PAT (Advanced Level) Practice 1001-1005

目录

• PAT (Advanced Level) Practice 1001-1005
  • PAT 计算机程序设计能力考试 甲级 练习题
  • 题库：PTA拼题A官网
  • 背景
  • 题目目录
  • 总结
  • 题目详解
    • 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)
  • 后记

PAT (Advanced Level) Practice 1001-1005

PAT 计算机程序设计能力考试 甲级 练习题

题库：PTA拼题A官网

背景

这是浙大背景的一个计算机考试

刷刷题练练手

在博客更新题解 每五题更新一次 共155题

题目目录

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)

总结

1001 签到题（格式）

1002 签到题（格式）

1003 多条最短路（Djikstra） 路径最大点权和

1004 DFS 统计树每层叶子节点个数

1005 签到题（格式）

题目详解

1001 A+B Format (20分)

Calculate a+b and output the sum in standard format -- that is, the digits must be separated into groups of three by commas (unless there are less than four digits).

Input Specification:

Each input file contains one test case. Each case contains a pair of integers a and b where −10^6 ≤ a, b ≤ 10^6. The numbers are separated by a space.

Output Specification:

For each test case, you should output the sum of a and b in one line. The sum must be written in the standard format.

Sample Input:

-1000000 9

Sample Output:

-999,991

这种题还用写么

需要注意的地方就是输出格式了

不过这种标准格式可以直接使用Java自带的Format库

AC代码：

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.text.NumberFormat;
import java.util.Scanner;

public class PAT1001 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

        int a = scanner.nextInt();
        int b = scanner.nextInt();
        int sum = a + b;

        NumberFormat format = NumberFormat.getInstance();

        String result = format.format(sum);
        System.out.println(result);
    }
}

1002 A+B for Polynomials (25分)

This time, you are supposed to find A+B where A and B are two polynomials.

Input

Each input file contains one test case. Each case occupies 2 lines, and each line contains the information of a polynomial: K N1 aN1 N2 aN2 … NK aNK, where K is the number of nonzero terms in the polynomial, Ni and aNi (i=1, 2, …, K) are the exponents and coefficients, respectively. It is given that 1 <= K <= 10，0 <= NK < … < N2 < N1 <=1000.

Output

For each test case you should output the sum of A and B in one line, with the same format as the input. Notice that there must be NO extra space at the end of each line. Please be accurate to 1 decimal place.

Sample Input

2 1 2.4 0 3.2
2 2 1.5 1 0.5

Sample Output

3 2 1.5 1 2.9 0 3.2

这题被坑了一会

因为答案输出要求始终精确到1位小数

即像1或者0都要输出成1.0和0.0这样

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.text.DecimalFormat;
import java.util.Scanner;

public class PAT1002 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

        double[] pol = new double[1001];

        for (int i = 0; i < 2; i++) {
            int k = scanner.nextInt();
            for (int j = 0; j < k; j++) {
                int n = scanner.nextInt();
                double a = scanner.nextDouble();
                pol[n] += a;
            }
        }

        DecimalFormat format = new DecimalFormat("0.0");

        int count = 0;

        StringBuilder result = new StringBuilder();

        for (int i = pol.length - 1; i >= 0; i--) {
            if (pol[i] == 0) {
                continue;
            }
            count++;
            result.append(" ").append(i).append(" ").append(format.format(pol[i]));
        }

        result.insert(0, count);

        System.out.println(result.toString());
    }
}

1003 Emergency (25分)

As an emergency rescue team leader of a city, you are given a special map of your country. The map shows several scattered cities connected by some roads. Amount of rescue teams in each city and the length of each road between any pair of cities are marked on the map. When there is an emergency call to you from some other city, your job is to lead your men to the place as quickly as possible, and at the mean time, call up as many hands on the way as possible.

Input

Each input file contains one test case. For each test case, the first line contains 4 positive integers: N (<= 500) – the number of cities (and the cities are numbered from 0 to N-1), M – the number of roads, C1 and C2 – the cities that you are currently in and that you must save, respectively. The next line contains N integers, where the i-th integer is the number of rescue teams in the i-th city. Then M lines follow, each describes a road with three integers c1, c2 and L, which are the pair of cities connected by a road and the length of that road, respectively. It is guaranteed that there exists at least one path from C1 to C2.

Output

For each test case, print in one line two numbers: the number of different shortest paths between C1 and C2, and the maximum amount of rescue teams you can possibly gather.
All the numbers in a line must be separated by exactly one space, and there is no extra space allowed at the end of a line.

Sample Input

5 6 0 2
1 2 1 5 3
0 1 1
0 2 2
0 3 1
1 2 1
2 4 1
3 4 1

Sample Output

2 4

总算来了一道不是签到题的题目

很容易看出来这是一个节点和边都带权的无向图

使用狄杰斯特拉（Djikstra）来求解最短路

话说每次用到这个算法就想到我同学用的狄杰特斯拉[手动狗头]

