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PAT (Advanced Level) Practice 甲级顶级题目_pta题库甲级

作者：菜鸟追梦旅行 | 2024-05-26 16:19:20

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pta题库甲级

PAT甲级题目

文章目录

1001 A+B Format (20 分)

题意：给定两个整数a和b，返回答案a+b要求千位分隔符

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;

string solve(string s)
{
    reverse(s.begin(), s.end());
    string ans = "";
    int n = s.size();
    int cnt = 0;
    for (int i = 0; i < n; i++)
    {
        ans += (s[i]);
        if (++cnt % 3 == 0 && i != (n - 1))
            ans += ",";
    }
    reverse(ans.begin(), ans.end());
    return ans;
}

signed main()
{
    int a, b;
    cin >> a >> b;
    if (a + b < 0)
        cout << "-";
    cout << solve(to_string(abs(a + b)));
}
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1002 A+B for Polynomials (25 分)

题意：多项式相加，共两行输入，每一行第一个数字是项数，之后每两个数字a和b构成一个a^b,然后要求输出两个多项式相加的答案

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;

struct cmp
{
    bool operator()(const int &x, const int &y)
    {
        return x > y;
    }
};
map<int, double, cmp> mp;

int n, x, Ans = 0;
double y;
void solve(int n)
{
    while (n--)
    {
        cin >> x >> y;
        mp[x] += y;
    }
}
signed main()
{
    cin >> n;
    solve(n);
    cin >> n;
    solve(n);
    for (auto t : mp)
        if (t.second)
            Ans++;
    cout << Ans;
    for (auto t : mp)
        if (t.second)
            printf(" %d %.1f", t.first, t.second);
}
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1003 Emergency (25 分)

题意：给定n，m，s，t分别为总点数，总边数，起点，终点，然后一行输入每一个城市的人数，之后m行输入每一条边，询问时s->t的最短路径的数量和在所有最短路径中可以召集的最多人数

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> PII;
const int N = 2e5 + 10, M = N * 2;
int dist[N], vis[N], peo[N], sum[N];
int ne[M], e[M], h[N], w[N], idx;
void add(int u, int v, int c)
{
    e[idx] = v;
    ne[idx] = h[u];
    w[idx] = c;
    h[u] = idx++;
}
int perple[N];
void dij(int s, int t)
{
    memset(dist, 0x3f, sizeof dist);
    dist[s] = 0;
    peo[s] = perple[s];
    sum[s] = 1;
    priority_queue<PII, vector<PII>, greater<PII>> que;
    que.push({dist[s], s});
    while (que.size())
    {
        PII q = que.top();
        que.pop();
        int ver = q.second, distance = q.first;
        if (vis[ver])
            continue;
        vis[ver] = 1;

        for (int i = h[ver]; i != -1; i = ne[i])
        {
            int j = e[i];

            if (dist[j] > dist[ver] + w[i])
            {
                dist[j] = dist[ver] + w[i];
                peo[j] = perple[j] + peo[ver];
                que.push(make_pair(dist[j], j));
                sum[j] = sum[ver];
            }
            else if ((dist[j] == dist[ver] + w[i]))
            {
                sum[j] += sum[ver];
                if (peo[j] < peo[ver] + perple[j])
                {
                    peo[j] = peo[ver] + perple[j];
                    que.push(make_pair(dist[j], j));
                }
            }
        }
    }
}

signed main()
{
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    for (int i = 0; i < n; i++)
    {
        h[i] = -1;
        cin >> perple[i];
    }
    while (m--)
    {
        int u, v, w;
        cin >> u >> v >> w;
        add(u, v, w);
        add(v, u, w);
    }
    dij(s, t);
    cout << sum[t] << " " << peo[t];
}
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1004 Counting Leaves (30 分)

题意：给定一棵树，输入n，m分别为树的总点数和非叶子节点数，之后m行输入第一个数是父亲节点编号，第二个数cnt为儿子节点，然后cnt个结点为儿子节点的编号，询问每一个深度的叶子节点数量

