概述
    熟悉常用的Linux操作
    实验一、词法分析实验
  • 原文地址:https://www.cnblogs.com/vocaloid01/p/9513990.html
Copyright © 2011-2022 走看看