AtCode ABC069 C-4-adjacent

N
a1 a2 ... aN

3
1 10 100

Yes

4
1 2 3 4

No

3
1 4 1

Yes

2
1 1

No

6
2 7 1 8 2 8

Yes

void coder_solution() {
    // 提升cin、cout效率
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    
    int n;
    cin >> n;
    vector<int> vi(n);
    
    int a = 0, b = 0;
    for(int i = 0; i < n; i++) {
        cin >> vi[i];
        if (vi[i] % 4 == 0) {
            a++;
        } else if ( vi[i] % 2 == 1) {
            b++;
        }
    }
    
    if(a + 1 > b) {
        cout << "Yes";
    } else {
        if(a + 1 == b) {
            if (a + b == n) {
                cout << "Yes";
            } else {
                cout << "No";
            }
        } else {
            cout << "No";
        }
    }
}

#define yes "Yes"
#define no "No"
int N, A[201010];
string solve() {
    int c2 = 0;
    int c4 = 0;
    int c = 0;

    for (int i = 0; i < N; i==) {
        if (A[i] % 4 == 0) c4++;
        else if (A[i] % 2 == 0) c2++;
        else c++;
    }

    if (c4 + 1 == c && N == (c4 + c)) return yes;
    if (c4 < c) return no;

    return yes;
}

void _main() {
    cin >> N;
    for (int i = 0; i < N; i++ ) cin >> A[i];
    cout << solve() << endl;
}