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, M = N;

int h[N], e[M], ne[M], idx;
void add(int u, int v)
{
    e[idx] = v;
    ne[idx] = h[u];
    h[u] = idx++;
}

int va[N], Max_depth;

void dfs(int u, int depth)
{
    Max_depth = max(Max_depth, depth);
    if (h[u] == -1)
    {
        va[depth]++;
        return;
    }
    for (int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        dfs(j, depth + 1);
    }
}

signed main()
{
    int n, m;
    cin >> n >> m;
    memset(h, -1, sizeof h);
    for (int i = 0; i < m; i++)
    {
        int u, v, num;
        cin >> u >> num;
        while (num--)
        {
            cin >> v;
            add(u, v);
        }
    }

    dfs(1, 0);
    cout << va[0];
    for (int i = 1; i <= Max_depth; i++)
        cout << " " << va[i];
}
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1005 Spell It Right (20 分)

题意：输入一个整数a<10¹⁰⁰,询问每一位的数字相加的和，用英文输出

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;

map<int, string> mp;
string s[N] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "zero"};
signed main()
{
    for (int i = 1; i <= 10; i++)
        mp[i % 10] = s[i - 1];
    string s;
    cin >> s;
    int ans = 0;
    for (int i = 0; i < s.size(); i++)
        ans += (s[i] - '0');

    s = to_string(ans);
    cout<<mp[s[0] - '0'];
    for (int i = 1; i < s.size(); i++)
        cout <<" "<< mp[s[i] - '0'];
}
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1006 Sign In and Sign Out (25 分)

题意：给定n行字符串，每一行分别是人名，到达时间，离开时间，询问最早到达和最晚离开的人名

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10;

string a, b;
string AA = "25", BB = "00:00:00";
signed main()
{
    int n;
    cin >> n;
    while (n--)
    {
        string name, time_arrive, time_leave;
        cin >> name >> time_arrive >> time_leave;
        if (AA > time_arrive)
        {
            AA = time_arrive;
            a = name;
        }
        if (BB < time_leave)
        {
            BB = time_leave;
            b = name;
        }
    }
    cout << a << " " << b;
}
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1007 Maximum Subsequence Sum (25 分)

题意：输入n个数，询问最大连续子序列的和，该序列的第一个数和最后一个数

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;
const int N = 1e4 + 10;

int a[N], dp[N], pre[N];
signed main()
{
    int n, f = 0;
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        cin >> a[i];
        if (a[i] >= 0)
            f = 1;
    }
    if (!f)
        return cout << 0 << " " << a[1] << " " << a[n], 0;

    for (int i = 1; i <= n; i++)
    {
        if (a[i] < dp[i - 1] + a[i])
        {
            dp[i] = dp[i - 1] + a[i];
            pre[i] = pre[i - 1];
        }
        else
        {
            dp[i] = a[i];
            pre[i] = i;
        }
    }
    int M = -1;
    for (int i = 1; i <= n; i++)
    {
        if (M < dp[i])
        {
            f = i;
            M = dp[i];
        }
    }
    cout << dp[f] << " " << a[pre[f]] << " " << a[f];
}
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1008 Elevator (20 分)

题意：给定一个整数n，每上一步需要6s 下一步需要4s 每一次停止再启动需要5s的时间，起始点是0，询问总共多少s

/*made in zzh*/
#include <bits/stdc++.h>
using namespace std;
signed main()
{
    int n, x, now, ans = 0;
    cin >> n;
    while (n--)
    {
        int x;
        cin >> x;
        if (x > now)
            ans += (x - now) * 6 + 5;
        else
            ans += (now - x) * 4 + 5;
        now = x;
    }
    cout << ans << endl;
}
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1009 Product of Polynomials 分数 25

题意：给定两个多项式，首先是整数n，后面n对（e，c）分别表示多项式的指数和系数。求解两个多项式相乘结果

#include <bits/stdc++.h>
using namespace std;
vector<double> a(1010, 0), b(1010, 0), c(2010, 0);
vector<pair<int, double>> ans;
signed main()
{
    int n, m;
    cin >> n;
    while (n--)
    {
        int e;
        double c;
        cin >> e >> c;
        a[e] = c;
    }
    cin >> m;
    while (m--)
    {
        int e;
        double c;
        cin >> e >> c;
        b[e] = c;
    }
    for (int i = 0; i <= 1000; i++)
        for (int j = 0; j <= 1000; j++)
            c[i + j] += (a[i] * b[j]);

    int sum = 0;
    for (int i = 0; i <= 2010; i++)
        if (c[i] != 0)
        {
            sum++;
            ans.push_back(make_pair(i, c[i]));
        }

    reverse(ans.begin(), ans.end());
    cout << sum;
    for (auto t : ans)
        printf(" %d %.1f", t.first, t.second);
}
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1010 Radix 分数 25

