Codeforces Round ＃770 (Div. 2) B. Fortune Telling

Title Description

Alice and Bob Fortune-telling , The process is as follows ：
Given array a contain n Number ,Alice The starting number of is x,Bob The starting number of is x+3, Two people on the array a Each number in performs one of the following operations , Give a number y, Ask who gets the last sum y identical , The two operations are as follows ：

• Put the current number d Turn into d+a_i
• Put the current number d Turn into d Exclusive or a_i

The two operations are addition and XOR , XOR is a bit operation , The algorithm is the same 0 different 1, namely
0 xor 0 = 0
0 xor 1 = 1
1 xor 0 = 1
1 xor 1 = 0
The operation of XOR is consistent with the operation of addition , The title says that there must be someone who can get y, So for each number in the array , Either XOR or add , Judge who gets the last number Last and y The same can be .

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 100010;

int a[N];

void solve()
{
     
    int n;
    long long x, y, cur;
    cin >> n >> x >> y;
    cur = x;
    for(int i = 0; i < n; i ++) 
    {
    
        int t; cin >> t;
        cur += t;
    }
    if(cur % 2 == y % 2) puts("Alice");
    else puts("Bob");
}

int main()
{
    
    int t; cin >> t;
    while(t --) solve();
    return 0;
}