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[Java] 1031. Hello World for U (20)-PAT甲级

1031. Hello World for U (20)
Given any string of N (>=5) characters, you are asked to form the characters into the shape of U. For example, “helloworld” can be printed as:
h d
e l
l r
lowo
That is, the characters must be printed in the original order, starting top-down from the left vertical line with n1 characters, then left to right along the bottom line with n2 characters, and finally bottom-up along the vertical line with n3 characters. And more, we would like U to be as squared as possible — that is, it must be satisfied that n1 = n3 = max { k| k <= n2 for all 3 <= n2 <= N } with n1 + n2 + n3 – 2 = N.

Input Specification:
Each input file contains one test case. Each case contains one string with no less than 5 and no more than 80 characters in a line. The string contains no white space.

Output Specification:
For each test case, print the input string in the shape of U as specified in the description.

Sample Input:
helloworld!

Sample Output:
h !
e d
l l
lowor

题目大意：用所给字符串按U型输出。n1和n3是左右两条竖线从上到下的字符个数，n2是底部横线从左到右的字符个数。
要求：
1. n1 == n3
2. n2 >= n1
3. n1为在满足上述条件的情况下的最大值

PS：感谢github用户@fs19910227提供的pull request～

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        final char[] chars = reader.readLine().toCharArray();
        final int length = chars.length + 2;
        final int rows = length / 3;
        final int columns = length / 3 + length % 3;
        final char[][] table = new char[rows][columns];
        int index = 0;
        //left
        for (int i = 0; i < rows; i++, index++) {
            table[i][0] = chars[index];
        }
        //bottom
        for (int i = 1; i < columns - 1; i++, index++) {
            table[rows - 1][i] = chars[index];
        }
        //right
        for (int i = rows - 1; i >= 0; i--, index++) {
            table[i][columns - 1] = chars[index];
        }
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < columns; j++) {
                char c = table[i][j] == 0 ? ' ' : table[i][j];
                System.out.printf("%c", c);
                if (j == columns - 1) {
                    System.out.println();
                }
            }
        }
    }
}

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

public class Main {

public static void main(String[] args) throws IOException {

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

final char[] chars = reader.readLine().toCharArray();

final int length = chars.length + 2;

final int rows = length / 3;

final int columns = length / 3 + length % 3;

final char[][] table = new char[rows][columns];

int index = 0;

//left

for (int i = 0; i < rows; i++, index++) {

table[i][0] = chars[index];

}

//bottom

for (int i = 1; i < columns - 1; i++, index++) {

table[rows - 1][i] = chars[index];

}

//right

for (int i = rows - 1; i >= 0; i--, index++) {

table[i][columns - 1] = chars[index];

}

for (int i = 0; i < rows; i++) {

for (int j = 0; j < columns; j++) {

char c = table[i][j] == 0 ? ' ' : table[i][j];

System.out.printf("%c", c);

if (j == columns - 1) {

System.out.println();

}

[Java] 1027. Colors in Mars (20)-PAT甲级

1027. Colors in Mars (20)
People in Mars represent the colors in their computers in a similar way as the Earth people. That is, a color is represented by a 6-digit number, where the first 2 digits are for Red, the middle 2 digits for Green, and the last 2 digits for Blue. The only difference is that they use radix 13 (0-9 and A-C) instead of 16. Now given a color in three decimal numbers (each between 0 and 168), you are supposed to output their Mars RGB values.

Input

Each input file contains one test case which occupies a line containing the three decimal color values.

Output

For each test case you should output the Mars RGB value in the following format: first output “#, then followed by a 6-digit number where all the English characters must be upper-cased. If a single color is only 1-digit long, you must print a “0” to the left.

