这是 meelo 原创的 IEEEXtreme极限编程大赛题解
Xtreme 10.0 - Game of Stones
题目来源 第10届IEEE极限编程大赛
https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/game-of-stones-1-1
Alice and Bob play a game. The game is turn based: Alice moves first, then Bob, and so on. There are N piles of stones; in every pile there is an odd number of stones. At every turn, the one to play must pick a pile and splits it into 3 piles with an odd number of stones each.
The player who cannot split any pile loses. As this game is too simple for both of them, they decided to play multiple games in parallel. The rules remain the same, but at every turn, the one to play must first pick a game and then split a pile only in that game. The one who loses is the one that can't split any pile in any game, i.e. all the piles in all the games have only 1 stone. Bob still thinks that he is at a disadvantage, since he is the second to move. Your task is to find the winner if both the players play optimally.
Input Format
The input begins with an integer T, giving the number of test cases in the input.
Each testcase begins with an integer G, on a line by itself, giving the number of games to be played in parallel.
The G games are then described in two lines as follows: The first line gives the number of piles in the game, and the second contains the number of stones in each of the piles.
Constraints
1 <= T <= 10
1 <= [Number of piles in all games in a testcase] <= 105
1 <= [Number of stones in a pile] <= 109
Output Format
For each testcase, output the winner, i.e. either Alice or Bob, on a line by itself.
Sample Input
2
2
3
1 3 5
2
3 7
1
5
1 3 5 7 9
Sample Output
Alice
Bob
Explanation
The sample input can be annotated as follows:
2 (the number of tests)
2 (the number of parallel games for the first test)
3 (the number of piles in the first game)
1 3 5
2 (the number of piles in the second game)
3 7
1 (the number of parallel games for the second test)
5 (the number of piles)
1 3 5 7 9题目解析
石子个数为N的堆，不论分堆的方式，总共有N//2（整除）次分堆的机会，
假设f(N)表示，石子个数为N的堆，总共分堆的次数，
可以验证：f(0)=0, f(1)=0, f(3)=1, f(5)=2, f(7)=3, f(9)=4…… 好心人可以证明一下。
举一个例子：
9=(1,1,7)=(1,1,(1,1,5))=(1,1,(1,1,(1,1,3)))=(1)*9
9=(1,1,7)=(1,1,(1,3,3))=(1,1,(1,(1,1,1),3))=(1)*9
9=(1,3,5)=(1,(1,1,1),5)=(1,(1,1,1),(1,1,3))=(1)*9
9=(1,3,5)=(1,3,(1,1,3))=(1,(1,1,1),(1,1,3))=(1)*9
9=(3,3,3)=((1,1,1),3,3)=((1,1,1),(1,1,1),3)=(1)*9
不论怎么分堆，9个石子最终有4次分堆的机会。
不同堆之间相互独立，所有堆的分堆次数，是每一个堆分堆次数的和；
不同游戏之间相互独立，所有游戏的分堆次数，是每一个游戏分堆次数的和；
其实问题的结果和不同的游戏根本没有关系，仔细想想其实这是出题者在引入额外的复杂性。
Alice或Bob走一步，即分堆一次，分堆次数则减少一次；
在Alice走第一步之前，如果所有游戏的所有堆的分堆次数为奇数则Alice赢，如果为偶数则Bob赢；
既然游戏的输赢只与和的奇偶性有关，对每一堆的分堆次数二进制末位做一位的二进制加法就好了；
然后，一位的二进制加法可以由异或实现。
程序
C++
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std; int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int T;
cin >> T;
for(int t=; t<T; t++){
int G;
cin >> G;
bool alice = false; // initial bob win
for(int g=; g<G; g++) {
int P;
cin >> P;
for(int p=; p<P; p++) {
int n;
cin >> n;
alice ^= (n >> ) & ; // one bit binary addition
}
}
if(alice) {
cout << "Alice" << endl;
}
else {
cout << "Bob" << endl;
}
}
return ;
}
Python3
T = int(input())
for test_case in range(T):
G = int(input())
pile = []
tot_pile =
string = ""
for i in range(G):
tot_pile += int(input())
string += " " + input()
pile = [int(x) for x in string.split()]
tot_turns =
for i in pile:
tot_turns += i//
if tot_turns % == :
print("Bob")
else:
print("Alice")
from: medium.com/xtremefun/xtreme-10-0-game-of-stones-c29aaa72ec1e