A

for i in range(int(input()): print(eval(input()))

B

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    ll k;
    cin >> n >> k;

    vector<ll> a(n + 1), pre(n + 1, 0);
    map<ll, int> cnt;
    ll ans = 0;
    cnt[k] = 1;
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        pre[i] = pre[i - 1] + a[i];
        ans += cnt[pre[i]];
        cnt[pre[i] + k]++;
    }
    cout << ans << "\n";
    
    return 0;
}

C

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

void Solve() {
    int n, m;
    cin >> n >> m;
    vector<string> a(n);
    int total = 0, sx = -1, sy = -1;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        for (int j = 0; j < m; j++) {
            if (a[i][j] == '1') {
                total++;
                if (sx == -1) {
                    sx = i;
                    sy = j;
                }
            }
        }
    }

    if (total == 0) {
        cout << "No" << endl;
        return;
    }

    vector<vector<int>> vis(n, vector<int>(m, 0));
    queue<pair<int,int>> q;
    q.push({sx, sy});
    vis[sx][sy] = true;
    int cnt = 0;

    int dx[] = {1, -1, 0, 0};
    int dy[] = {0, 0, 1, -1};

    while (!q.empty()) {
        auto [x, y] = q.front(); q.pop();
        cnt++;
        for (int d = 0; d < 4; d++) {
            int nx = x + dx[d], ny = y + dy[d];
            if (nx >= 0 && nx < n && ny >= 0 && ny < m
                && !vis[nx][ny] && a[nx][ny] == '1') {
                vis[nx][ny] = true;
                q.push({nx, ny});
            }
        }
    }

    cout << (cnt == total ? "Yes" : "No") << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int T = 1;
    cin >> T;
    while (T--) {
        Solve();
    }
    return 0;
}

D

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

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n;
    cin >> n;
    vector<int> a(n);
    int mx = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        mx = max(mx, a[i]);
    }

    ll ans = 0;
    for (int g = 1; g <= mx; g++) {
        ll s = 0;
        for (int x : a) {
            if (x % g == 0) s += x;
        }
        ans = max(ans, s * g);
    }
    cout << ans << "\n";
    
    return 0;
}

E

#include<bits/stdc++.h>
#define endl '\n'
#define ll long long
using namespace std;

const int MAXP = 10000;

vector<int> primes;

void init() {
    vector<bool> is_p(MAXP + 1, true);
    is_p[0] = is_p[1] = false;
    for (int i = 2; i <= MAXP; i++) {
        if (is_p[i]) {
            primes.push_back(i);
            for (ll j = (ll)i * i; j <= MAXP; j += i) is_p[j] = false;
        }
    }
}

void Solve() {
    int n;
    cin >> n;
    int s = 0;
    for (int p : primes) {
        if ((ll)p * p > n) break;
        while (n % p == 0) {
            n /= p;
            s++;
        }
    }
    if (n > 1) s++;
    cout << (s & 1 ? "Alice" : "Bob") << endl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    init();
    int T = 1;
    cin >> T;
    while (T--) Solve();
    return 0;
}

F

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

const ll MOD = 998244353;

ll norm(ll x) {
    x %= MOD;
    if (x < 0) x += MOD;
    return x;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;

    vector<ll> a(n + 1), b(n + 1);
    for (int i = 1; i <= n; i++) {
        cin >> a[i];
        a[i] = norm(a[i]);
    }
    for (int i = 1; i <= n; i++) {
        cin >> b[i];
        b[i] = norm(b[i]);
    }

    // E = sum_{i=1}^{n-1} b_i * a_i - (sum_{i=1}^{n-1} b_i) * a_n
    ll P = 0, T = 0;
    for (int i = 1; i < n; i++) {
        P = (P + b[i] * a[i]) % MOD;
        T = (T + b[i]) % MOD;
    }

    int q;
    cin >> q;
    while (q--) {
        int type;
        cin >> type;
        if (type == 1) {
            int x;
            ll v;
            cin >> x >> v;
            v = norm(v);
            if (x < n) {
                P = (P + b[x] * v) % MOD;
            }
            a[x] = norm(a[x] + v);
        } else if (type == 2) {
            int x;
            ll v;
            cin >> x >> v;
            v = norm(v);
            if (x < n) {
                P = (P + a[x] * v) % MOD;
                T = (T + v) % MOD;
            }
            b[x] = norm(b[x] + v);
        } else {
            ll v;
            cin >> v;
            v = norm(v);
            ll E = norm(P - T * a[n] % MOD);
            cout << norm(E + v) << "\n";
        }
    }

    return 0;
}