PAT（乙级）2020年秋季考试

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7-1 多二了一点 (15分)

题解

模拟。

代码

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

int sum(string s) {
    int res = 0;
    for (char c : s) res += c - '0';
    return res;
}

int main() {
    string s;
    cin >> s;
    if (s.size() & 1) {
        cout << "Error: " << s.size() << " digit(s)" << "
";
    } else {
        string s_x = s.substr(s.size() / 2);
        string s_y = s.substr(0, s.size() / 2);
        if (sum(s_x) - sum(s_y) == 2)
            cout << "Yes: " << s_x << " - " << s_y << " = 2" << "
";
        else
            cout << "No: " << s_x << " - " << s_y << " != 2" << "
";
    }
}

7-2 数字之王 (20分)

题解

模拟。

代码

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

int func(int n) {
    if (n == 0) return 0;
    int mul = 1;
    while (n) {
        int x = n % 10;
        mul *= x * x * x;
        n /= 10;
    }
    int res = 0;
    while (mul) {
        res += mul % 10;
        mul /= 10;
    }
    return res;
}

int main() {
    int N1, N2;
    cin >> N1 >> N2;
    vector<int> a(N2 - N1 + 1);
    iota(a.begin(), a.end(), N1);
    while (any_of(a.begin(), a.end(), [](int x) { return x >= 10; })) {
        for (auto &x : a)
            x = func(x);
    }
    int mx = 0;
    map<int, int> cnt;
    for (auto x : a) {
        if (++cnt[x] > mx)
            mx = cnt[x];
    }
    vector<int> ans;
    for (auto i : cnt) {
        if (i.second == mx)
            ans.push_back(i.first);
    }
    sort(ans.begin(), ans.end());
    cout << mx << "
";
    for (int i = 0; i < ans.size(); i++)
        cout << ans[i] << " 
"[i == ans.size() - 1];
}

7-3 如需挪车请致电 (20分)

题解

有点麻烦的模拟。

代码

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

map<string, int> mp{
    {"ling", 0},
    {"yi", 1},
    {"er", 2},
    {"san", 3},
    {"si", 4},
    {"wu", 5},
    {"liu", 6},
    {"qi", 7},
    {"ba", 8},
    {"jiu", 9}
};

int to_int(string s) {
    if (mp.count(s)) {
        return mp[s];
    } else {
        int res = 0;
        for (int i = 0; i < int(s.size()); i++)
            res = res * 10 + s[i] - '0';
        return res;
    }
}

int func(string s) {
    if (s.find("sqrt") != string::npos) {
        return sqrt(to_int(s.substr(4)));
    }
    for (char oper : string("+-*/%^")) {
        int oper_pos = s.find(oper);
        if (oper_pos != string::npos) {
            int a = to_int(s.substr(0, oper_pos));
            int b = to_int(s.substr(oper_pos + 1));
            if (oper == '+') return a + b;
            if (oper == '-') return a - b;
            if (oper == '*') return a * b;
            if (oper == '/') return a / b;
            if (oper == '%') return a % b;
            if (oper == '^') return pow(a, b);
        }
    }
    return to_int(s);
}

int main() {
    for (int i = 0; i < 11; i++) {
        string s;
        cin >> s;
        cout << func(s);
    }
}

7-4 胖达与盆盆奶 (20分)

题解

将熊猫的编号视作横坐标，体重视作纵坐标，将纵坐标相连即得一条凹凸的曲线，每次从曲线的凹处向两边延伸，两个凹处的相接点取较大值。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n;
    cin >> n;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    vector<int> p(n);
    iota(p.begin(), p.end(), 0);
    sort(p.begin(), p.end(), [&](int x, int y) {
        return a[x] < a[y];
    });
    vector<int> milk(n);
    vector<bool> vis(n);
    for (auto i : p) {
        if (vis[i]) continue;
        milk[i] = 200;
        vis[i] = true;
        for (int j = i - 1; j >= 0 and a[j] >= a[j + 1]; j--) {
            milk[j] = max(milk[j], milk[j + 1] + (a[j] > a[j + 1] ? 100 : 0));
            vis[j] = true;
        }
        for (int j = i + 1; j < n and a[j] >= a[j - 1]; j++) {
            milk[j] = max(milk[j], milk[j - 1] + (a[j] > a[j - 1] ? 100 : 0));
            vis[j] = true;
        }
    }
    cout << accumulate(milk.begin(), milk.end(), 0) << "
";
}

7-5 买地攻略 (25分)

题解一

$O_{(n^2)}$，枚举区间的左右端点，但理论上来讲应该过不了。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; i++)
        cin >> a[i];
    int ans = 0;
    for (int i = 0; i < n; i++) {
        int sum = 0;
        for (int j = i; j < n; j++) {
            sum += a[j];
            if (sum <= m) ++ans;
        }
    }
    cout << ans << "
";
}

题解二

$O_{(nlog_n)}$，枚举区间左端点，利用前缀和二分查找右端点。

代码

#include <bits/stdc++.h>
using namespace std;
int main() {
    int n, m;
    cin >> n >> m;
    vector<int> a(n);
    for (int i = 0; i < n; i++) 
        cin >> a[i];
    vector<int> pre(n);
    for (int i = 0; i < n; i++) {
        if (i == 0) pre[i] = a[i];
        else pre[i] = pre[i - 1] + a[i];
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
        int j = upper_bound(pre.begin(), pre.end(), (i == 0 ? 0 : pre[i - 1]) + m) - pre.begin();
        ans += j - i;
    }
    cout << ans << "
";
}