题意：给定两个字符串，如果op为1，最后一个数就是第一个字符串的进制数，否则是第二个进制数，问是否存在最小的可以使s1 == s2的进制数

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

#define int long long
const int INF = 1e18;

int solve(string s, int j)
{
    int sum = 0;
    int len = s.size();
    for (int i = 0; i < len; i++)
    {
        if (sum >= INF / j)
            return INF;
        if (s[i] <= '9' && s[i] >= '0')
            sum = sum * j + s[i] - '0';
        else
            sum = sum * j + s[i] - 'a' + 10;
    }
    if (sum < 0)
        return INF;
    return sum;
}

int wor(string s)
{
    int ans = 0;
    int len = s.size();
    for (int i = 0; i < len; i++)
    {
        if (s[i] <= '9' && s[i] >= '0')
            ans = max(ans, (int)s[i] - '0');
        else
            ans = max(ans, (int)s[i] - 'a' + 10);
    }
    return ans + 1;
}

signed main()
{
    int b = 0;
    string s1, s2;
    int op, tag;
    cin >> s1 >> s2 >> op >> tag;
    if (op == 2)
        swap(s1, s2);

    int a = solve(s1, tag);

    int l = wor(s2);
    int r = max(l, a + 1);
    while (l < r)
    {
        int mid = (l + r) / 2;
        if (solve(s2, mid) >= a)
            r = mid;
        else
            l = mid + 1;
    }
    if (solve(s2, l) == a)
        cout << l << endl;
    else
        cout << "Impossible\n";
}
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1011 World Cup Betting 分数 20

题意：公式(max×65%−1)×2，输入就是三行，找每一行最大值，*起来带入max输出就行，数字顺序是W T L

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

const int N = 2e3 + 10;

int cnt[5];
signed main()
{
    map<int, string> mp;
    mp[0] = "W";
    mp[1] = "T";
    mp[2] = "L";
    double ans = 1;
    for (int i = 0; i < 3; i++)
    {
        double c = -10000;
        for (int j = 0; j < 3; j++)
        {
            double x;
            cin >> x;
            c = max(c, x);
            if (c == x)
                cnt[i] = j;
        }
        ans *= c;
    }
    for (int i = 0; i < 3; i++)
        cout << mp[cnt[i]] << " ";
    printf("%.2f", (ans * 0.65 - 1) * 2);
}
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1012 The Best Rank 分数 25

题意：给定n行ID名，C，M，E。A定义为C，M和E的平均值，排序的优先级为A>C>M>E，然后查找对应ID的最高排名，有并列特判。

#include <bits/stdc++.h>
using namespace std;
const int N = 2010;
struct node
{
    int id;
    double score[5];
    int ans = 10;
    string S;
} a[N];
map<int, string> mp = {
    {0, "A"},
    {1, "C"},
    {2, "M"},
    {3, "E"}};
int now = 0;

bool cmp(node a, node b)
{
    return a.score[now] > b.score[now];
}

signed main()
{
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].id >> a[i].score[1] >> a[i].score[2] >> a[i].score[3];
        a[i].score[0] = (a[i].score[1] + a[i].score[2] + a[i].score[3]) / 3.0;
    }

    for (; now < 4; now++)
    {
        sort(a, a + n, cmp);
        for (int i = 0; i < n; i++)
        {
            // 可以直接将前面的继承过来
            int rank = i;
            while (rank--)
                if (a[i].score[now] != a[rank].score[now])
                    break;

            if (rank + 2 < a[i].ans)
            {
                a[i].ans = rank + 2;
                a[i].S = mp[now];
            }
        }
    }
    while (m--)
    {
        int id, flag = 0;
        cin >> id;
        for (int i = 0; i < n; i++)
            if (a[i].id == id)
            {
                cout << a[i].ans << " " << a[i].S << endl;
                flag = 1;
                break;
            }

        if (!flag)
            cout << "N/A\n";
    }
}
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1013 Battle Over Cities 分数 25

