hdu 1907 John
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John
Time Limit: 5000/1000 MS (Java/Others) Memory Limit: 65535/32768 K (Java/Others)Total Submission(s): 5305 Accepted Submission(s): 3061
Problem Description
Little John is playing very funny game with his younger brother. There is one big box filled with M&Ms of different colors. At first John has to eat several M&Ms of the same color. Then his opponent has to make a turn. And so on. Please note that each player has to eat at least one M&M during his turn. If John (or his brother) will eat the last M&M from the box he will be considered as a looser and he will have to buy a new candy box.
Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.
Both of players are using optimal game strategy. John starts first always. You will be given information about M&Ms and your task is to determine a winner of such a beautiful game.
Input
The first line of input will contain a single integer T – the number of test cases. Next T pairs of lines will describe tests in a following format. The first line of each test will contain an integer N – the amount of different M&M colors in a box. Next line will contain N integers Ai, separated by spaces – amount of M&Ms of i-th color.
Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747
Constraints:
1 <= T <= 474,
1 <= N <= 47,
1 <= Ai <= 4747
Output
Output T lines each of them containing information about game winner. Print “John” if John will win the game or “Brother” in other case.
Sample Input
233 5 111
Sample Output
JohnBrother
题意:有n个不用的糖果盒子,每个盒子里有相对应的糖果的数量,两个人的博弈,两人轮流拿,最少拿一个最多把这个盒子的全部拿走,拿走最后一个的输,基于Nim博弈的变形即anti-Nim。
任给N堆石子,两人轮流从任意一堆中取,至少一个,至多这堆的全部,取到最后一颗石子的人获胜,先取的人如何获胜?
T=a^b^c...(n个数异或即可),若开始的时候T=0,则先取者必败,如果开始的时候T>0,那么只要每次取的石子使T=0,即先取者有获胜的方法。
在这个题中要注意盒子中只有1的个数。
#include<iostream>#include<cstdio>using namespace std;int main(){int ca;int n,x;scanf("%d",&ca);while(ca--){int t=0,ans=0;scanf("%d",&n);for(int i=0;i<n;i++){scanf("%d",&x);t^=x;if(x>1)ans++;}if((!ans&&n&1)||(!t&&ans>1))printf("Brother\n");elseprintf("John\n");}return 0;}
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