Being a Good Boy in Spring Festival HDU-1850 Game theory, using the nature of XOR

• Since any card can be taken in a pile, the number of cards in a pile is $x$ , there is$sg[x]=x$
• First find the total game $SG$值$ans=sg[a_1]\oplus sg\left[a_2\right]\oplus ...\oplus sg\left[a_n\right]$
• 若$ans=0$ means the first move is defeated.
• On the contrary, the first mover will win. Then the path corresponding to the first game: Suppose there are two piles of cards $x,y$ , assuming$y$ in the card, is taken after$Y^{\prime }$ , It must be$P$ point, which satisfies$x\oplus Y^{\prime }=0$ . According to the nature of XOR, for a non-zero number$x$ ，有 $x\oplus x=0$。
• It is assumed to be the first $To$ operate the$i$ heap, first set$an{s}^{\prime }=ans\oplus sg[a_i]\; The$ result is in addition to theiiThe totalSG of theremaining piles outside the$i$ pile$SG$ value, if the$an{s}^{\prime }$ ≤ $a_i$It means that ans ′ ans’ can be taken from the pile$an{s}^{\text{'}}$ Cards, making the game’s new$The\; SG$ value is 0, which meansPPwhen it corresponds to the backhand state$P$ point.

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e2+10;
int a[N];
int main(){
    
    
	int m;
	while(scanf("%d",&m),m){
    
    
		int ans=0;
		for(int i=1;i<=m;++i){
    
    
			scanf("%d",&a[i]);
			ans^=a[i];
		}
		if(!ans) puts("0");
		else{
    
    
			int cnt=0;
			for(int i=1;i<=m;++i){
    
    
				if((ans^a[i])<=a[i]) ++cnt;
			}
			printf("%d\n",cnt);
		}
	}
	return 0;
}