A题：

输出-1，当且仅当数组大小为2，并且不递增。

如果已经递增，输出0。

否则至多三次完成操作，如果最小或最大在开头或者结尾，可证只需一次操作。

如果两者出现在中间，可证只需二次操作。

不然，要求三次操作。

#include <bits/stdc++.h>
using namespace std;
int main() {
    int tt; cin >> tt;
    while (tt -- ) {
        int n; cin >> n;
        vector<int> a(n);
        for (int i = 0; i < n; i ++ ) cin >> a[i];
        if (is_sorted(a.begin(), a.end())) cout << 0 << endl;
        else {
            if (n == 2) cout << -1 << endl;
            else {
                auto [mn, mx] = minmax_element(a.begin(), a.end());
                if (*mn == a[0] || *mx == a[n - 1]) cout << 1 << endl;
                else {
                    bool ok = false;
                    for (int i = 1; i < n; i ++ ) {
                        if (a[i] == *mx) ok = true;
                    }
                    for (int i = 0; i < n - 1; i ++ ) {
                        if (a[i] == *mn) ok = true;
                    }
                    if (ok) cout << 2 << endl;
                    else {
                        cout << 3 << endl;
                    }
                }
            }
        }
    }
    return 0;
}

b题：

题目要求的是变换最少使得单增，转一下思路。

找到最长上升子序列。

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n; cin >> n;
    vector<int> dp(n + 1, 1e9);
    int len = 0;
    for (int i = 1; i <= n; i ++ ) {
        int x; cin >> x;
        if (len == 0 || x >= dp[len]) {
            dp[++ len] = x;
        }else {
            *upper_bound(dp.begin() + 1, dp.end(), x) = x;
        }
    }
    cout << n - len << endl;
    return 0;
}

c题：

先处理[1, r]

后面这就可以当作[1, l - 1]放到dp里了

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    string s; cin >> s;
    int n = s.size();
    s = " " + s;
    int c0 = 0, c1 = 0, c2 = 0;
    int dp[2][2][2]{};
    dp[0][0][0] = 1;
    ll ans = 0;
    for (int i = 1; i <= n; i ++ ) {
        if (s[i] == 'j') {
            c0 ++;
        }else if (s[i] == 'q') {
            c1 ++;
            c2 += c0;
        }
        c0 %= 2, c1 %= 2, c2 %= 2;
        for (int a = 0; a < 2; a ++ ) {
            for (int b = 0; b < 2; b ++ ) {
                for (int c = 0; c < 2; c ++ ) {
                    int t = (c2 - c + 2) % 2;
                    t -= a * (c1 - b + 2) % 2;
                    t = (t % 2 + 2) % 2;
                    if (t == 0) ans += dp[a][b][c];
                }
            }
        }
        dp[c0][c1][c2] ++;
    }
    cout << ans << endl;
    return 0;
}

d题：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
    int n; cin >> n;
    ll cnt0, cnt1 = 0;
    ll ans = 0;
    for (int i = 1; i <= n; i ++ ) {
        int x; cin >> x;
        if (x == 0) continue;
        if (x > 0) {
            if (cnt0 >= x) {
                cnt0 -= x;
                cnt1 += x;
            }else {
                ans += x - cnt0;
                cnt0 = 0;
                cnt1 += x;
            }
        }else {
            x = -x;
            if (cnt1 >= x) {
                cnt1 -= x;
                cnt0 += x;
            }else {
                ans += x - cnt1;
                cnt1 = 0;
                cnt0 += x;
            }
        }
    }
    cout << ans << endl;
    
    return 0;
}

e题：

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 998244353;
const int N = 3e5 + 9;
ll f[N];
ll qmi(ll a, ll b) {
    ll res = 1;
    while (b) {
        if (b & 1) res = res * a % mod;
        b >>= 1;
        a = a * a % mod;
    }
    return res;
}
ll inv(ll x) {
    return qmi(x, mod - 2);
}
void iniv() {
    f[0] = f[1] = 1;
    for (int i = 2; i <= 3e5; i ++ ) {
        f[i] = f[i - 1] * i % mod;
    }
}
ll C(ll a, ll b) {
    return f[a] * inv(f[b]) % mod * inv(f[a - b]) % mod;
}
int main() {
    iniv();
    int n; cin >> n;
    ll ans = 1;
    for (int i = 1; i <= n / 3; i ++ ) {
        int a[3]{};
        int minv = 1e9;
        for (int j = 0; j < 3; j ++ ) {
            cin >> a[j];
            minv = min(minv, a[j]);
        }
        int cnt = 0;
        for (int j = 0; j < 3; j ++ ) {
            if (minv == a[j]) cnt ++;
        }
        ans = ans * cnt % mod;
    }
    ans = ans * C(n / 3, n / 6) % mod;
    cout << ans << endl;
    return 0;
}

g题：

#include <bits/stdc++.h>
using namespace std;
int main() {
    int tt; cin >> tt;
    while (tt -- ) {
        long long n; cin >> n;
        if (n == 1 || n == 5 || n == 6) cout << "WIN\n";
        else cout << "LOSE\n";
    }
    return 0;
}