温州大学OJ二叉树系列妙解

如何在OJ只支持c++98的苛刻条件下写下优雅的代码

#include 

void solve() {
    char c = getchar();
    if (c == '.') return;
    solve();
    std::cout << c;
    solve();
}

signed main() {
    solve();
    std::cout << "n";
    return 0;
}

#include 

std::string s;

void solve(std::string k) {
    char c = getchar();
    if (c == '.') return;
    solve(k + "-");
    s = k + c + "n" + s;
    solve(k + "-");
}

signed main() {
    solve("");
    std::cout << s;
    return 0;
}

#include 

const int MX = 1e4 + 100;

std::string s;
std::vector v(MX);

int n, cnt;
void solve(int x) {
    char c = getchar();
    if (c == '.') return;
    solve(x << 1);
    v[x] = s + c;
    n = std::max(n, x);
    s += "-";
    solve(x << 1 | 1);
}

signed main() {
    solve(1);
    for (int i = 1; i <= n; ++i) {
        if (v[i].size()) {
            std::cout << v[i].substr(cnt);
            cnt = v[i].size();           
        }
        if (!(i & (i + 1)) || i == n) {
            std::cout << "n";
            cnt = 0;
        }
    }
    return 0;
}

#include 

std::string s;

int solve() {
    char c = getchar();
    if (c == '.') return 0;
    s += c + std::string("[1](");
    const int p = s.size() - 3;
    s[p] += solve();
    s += ",";
    s[p] += solve();
    s += ")";
    return s[p] - 48;
}

signed main() {
    solve();
    const int n = s.size();
    for (int i = 0; i < n; ++i) {
        if (i + 3 < n && s.substr(i, 3) == "(,)") i += 3;
        if (i > 0 && s[i - 1] == '[') std::cout << s[i] - 48;
        else std::cout << s[i];
    }
    std::cout << "n";
    return 0;
}

#include 

std::string a, b;
std::mapmp;

void solve(int pl, int pr, int il) {
    if (pl > pr) return;
    int ipos = mp[a[pl]];
    int len = ipos - il;
    solve(pl + 1, pl + len, il);
    solve(pl + len + 1, pr, ipos + 1);
    std::cout << a[pl];
}

signed main() {
    std::cin >> a >> b;
    for (int i = 0; i < a.size(); ++i) mp[b[i]] = i;
    solve(0, a.size() - 1, 0);
    putchar('n');
    return 0;
}

#include 

std::string s, a, b;

int k;
void solve(char flag) {
    if (isalpha(flag)) return;
    char c = s[s.size() - ++k];
    solve(c);
    b = c + b;
    solve(c); 
    a = c + a;
}

signed main() {
    std::cin >> s;
    solve('#');
    std::cout << a + "n" + b + "n";
    return 0;
}