题意：给定n个点，m条路，k个询问，询问是删除编号id的点，为了使图联通需要新增几条路

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

const int N = 1e3 + 10, M = 2 * N * N;

int h[N], e[M], ne[M], idx;
int st[N];

void add(int u, int v)
{
    e[idx] = v;
    ne[idx] = h[u];
    h[u] = idx++;
}

void dfs(int u)
{
    for (int i = h[u]; i != -1; i = ne[i])
    {
        int j = e[i];
        if (st[j] == -1 || st[j] == 1)
            continue;
        st[j] = 1;
        dfs(j);
    }
}

signed main()
{
    int n, m, k;
    cin >> n >> m >> k;
    for (int i = 1; i <= n; i++)
        h[i] = -1;
    for (int i = 0; i < m; i++)
    {
        int u, v;
        cin >> u >> v;
        add(u, v);
        add(v, u);
    }
    while (k--)
    {
        int ans = 0;
        memset(st, 0, sizeof st);
        int x;
        cin >> x;
        st[x] = -1;
        for (int i = 1; i <= n; i++)
        {
            if (i == x)
                continue;
            if (st[i] == 0)
            {
                dfs(i);
                ans++;
            }
        }
        cout << ans - 1 << "\n";
    }
}
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1014 Waiting in Line 分数 30

题意：n个窗口，每一个窗口可以排队m人，总共k个人，每一个人需求时是正整数，然后q次询问。排队规则是队伍人数一样长的时候，排编号最小的队伍，否则去队伍人数最少的队伍排队。然后对于17：00之后的请求不再处理，只处理[8：00，17：00）之间的请求，而且如果处理完成时间大于17：00，也要给出一个目标时间（一定合法，不会超级大）。

1015 Reversible Primes 分数 20

题意：给定10进制数字n,d，判断10进制n和反转后的d进制的n是否为素数。

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

int solve(string a, int b) // 将b进制的a 转化成10进制的now
{
    int now = 0;
    for (int i = 0; i < (int)a.size(); i++)
        now = now * b + a[i] - '0';
    return now;
}

string solve_tran(int a, int b) // 将10进制的a 转化成b进制的now
{
    string now = "";
    while (a)
    {
        now += to_string(a % b);
        a /= b;
    }
    return now;
}

int isprime(int a)
{
    int m = sqrt(a);
    for (int i = 2; i <= m; i++)
        if (a % i == 0)
            return 0;
    return !(a == 1);
}

signed main()
{
    string n;
    int d;
    while (cin >> n >> d)
    {
        if (n[0] == '-')
            break;
        string c = solve_tran(atoll(n.c_str()), d);
        int p = solve(c, d);
        if (isprime(atoll(n.c_str())) && isprime(p))
            cout << "Yes\n";
        else
            cout << "No\n";
    }
}
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1016 Phone Bills 分数 25

题意：给定24小时的价格，单位是美分/分钟，然后给定n个任务（无序），首先是任务的受理人，之后是任务的发生时间，最后是任务的类型，on-line是开始，off-line是结束，同时会存在非法任务，对于多次任务的开始，只记录最后一次，对于多次任务的结束，只记录最初一次。问每一个受理人的使用时间和价钱，价钱为0的受理人不输出。

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 5e4 + 10;
int money[30], SUM[N], n, month;
map<string, vector<string>> mp;

int solve(string a, string b)
{
    int d = atoll(a.substr(0, 2).c_str()), d1 = atoll(b.substr(0, 2).c_str());
    int h = atoll(a.substr(0 + 3, 2).c_str()), h1 = atoll(b.substr(0 + 3, 2).c_str());
    int m = atoll(a.substr(0 + 6, 2).c_str()), m1 = atoll(b.substr(0 + 6, 2).c_str());
    return ((d - d1) * 60 * 24 + (h - h1) * 60 + (m - m1));
}
double work(string a, string b)
{
    int d = atoll(a.substr(0, 2).c_str()), d1 = atoll(b.substr(0, 2).c_str());
    int h = atoll(a.substr(0 + 3, 2).c_str()), h1 = atoll(b.substr(0 + 3, 2).c_str());
    int m = atoll(a.substr(0 + 6, 2).c_str()), m1 = atoll(b.substr(0 + 6, 2).c_str());
    return (double)(SUM[(d)*60 * 24 + (h)*60 + m] - SUM[(d1)*60 * 24 + (h1)*60 + m1]);
}

struct node
{
    string name, val, op;
} a[N];
bool cmp(node a, node b)
{
    return a.val < b.val;
}

signed main()
{
    for (int i = 1; i <= 24; i++)
        cin >> money[i];
    for (int i = 1; i <= 31; i++) // 利用前缀和进行预处理
        for (int j = 1; j <= 24; j++)
            for (int k = 1; k <= 60; k++)
                SUM[(i - 1) * 60 * 24 + (j - 1) * 60 + k] = SUM[(i - 1) * 60 * 24 + (j - 1) * 60 + k - 1] + money[j];
    cin >> n;
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].name;
        scanf("%lld:", &month);
        cin >> a[i].val >> a[i].op;
    }
    sort(a, a + n, cmp);

