PAT (Advanced Level) Practice 1001-1005
PAT computer programming exercises Grade Proficiency Test
Questions and Answers: PTA official website A title fight
background
This is a computer exam Zhejiang University background
Brush the question practice your hand
Update solution to a problem in the blog title is updated once every five total 155 questions
Directory title
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)
to sum up
1001 attendance title (format)
1002 attendance title (format)
over 1003 pieces shortest (Djikstra) maximum point right path and
1004 DFS count the number of each tree leaf nodes
1005 to sign the title (format)
Detailed topics
1001 A + B Format (20 minutes)
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
This problem also used to write it
Needs attention is the output format
However, this standard format can be used directly Java library that comes with Format
AC Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.text.NumberFormat;
import java.util.Scanner;
public class PTA1001 {
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 3.2This question is a pit for a while
Because the answer is always accurate output requirements to a decimal
I.e. like 1or 0be exported into 1.0and 0.0so
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.text.DecimalFormat;
import java.util.Scanner;
public class PTA1002 {
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 4Finally came a sign question is not the subject of
It is easy to see that this is not a nodes and edges are weighted directed graph
Use 狄杰斯特拉（Djikstra）to solve the shortest
words every time I think of the algorithm used by the students of狄杰特斯拉[Manual Kobold]
You need to save each
road_count [i] reaches the same edge weight how many nodes i There are currently road
can gather after people [i] reaches the node i largest rescue team number
dis [i] reaches the node i of the shortest length of the
pre [i] access pre-node (which node to which node from i) for node i
VIS [i] whether the accessed node i
This title is stored using the adjacency matrix structure of FIG.
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Arrays;
import java.util.Scanner;
public class PTA1003 {
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 1This problem is traversing the tree
Look meaning of the questions is easy to think traverse the level, but found that this tree is not very good build according to the input
The beginning is to do so: the establishment of this side edge reading a binary tree, then the sequence traversal. There is no order to find out the title given node, it can not reasonably achievements, built in the forest and then connect, then this problem will only increase the difficulty.
So changing ideas: deep search about this tree
Using the ArrayList [] saves all input result, a further subject began two-digit number to mislead, in fact, as a direct line int read, after reading all of the corresponding input ArrayList [i] represents a node, wherein if no children, it was recorded as a leaf node, an update can be made to answer.
In fact, the title of count values to only be 1 and 2, because the non-leaf nodes at most two children, also can be optimized accordingly.
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Scanner;
public class PTA1004 {
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 fiveVisible sign title, a start to see 10 to the 100th, hand and looked at the title and deleted after play BigInteger.
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;
public class PTA1005 {
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']);
}
}
}
postscript
Brush title frequency is not fixed daily series therefore also possible Gugu Gu's blog