Sample Input
15 43 71
Sample Output
#123456

题目大意：给三个十进制的数，把它们转换为十三进制的数输出。要求在前面加上一个”#”号
分析：因为0~168的十进制转换为13进制不会超过两位数，所以这个两位数为(num / 13)(num % 13)构成的数字

PS：感谢github用户@fs19910227提供的pull request～

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
    private static char[] signTables = "0123456789ABC".toCharArray();

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] split = reader.readLine().split(" ");
        System.out.print("#");
        for (String aSplit : split) {
            Integer integer = Integer.valueOf(aSplit);
            System.out.printf("%c%c", signTables[integer / 13], signTables[integer % 13]);
        }
        System.out.println();
    }

}

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

public class Main {

private static char[] signTables = "0123456789ABC".toCharArray();

public static void main(String[] args) throws IOException {

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

String[] split = reader.readLine().split(" ");

System.out.print("#");

for (String aSplit : split) {

Integer integer = Integer.valueOf(aSplit);

System.out.printf("%c%c", signTables[integer / 13], signTables[integer % 13]);

}

System.out.println();

}

[Java] 1022. Digital Library (30)-PAT甲级

1022. Digital Library (30)
A Digital Library contains millions of books, stored according to their titles, authors, key words of their abstracts, publishers, and published years. Each book is assigned an unique 7-digit number as its ID. Given any query from a reader, you are supposed to output the resulting books, sorted in increasing order of their ID’s.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=10000) which is the total number of books. Then N blocks follow, each contains the information of a book in 6 lines:

Line #1: the 7-digit ID number;
Line #2: the book title — a string of no more than 80 characters;
Line #3: the author — a string of no more than 80 characters;
Line #4: the key words — each word is a string of no more than 10 characters without any white space, and the keywords are separated by exactly one space;
Line #5: the publisher — a string of no more than 80 characters;
Line #6: the published year — a 4-digit number which is in the range [1000, 3000].
It is assumed that each book belongs to one author only, and contains no more than 5 key words; there are no more than 1000 distinct key words in total; and there are no more than 1000 distinct publishers.

After the book information, there is a line containing a positive integer M (<=1000) which is the number of user’s search queries. Then M lines follow, each in one of the formats shown below:

1: a book title
2: name of an author
3: a key word
4: name of a publisher
5: a 4-digit number representing the year
Output Specification:

For each query, first print the original query in a line, then output the resulting book ID’s in increasing order, each occupying a line. If no book is found, print “Not Found” instead.

Sample Input:
3
1111111
The Testing Book
Yue Chen
test code debug sort keywords
ZUCS Print
2011
3333333
Another Testing Book
Yue Chen
test code sort keywords
ZUCS Print2
2012
2222222
The Testing Book
CYLL
keywords debug book
ZUCS Print2
2011
6
1: The Testing Book
2: Yue Chen
3: keywords
4: ZUCS Print
5: 2011
3: blablabla
Sample Output:
1: The Testing Book
1111111
2222222
2: Yue Chen
1111111
3333333
3: keywords
1111111
2222222
3333333
4: ZUCS Print
1111111
5: 2011
1111111
2222222
3: blablabla
Not Found

题目大意：模拟数字图书馆的查询功能。会给出n本书的信息，以及m个需要查询的命令，数字标号对应相应的命令，数字编号后面的字符串是查询的搜索词，要求输出这行命令以及输出满足条件的书的id，如果一个都没有找到，输出Not Found

PS：感谢github用户@fs19910227提供的pull request～

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.TreeSet;

public class Main {
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 6 * 1024 * 1024);

    static class Library {
        HashMap<String, TreeSet<String>>[] allMaps = new HashMap[5];

        Library() throws IOException {
            for (int i = 0; i < 5; i++) {
                HashMap<String, TreeSet<String>> map = new HashMap<>(1000, 1);
                allMaps[i] = map;
            }
            int record_size = Integer.valueOf(reader.readLine());
            for (int i = 0; i < record_size; i++) {
                String id = reader.readLine();
                for (int j = 0; j < allMaps.length; j++) {
                    HashMap<String, TreeSet<String>> map = allMaps[j];
                    if (j == 2) {
                        String[] keys = reader.readLine().split(" ");
                        for (String key : keys) {
                            putValue(map, key, id);
                        }
                    } else {
                        String key = reader.readLine();
                        putValue(map, key, id);
                    }
                }
            }
        }

