HDU-1907 John (Anti-SG)

Myth丶恋晨 2022-06-10 10:36 80阅读 0赞

##                                               John       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.  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**  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  2 3 3 5 1 1 1  Sample Output  John Brother       思路：这是一道典型的Anti-SG问题，颜色的种数就相当于堆的个数，每个颜色的巧克力豆的个数就是这一堆石子的个数，拿到最后一个巧克力豆的人会输。对于求解Anti-SG这里有教学，要用到SJ定理http://blog.csdn.net/kele52he/article/details/77009784       `#include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; int main(){ int t; scanf("%d",&t); while(t--){ int n; scanf("%d",&n); int ans=0; int flag=0; for(int i=0;i<n;i++){ int x; scanf("%d",&x); ans^=x; if(x>1) flag=1; } if((ans&&!flag)||(!ans&&flag)) printf("Brother\n"); else printf("John\n"); } return 0; }`              ##