wuzhibin

摘要： (一)巴什博奕(Bash Game):只有一堆n个物品,两个人轮流从这堆物品中取物,规定每次至少取一个,最多取m个.最后取光者得胜.n = (m+1)r+s , (r为任意自然数,s≤m),即n%(m+1) != 0,则先取者肯定获胜(二)威佐夫博奕(Wythoff Game):有两堆各若干个物品,两个人轮流从某一堆或同时从两堆中取同样多的物品,规定每次至少取一个,多者不限,最后取光者得胜.(ak,bk)(ak ≤ bk ,k=0,1,2,...,n)表示奇异局势求法：ak =[k(1+√5)/2], bk= ak + k (k=0,1,2,...,n方括号表示取整函数)判断：Gold=(1+ 阅读全文

摘要： Source CodeProblem:2234User:billforumMemory:192KTime:16MSLanguage:C++Result:AcceptedSource Code#include <iostream>#include <cstdio>using namespace std;int main(){ int n,ans=0,tmp; while(cin>>n) { ans=0; for(int i=0;i<n;i++) { cin>>tmp; ans^=tmp; } if(ans==0) printf("N 阅读全文