        TreeSet<String> query(int id, String queryString) {
            HashMap<String, TreeSet<String>> allMap = allMaps[id - 1];
            TreeSet<String> integers = allMap.get(queryString);
            return integers;
        }

        private void putValue(HashMap<String, TreeSet<String>> map, String key, String id) {
            TreeSet<String> integers = map.computeIfAbsent(key, k -> new TreeSet<>());
            integers.add(id);
        }
    }

    public static void main(String[] args) throws IOException {
        Library library = new Library();
        int query_size = Integer.valueOf(reader.readLine());
        for (int i = 0; i < query_size; i++) {
            String line = reader.readLine();
            System.out.println(line);
            int id = line.charAt(0) - 48;
            String query = line.substring(3);
            TreeSet<String> queryResult = library.query(id, query);
            if (queryResult == null) {
                System.out.println("Not Found");
            } else {
                for (String book_id : queryResult) {
                    System.out.println(book_id);
                }
            }
        }
    }
}

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.HashMap;

import java.util.TreeSet;

public class Main {

static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in), 6 * 1024 * 1024);

static class Library {

HashMap<String, TreeSet<String>>[] allMaps = new HashMap[5];

Library() throws IOException {

for (int i = 0; i < 5; i++) {

HashMap<String, TreeSet<String>> map = new HashMap<>(1000, 1);

allMaps[i] = map;

}

int record_size = Integer.valueOf(reader.readLine());

for (int i = 0; i < record_size; i++) {

String id = reader.readLine();

for (int j = 0; j < allMaps.length; j++) {

HashMap<String, TreeSet<String>> map = allMaps[j];

if (j == 2) {

String[] keys = reader.readLine().split(" ");

for (String key : keys) {

putValue(map, key, id);

}

} else {

String key = reader.readLine();

putValue(map, key, id);

}

TreeSet<String> query(int id, String queryString) {

HashMap<String, TreeSet<String>> allMap = allMaps[id - 1];

TreeSet<String> integers = allMap.get(queryString);

return integers;

}

private void putValue(HashMap<String, TreeSet<String>> map, String key, String id) {

TreeSet<String> integers = map.computeIfAbsent(key, k -> new TreeSet<>());

integers.add(id);

}

public static void main(String[] args) throws IOException {

Library library = new Library();

int query_size = Integer.valueOf(reader.readLine());

for (int i = 0; i < query_size; i++) {

String line = reader.readLine();

System.out.println(line);

int id = line.charAt(0) - 48;

String query = line.substring(3);

TreeSet<String> queryResult = library.query(id, query);

if (queryResult == null) {

System.out.println("Not Found");

} else {

for (String book_id : queryResult) {

System.out.println(book_id);

}

[Java] 1020. Tree Traversals (25)-PAT甲级

1020. Tree Traversals (25)
Suppose that all the keys in a binary tree are distinct positive integers. Given the postorder and inorder traversal sequences, you are supposed to output the level order traversal sequence of the corresponding binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=30), the total number of nodes in the binary tree. The second line gives the postorder sequence and the third line gives the inorder sequence. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding binary tree. All the numbers in a line must be separated by exactly one space, and there must be no extra space at the end of the line.

Sample Input:
7
2 3 1 5 7 6 4
1 2 3 4 5 6 7
Sample Output:
4 1 6 3 5 7 2

题目大意：给定一棵二叉树的后序遍历和中序遍历，请你输出其层序遍历的序列。这里假设键值都是互不相等的正整数。

PS：感谢github用户@fs19910227提供的pull request～

import java.util.LinkedList;
import java.util.Scanner;

public class Main {
    static Scanner scanner = new Scanner(System.in);

    static class Node {
        int value;
        Node left;
        Node right;

        Node(int value) {
            this.value = value;
        }

        @Override
        public String toString() {
            return "Node{" +
                    "value=" + value +
                    ", left=" + left +
                    ", right=" + right +
                    '}';
        }
    }

    private static Node bulidTree() {
        int size = scanner.nextInt();
        int[] postOrder = new int[size];
        int[] inOrder = new int[size];
        for (int i = 0; i < size; i++) {
            postOrder[i] = scanner.nextInt();
        }
        for (int i = 0; i < size; i++) {
            inOrder[i] = scanner.nextInt();
        }
        Node root = build(postOrder, inOrder,
                0, size - 1,
                0, size - 1);
        return root;
    }

    private static Node build(int[] postOrder, int[] inOrder, int postStart, int postEnd, int inStart, int inEnd) {
        if (postStart > postEnd) {
            return null;
        }
        if (postStart == postEnd) {
            return new Node(postOrder[postStart]);
        }
        int root = postOrder[postEnd--];

