C. Make Good

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whiteawll 2022-08-16 14:49:20 博主文章分类：练习题目 ©著作权

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// Problem: C. Make Good
// Contest: Codeforces - Good Bye 2019
// URL: https://codeforces.com/problemset/problem/1270/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 2022-03-03 17:35:33
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair<int, int>
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
#define int long long
const int mod1 = 1e18;
void solve() {
    int n;
    cin >> n;
    vector<int> a(n);
    int l = 0, r = 0;
    for (int i = 0; i < n; i ++) {
        cin >> a[i];
        l += a[i];
        r ^= a[i];
    }
    int dif = r * 2 - l;
    if (dif == 0) {
        cout << 0 << endl;
        cout << endl;
    }
    else {
        if (dif <= 0) {
            if (l % 2) {
                l += (mod1 - 1), r ^= (mod1 - 1);
                dif = 2 * r - l;
                cout << 3 << endl;
                cout << mod1 - 1 <<  ' ' << dif / 2 << ' ' << dif / 2 << endl;
            }
            else {
                 l += mod1, r ^= mod1;
                 dif = 2 * r - l;
                 cout << 3 << endl;
                 cout << mod1 << ' ' << dif / 2 << " " << dif / 2 << endl;
            }
            return ;
        }
        if (dif % 2 == 0) {
            cout << 2 << endl;
            cout << dif / 2 << ' ' << dif / 2 << endl;
        }
        else {
            if (l % 2) l += (mod1 - 1), r ^=(mod1 - 1);
            
            dif = 2 * r - l;
            cout << 3 << endl;
            cout << mod1 - 1 << " " << dif / 2 << " " << dif / 2 << endl;
        }
    }
}
signed main () {
    int t;
    t =1;
    cin >> t;
    while (t --) solve();
    return 0;
}