    for (int i = 0; i < n; i++)
        if (a[i].op == "on-line")
        {
            if (mp[a[i].name].size() % 2)
                mp[a[i].name].back() = a[i].val;
            else
                mp[a[i].name].push_back(a[i].val);
        }
        else if (a[i].op == "off-line" && (mp[a[i].name].size() % 2))
            mp[a[i].name].push_back(a[i].val);
    for (auto t : mp)
    {
        vector<string> now = t.second;
        double sum = 0;
        for (int i = 0; i + 1 < (int)now.size(); i += 2)
            sum += work(now[i + 1], now[i]);
        if (sum > 0)
        {
            cout << t.first << " ";
            if (month < 10)
                cout << "0";
            cout << month << endl;
            for (int i = 0; i + 1 < (int)now.size(); i += 2)
            {
                cout << now[i] << " " << now[i + 1] << " " << solve(now[i + 1], now[i]) << " ";
                printf("$%.2f\n", work(now[i + 1], now[i]) / 100.00);
            }
            printf("Total amount: $%.2f\n", sum / 100.00);
        }
    }
}
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1017 Queueing at Bank 分数 25

题意：给定n个任务，每一行给出开始时间，持续时间，仅有k个机器可以运行任务，询问平均等待时间。机器工作时间是8点到17点，17点之后到达的任务不再运行。

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e4 + 10;

int sub(string a, string b)
{
    int d = atoll(a.substr(0, 2).c_str()), d1 = atoll(b.substr(0, 2).c_str());
    int h = atoll(a.substr(0 + 3, 2).c_str()), h1 = atoll(b.substr(0 + 3, 2).c_str());
    int m = atoll(a.substr(0 + 6, 2).c_str()), m1 = atoll(b.substr(0 + 6, 2).c_str());
    return ((d - d1) * 60 * 60 + (h - h1) * 60 + (m - m1));
}
string To_string(int s)
{
    if (s < 10ll)
        return ("0" + to_string(s));
    return to_string(s);
}
string solve(string a, int b)
{
    int h = atoll(a.substr(0, 2).c_str());
    int m = atoll(a.substr(0 + 3, 2).c_str());
    int s = atoll(a.substr(0 + 6, 2).c_str());
    h += ((m + b) / 60);
    m = (m + b) % 60;
    return (To_string(h) + ":" + To_string(m) + ":" + To_string(s));
}

struct node
{
    string t;
    int p;
} a[N];
bool cmp(node a, node b)
{
    return a.t < b.t;
}

signed main()
{
    int n, k, ans = 0;
    cin >> n >> k;
    for (int i = 0; i < n; i++)
        cin >> a[i].t >> a[i].p;
    sort(a, a + n, cmp);

    priority_queue<string, vector<string>, greater<string>> que;
    for (int i = 0; i < n; i++)
    {
        if (a[i].t < "08:00:00")
        {
            ans += sub("08:00:00", a[i].t);
            a[i].t = "08:00:00";
        }
        if (a[i].t >= "17:00:01")
        {
            n = i;
            break;
        }
        if (que.size() < k)
            que.push(solve(a[i].t, a[i].p));
        else
        {
            string now = que.top();
            que.pop();
            if (now > a[i].t)
                ans += (sub(now, a[i].t));
            que.push(solve(max(now, a[i].t), a[i].p));
        }
    }
    printf("%.1f\n", round(ans / 6.0 / n) / 10.0);
}
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PAT (Advanced Level) Practice 甲级 顶级题目_pta题库甲级

文章目录

1001 A+B Format (20 分)

1002 A+B for Polynomials (25 分)

1003 Emergency (25 分)

1004 Counting Leaves (30 分)

1005 Spell It Right (20 分)

1006 Sign In and Sign Out (25 分)

1007 Maximum Subsequence Sum (25 分)

1008 Elevator (20 分)

1009 Product of Polynomials 分数 25

1010 Radix 分数 25

1011 World Cup Betting 分数 20

1012 The Best Rank 分数 25

1013 Battle Over Cities 分数 25

1014 Waiting in Line 分数 30

1015 Reversible Primes 分数 20

1016 Phone Bills 分数 25

1017 Queueing at Bank 分数 25

PAT (Advanced Level) Practice 甲级顶级题目_pta题库甲级