PAT（乙级）2020年春季考试

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7-1 对称日

题解

模拟，注意年月日不足位在前面补零。

代码

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

map<string, string> mp{
    {"Jan", "1"},
    {"Feb", "2"},
    {"Mar", "3"},
    {"Apr", "4"},
    {"May", "5"},
    {"Jun", "6"},
    {"Jul", "7"},
    {"Aug", "8"},
    {"Sep", "9"},
    {"Oct", "10"},
    {"Nov", "11"},
    {"Dec", "12"}
};

bool is_palindrome(const string &s) {
    string t(s);
    reverse(t.begin(), t.end());
    return s == t;
}

void solve() {
    string m, d, y; cin >> m >> d >> y;
    d.pop_back();
    if (y.size() < 4) y = string(4 - y.size(), '0') + y;
    if (m.size() < 2) m = '0' + m;
    if (d.size() < 2) d = '0' + d;
    string s = y + mp[m] + d;
    cout << (is_palindrome(s) ? 'Y' : 'N') << ' ' << s << "\n";
}

int main() {
    int t; cin >> t;
    while (t--) solve();
}

7-2 超标区间

题解

维护一个元素超过阈值的区间，然后以小于阈值的数为分割点。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n, t; cin >> n >> t;
    int a[n + 1] = {};
    for (int i = 0; i < n; ++i) {
        cin >> a[i];
    }
    vector<pair<int, int>> res;
    int l = 0, r = 0;
    for (int i = 0; i <= n; ++i) {
        if (a[i] > t) {
            if (a[r] > t) {
                r = i;
            } else {
                l = i;
                r = i;
            }
        } else {
            if (a[r] > t) {
                res.emplace_back(l, r);
            }
            l = i;
            r = i;
        }
    }
    if (!res.empty()) {
        for (auto pr : res)
            cout << "[" << pr.first << ", " << pr.second << "]" << "\n";
    } else {
        cout << *max_element(a, a + n) << "\n";
    }
}

7-3 钱串子的加法

题解

将两个数去除前导零后统一为同一长度，模拟相加即可，注意进位。

Tips

这题数据似乎卡得不是很严，像去除前导零和计算两个数的和如果用 $O_{(n)}$ 的  s.erase(s.begin())  和  ans = ch + ans 即使总时间复杂度会达到 $O_{(n^2)}$ 也不会超时。只会比下面 $O_{(n)}$ 的代码慢十倍左右。

代码

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

map<char, int> mp1;
map<int, char> mp2;

void init() {
    int num = 0;
    for (char c : string("0123456789abcdefghijklmnopqrst")) mp1[c] = num++;
    for (auto pr : mp1) mp2[pr.second] = pr.first;
}

void del_lead_zero(string &s) {
    reverse(s.begin(), s.end());
    while (s.size() > 1 and s.back() == '0') s.pop_back();
    reverse(s.begin(), s.end());
}

void sum(string &s1, string &s2) {
    del_lead_zero(s1);
    del_lead_zero(s2);
    if (s1.size() < s2.size()) {
        swap(s1, s2);
    }
    s2 = string(s1.size() - s2.size(), '0') + s2;
    bool carry = false;
    for (int i = s1.size() - 1; i >= 0; --i) {
        int sum = mp1[s1[i]] + mp1[s2[i]] + carry;
        s1[i] = mp2[sum % 30];
        carry = (sum >= 30);
    }
    if (carry) s1 = "1" + s1;
}

int main() {
    init();
    string s1, s2; cin >> s1 >> s2;
    sum(s1, s2);
    cout << s1 << "\n";
}

7-4 全素日

题解

模拟。

代码

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

bool isprime(int n) {
    if (n <= 1) return false;
    for (int i = 2; i * i <= n; ++i)
        if (n % i == 0) return false;
    return true;
}

int main() {
    string s; cin >> s;
    bool all_prime = true;
    for (int i = 0; i < s.size(); ++i) {
        string t = s.substr(i);
        bool is_prime = isprime(stoi(t));
        if (not is_prime) all_prime = false;
        cout << t << ' ' << (is ? "Yes" : "No") << "\n";
    }
    if (all_prime) cout << "All Prime!" << "\n";
}

7-5 裁判机

题解

因为时间复杂度为 $O_{(n^2)}$ 且最多的情况有 $10^4$ 个数，所以不能再用  map 或 set 这类每次操作时间复杂度为 $O_{(log_n)}$ 的容器了，可以使用 $O_{(1)}$ 的散列（数组映射）。

代码

#include <bits/stdc++.h>
using namespace std;
constexpr int N = 1e5 + 100;

bool appear[N], sub[N];

int main() {
    int t1, t2; cin >> t1 >> t2;
    appear[t1] = appear[t2] = true;
    sub[abs(t1 - t2)] = true;
    int n, m; cin >> n >> m;
    int a[n][m] = {};
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cin >> a[i][j];
        }
    }
    vector<int> res = {t1, t2};
    bool out[n] = {};
    for (int j = 0; j < m; ++j) {
        for (int i = 0; i < n; ++i) {
            if (out[i]) continue;
            if (appear[a[i][j]] or !sub[a[i][j]]) {
                cout << "Round #" << j + 1 << ": " << i + 1 << " is out.\n";
                out[i] = true;
                continue;
            }
            for (auto ele : res) sub[abs(a[i][j] - ele)] = true;
            res.push_back(a[i][j]);
            appear[a[i][j]] = true;
        }
    }
    if (find(out, out + n, false) - out == n) {
        cout << "No winner.";
    } else {
        cout << "Winner(s):";
        for (int i = 0; i < n; ++i)
            if (not out[i]) cout << ' ' << i + 1;
    }
}