        //find root in inOrder
        int inIndex = -1;
        for (int i = inStart; i <= inEnd; i++) {
            if (root == inOrder[i]) {
                inIndex = i;
                break;
            }
        }
        //recursion build
        int leftSize = inIndex - inStart;
        int rightSize = inEnd - inIndex;
        Node rootNode = new Node(root);
        rootNode.left = build(postOrder, inOrder,
                postStart, postStart + leftSize - 1,
                inStart, inIndex - 1);
        rootNode.right = build(postOrder, inOrder, postEnd - rightSize + 1, postEnd, inIndex + 1, inEnd);
        return rootNode;
    }

    public static void main(String[] args) {
        Node root = bulidTree();
        LinkedList<Node> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            Node poll = queue.poll();
            if (poll.left != null) {
                queue.add(poll.left);
            }
            if (poll.right != null) {
                queue.add(poll.right);
            }
            System.out.printf("%d%s", poll.value, queue.isEmpty() ? "\n" : " ");
        }
    }
}

import java.util.LinkedList;

import java.util.Scanner;

public class Main {

static Scanner scanner = new Scanner(System.in);

static class Node {

int value;

Node left;

Node right;

Node(int value) {

this.value = value;

}

@Override

public String toString() {

return "Node{" +

"value=" + value +

", left=" + left +

", right=" + right +

'}';

}

private static Node bulidTree() {

int size = scanner.nextInt();

int[] postOrder = new int[size];

int[] inOrder = new int[size];

for (int i = 0; i < size; i++) {

postOrder[i] = scanner.nextInt();

}

for (int i = 0; i < size; i++) {

inOrder[i] = scanner.nextInt();

}

Node root = build(postOrder, inOrder,

0, size - 1,

0, size - 1);

return root;

}

private static Node build(int[] postOrder, int[] inOrder, int postStart, int postEnd, int inStart, int inEnd) {

if (postStart > postEnd) {

return null;

}

if (postStart == postEnd) {

return new Node(postOrder[postStart]);

}

int root = postOrder[postEnd--];

//find root in inOrder

int inIndex = -1;

for (int i = inStart; i <= inEnd; i++) {

if (root == inOrder[i]) {

inIndex = i;

break;

}

//recursion build

int leftSize = inIndex - inStart;

int rightSize = inEnd - inIndex;

Node rootNode = new Node(root);

rootNode.left = build(postOrder, inOrder,

postStart, postStart + leftSize - 1,

inStart, inIndex - 1);

rootNode.right = build(postOrder, inOrder, postEnd - rightSize + 1, postEnd, inIndex + 1, inEnd);

return rootNode;

}

public static void main(String[] args) {

Node root = bulidTree();

LinkedList<Node> queue = new LinkedList<>();

queue.add(root);

while (!queue.isEmpty()) {

Node poll = queue.poll();

if (poll.left != null) {

queue.add(poll.left);

}

if (poll.right != null) {

queue.add(poll.right);

}

System.out.printf("%d%s", poll.value, queue.isEmpty() ? "\n" : " ");

}

[Java] 1019. General Palindromic Number (20)-PAT甲级

1019. General Palindromic Number (20)
A number that will be the same when it is written forwards or backwards is known as a Palindromic Number. For example, 1234321 is a palindromic number. All single digit numbers are palindromic numbers.

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number N > 0 in base b >= 2, where it is written in standard notation with k+1 digits ai as the sum of (aibi) for i from 0 to k. Here, as usual, 0 <= ai < b for all i and ak is non-zero. Then N is palindromic if and only if ai = ak-i for all i. Zero is written 0 in any base and is also palindromic by definition.

Given any non-negative decimal integer N and a base b, you are supposed to tell if N is a palindromic number in base b.

Input Specification:

Each input file contains one test case. Each case consists of two non-negative numbers N and b, where 0 <= N <= 109 is the decimal number and 2 <= b <= 109 is the base. The numbers are separated by a space.

Output Specification:

For each test case, first print in one line “Yes” if N is a palindromic number in base b, or “No” if not. Then in the next line, print N as the number in base b in the form “ak ak-1 … a0”. Notice that there must be no extra space at the end of output.

