Be the Winner（结论：反nim博弈）

Be the Winner

结论

记一个结论：反nim博弈，先手必胜1：尼姆和为零，所有值为1。2：尼姆和不为零，有一个大于1的数。

代码

/*
  Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>

#define mp make_pair
#define pb push_back
#define endl '\n'
#define mid (l + r >> 1)
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define ls rt << 1
#define rs rt << 1 | 1

using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;

const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;

inline ll read() {
    ll f = 1, x = 0;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-')    f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = (x << 1) + (x << 3) + (c ^ 48);
        c = getchar();
    }
    return f * x;
}

int main() {
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    // ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    int n;
    while(cin >> n) {
        int num = 0, ans = 0, x;
        for(int i = 1; i <= n; i++) {
            cin >> x;
            ans ^= x;
            if(x > 1) num++;
        }
        if((num && ans) || (!ans && !num)) puts("Yes");
        else puts("No");
    }
    return 0;
}