鱨鱳集训第榹天晚上

T1：
打了接近两个小时，结果不知道为什么RE了，还不如直接输出V或F。

#include 
using namespace std;
string s;
int len;
bool ans, cnt1, cnt2;
char op;
stack s1;
stack s2;
int youxian(char op) {
    if (op == '!')
        return 4;
    if (op == '&')
        return 3;
    if (op == '|')
        return 2;
    if (op == '(')
        return 1;
    return 0;
}
void xiaojisuan() {
    op = s2.top();
    s2.pop();
    if (op == '!') {
        cnt1 = s1.top();
        s1.pop();
        s1.push(!cnt1);
    } else if (op == '&') {
        cnt1 = s1.top();
        s1.pop();
        cnt2 = s1.top();
        s1.pop();
        s1.push(cnt1 && cnt2);
    } else if (op == '|') {
        cnt1 = s1.top();
        s1.pop();
        cnt2 = s1.top();
        s1.pop();
        s1.push(cnt1 || cnt2);
    }
}
bool dajisuan() {
    for (int i = 0; i = youxian(ch)) xiaojisuan();
            s2.push(ch);
        } else if (ch == '(')
            s2.push(ch);
        else if (ch == ')') {
            while (!s2.empty() && s2.top() != '(') xiaojisuan();
            s2.pop();
        }
    }
    while (!s2.empty()) xiaojisuan();
    return s1.top();
}
int main() {
    freopen("boolean.in", "r", stdin);
    freopen("boolean.out", "w", stdout);
    getline(cin, s);
    len = s.size();
    ans = dajisuan();
    cout << (ans ? "V" : "F");
    return 0;
}

T2：
暴力代码，直接AC

#include 
using namespace std;
string s1, s2;
int n, k, cnt, idx, ans;
int main() {
    freopen("master.in", "r", stdin);
    freopen("master.out", "w", stdout);
    cin >> n >> k >> s1 >> s2;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j <= n; j++) {
            cnt = idx = 0;
            for (int l = 1; l = k)
                        break;
                    idx++;
                }
                cnt++;
            }
            ans = max(ans, cnt);
        }
    }
    cout << ans;
    return 0;
}

T3：
骗分，一分不得

#include 
using namespace std;
int n, m;
int main() {
    freopen("starless.in", "r", stdin);
    freopen("starless.out", "w", stdout);
    cin >> n >> m;
    cout << n << " " << n * (n - 1);
    return 0;
}

《玄学骗分》
T4：
只为第一个子任务，拿下40分

#include 
using namespace std;
int f[5005][5005], len1, len2;
string s1, s2;
int main() {
    freopen("lcs.in", "r", stdin);
    freopen("lcs.out", "w", stdout);
    cin >> s1 >> s2;
    len1 = s1.size(), len2 = s2.size();
    for (int i = 1; i < len1; i++) {
        for (int j = 1; j < len2; j++) {
            f[i][j] = max(f[i - 1][j], f[i][j - 1]);
            if (s1[i - 1] == s2[j - 1])
                f[i][j] = max(f[i][j], f[i - 1][j - 1] + 1);
        }
    }
    cout << f[len1 - 1][len2 - 1] << '\n';
    cout << 88;
    return 0;
}

0+100+0+40=140分