需要分别保存

road_count[i] 到达节点i当前共有多少条相同边权的路

people[i] 到达节点i后可聚集的最大救援队人数

dis[i] 到达节点i的最短路长度

pre[i] 访问节点i的前置节点（从哪个节点到的节点i）

vis[i] 节点i是否被访问过

这题使用邻接矩阵存储了图结构

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;

public class PAT1003 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

        int N = scanner.nextInt();
        int M = scanner.nextInt();
        int start = scanner.nextInt();
        int end = scanner.nextInt();

        int[] vertex_team = new int[N];
        for (int i = 0; i < N; i++) {
            vertex_team[i] = scanner.nextInt();
        }

        int[][] map = new int[N][N];
        for (int i = 0; i < N; i++) {
            Arrays.fill(map[i], Integer.MAX_VALUE);
        }
        for (int i = 0; i < M; i++) {
            int from = scanner.nextInt();
            int to = scanner.nextInt();
            int dis = scanner.nextInt();
            map[from][to] = map[to][from] = dis;
        }

        int[] road_count = new int[N];
        road_count[start] = 1;

        int[] people = new int[N];
        people[start] = vertex_team[start];

        int[] dis = new int[N];
        Arrays.fill(dis, Integer.MAX_VALUE);
        dis[start] = 0;

        int[] pre = new int[N];
        Arrays.fill(pre, -1);
        pre[start] = start;

        boolean[] vis = new boolean[N];

        for (int i = 0; i < N; i++) {
            int now = -1, min = Integer.MAX_VALUE;
            for (int j = 0; j < N; j++) {
                if (!vis[j] && dis[j] < min) {
                    now = j;
                    min = dis[j];
                }
            }
            if (now == -1) {
                break;
            }
            vis[now] = true;
            for (int j = 0; j < N; j++) {
                if (!vis[j] && map[now][j] != Integer.MAX_VALUE) {
                    if (dis[now] + map[now][j] < dis[j]) {
                        dis[j] = dis[now] + map[now][j];
                        pre[j] = pre[now];
                        people[j] = people[now] + vertex_team[j];
                        road_count[j] = road_count[now];
                    } else if (dis[now] + map[now][j] == dis[j]) {
                        road_count[j] += road_count[now];
                        if (people[now] + vertex_team[j] > people[j]) {
                            people[j] = people[now] + vertex_team[j];
                        }
                    }
                }
            }
        }
        System.out.printf("%d %d", road_count[end], people[end]);
    }
}

1004 Counting Leaves (30分)

A family hierarchy is usually presented by a pedigree tree. Your job is to count those family members who have no child.

Input

Each input file contains one test case. Each case starts with a line containing 0 < N < 100, the number of nodes in a tree, and M (< N), the number of non-leaf nodes. Then M lines follow, each in the format:

ID K ID[1] ID[2] … ID[K]
where ID is a two-digit number representing a given non-leaf node, K is the number of its children, followed by a sequence of two-digit ID’s of its children. For the sake of simplicity, let us fix the root ID to be 01.

Output

For each test case, you are supposed to count those family members who have no child for every seniority level starting from the root. The numbers must be printed in a line, separated by a space, and there must be no extra space at the end of each line.

The sample case represents a tree with only 2 nodes, where 01 is the root and 02 is its only child. Hence on the root 01 level, there is 0 leaf node; and on the next level, there is 1 leaf node. Then we should output “0 1” in a line.

Sample Input

2 1
01 1 02

Sample Output

0 1

这题是树的遍历

看题意很容易想到层次遍历，却发现根据输入不太好构建这棵树

一开始是想这样做的：边读取边建立这一棵二叉树，然后层序遍历。发现题目里没有给明节点给出的顺序，也就无法合理的建树，建森林再连接的话只会增加此题难度。

于是改变思路：深搜一下这棵树

利用ArrayList[]保存所有输入结果，一开始还被题目的two-digit number给误导了，其实直接照int读入就行，读取全部输入之后对应的ArrayList[i]就代表了一个节点，如果其中没有孩子，就记录为一个叶子节点，对答案做一次更新即可。

其实题目给的count的取值只会是1和2，因为非叶子节点至多两个孩子，还可以据此优化。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;

public class PAT1004 {
    static ArrayList<Integer>[] tree = new ArrayList[128];
    static int[] ans = new int[128];
    static int max_depth = -1;

    static void dfs(int index, int depth) {
        if (tree[index].size() == 0) {
            ans[depth]++;
            max_depth = Math.max(max_depth, depth);
        }
        for (int child : tree[index]) {
            dfs(child, depth + 1);
        }
    }

    public static void main(String[] args) {
        for (int i = 0; i < tree.length; i++) {
            tree[i] = new ArrayList<>();
        }

        Scanner scanner = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

        int N = scanner.nextInt();
        if (N == 0) {
            System.out.println(N);
            System.exit(0);
        }

        int M = scanner.nextInt();
        for (int i = 0; i < M; i++) {
            int root = scanner.nextInt();
            int count = scanner.nextInt();
            for (int j = 0; j < count; j++) {
                int child = scanner.nextInt();
                tree[root].add(child);
            }
        }

        dfs(1, 0);

        System.out.print(ans[0]);
        for (int i = 1; i <= max_depth; i++) {
            System.out.print(" " + ans[i]);
        }
    }
}

1005 Spell It Right (20分)

Given a non-negative integer N, your task is to compute the sum of all the digits of N, and output every digit of the sum in English.

Input Specification:

Each input file contains one test case. Each case occupies one line which contains an N (<= 10^100).

Output Specification:

For each test case, output in one line the digits of the sum in English words. There must be one space between two consecutive words, but no extra space at the end of a line.

Sample Input:

12345

Sample Output:

one five

肉眼可见的签到题，一开始看到10的100次方，手打出BigInteger之后看了看题又删掉了。

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public class PAT1005 {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(new BufferedReader(new InputStreamReader(System.in)));

        String input = scanner.nextLine();

        int sum = 0;
        int len = input.length();
        for (int i = 0; i < len; i++) {
            sum += input.charAt(i) - '0';
        }

        String[] words = {"zero", "one", "two", "three", "four",
                "five", "six", "seven", "eight", "nine"};

        String output = Integer.toString(sum);

        len = output.length();
        System.out.print(words[output.charAt(0) - '0']);
        for (int i = 1; i < len; i++) {
            System.out.print(" " + words[output.charAt(i) - '0']);
        }
    }
}

后记

每天刷题频率不固定 故此系列博客也有咕咕咕的可能