# PAT (Advanced Level) Practice 1001-1005

## 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



AC代码：

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

public class PAT1001 {
public static void main(String[] args) {

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


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

public class PAT1002 {
public static void main(String[] args) {

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


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

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

public class PAT1003 {
public static void main(String[] args) {

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[] 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];
} else if (dis[now] + map[now][j] == dis[j]) {
if (people[now] + vertex_team[j] > people[j]) {
people[j] = people[now] + vertex_team[j];
}
}
}
}
}
}
}



### 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


import java.io.BufferedReader;
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<>();
}

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();
}
}

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


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

public class PAT1005 {
public static void main(String[] args) {

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

posted on 2020-02-09 21:39  Licsber  阅读(248)  评论(0编辑  收藏

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