Sample Input 1:
27 2
Sample Output 1:
Yes
1 1 0 1 1
Sample Input 2:
121 5
Sample Output 2:
No
4 4 1

题目大意：给出两个整数a和b，问十进制的a在b进制下是否为回文数。是的话输出Yes，不是输出No。并且输出a在b进制下的表示，以空格隔开

PS：感谢github用户@fs19910227提供的pull request～

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

public class Main {
    private static ArrayList<Integer> toPalindromic(int number, int base) {
        ArrayList<Integer> list = new ArrayList<>(31);
        if (number == 0) {
            list.add(0);
            return list;
        }
        while (number != 0) {
            int bit = number % base;
            list.add(bit);
            number = number / base;
        }
        return list;
    }

    static boolean isPalindromic(ArrayList<Integer> list) {
        int length = list.size();
        for (int i = 0, j = length - 1; i < length / 2; i++, j--) {
            if (!list.get(i).equals(list.get(j))) {
                return false;
            }
        }
        return true;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] split = reader.readLine().split(" ");
        int number = Integer.valueOf(split[0]);
        int base = Integer.valueOf(split[1]);
        ArrayList<Integer> builder = toPalindromic(number, base);
        System.out.println(isPalindromic(builder) ? "Yes" : "No");
        for (int i = builder.size() - 1; i >= 0; i--) {
            System.out.printf("%d%s", builder.get(i), i == 0 ? "\n" : " ");
        }
    }
}

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.util.ArrayList;

public class Main {

private static ArrayList<Integer> toPalindromic(int number, int base) {

ArrayList<Integer> list = new ArrayList<>(31);

if (number == 0) {

list.add(0);

return list;

}

while (number != 0) {

int bit = number % base;

list.add(bit);

number = number / base;

}

return list;

}

static boolean isPalindromic(ArrayList<Integer> list) {

int length = list.size();

for (int i = 0, j = length - 1; i < length / 2; i++, j--) {

if (!list.get(i).equals(list.get(j))) {

return false;

}

return true;

}

public static void main(String[] args) throws IOException {

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

String[] split = reader.readLine().split(" ");

int number = Integer.valueOf(split[0]);

int base = Integer.valueOf(split[1]);

ArrayList<Integer> builder = toPalindromic(number, base);

System.out.println(isPalindromic(builder) ? "Yes" : "No");

for (int i = builder.size() - 1; i >= 0; i--) {

System.out.printf("%d%s", builder.get(i), i == 0 ? "\n" : " ");

}

[Java] 1015. Reversible Primes (20)-PAT甲级

1015. Reversible Primes (20)
A reversible prime in any number system is a prime whose “reverse” in that number system is also a prime. For example in the decimal system 73 is a reversible prime because its reverse 37 is also a prime.

Now given any two positive integers N (< 105) and D (1 < D <= 10), you are supposed to tell if N is a reversible prime with radix D.

Input Specification:

The input file consists of several test cases. Each case occupies a line which contains two integers N and D. The input is finished by a negative N.

Output Specification:

For each test case, print in one line “Yes” if N is a reversible prime with radix D, or “No” if not.

Sample Input:
73 10
23 2
23 10
-2
Sample Output:
Yes
Yes
No
题目大意：如果一个数本身是素数，而且在d进制下反转后的数在十进制下也是素数，就输出Yes，否则就输出No

PS：感谢github用户@fs19910227提供的pull request～

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;

public class Main {
    private boolean isPrime(String number) {
        int value = Integer.valueOf(number);
        if (value <= 1) return false;
        if (value == 2) return true;
        if (value % 2 == 0) return false;
        int maxLimit = (int) Math.sqrt(value);
        for (int i = 3; i <= maxLimit; i += 2) {
            if (value % i == 0) {
                return false;
            }
        }
        return true;
    }

    private String reversePrime(String number, int radix) {
        int value = Integer.valueOf(number);
        StringBuilder builder = new StringBuilder();
        for (; value != 0; value = value / radix) {
            int bit = value % radix;
            builder.append(bit);
        }
        return new BigInteger(builder.toString(), radix).toString();
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        Main rp = new Main();
        String[] split = reader.readLine().split(" ");
        for (; split.length > 1; split = reader.readLine().split(" ")) {
            String number = split[0];
            int radix = Integer.valueOf(split[1]);
            if (rp.isPrime(number) &&
                    rp.isPrime(rp.reversePrime(number, radix))) {
                System.out.println("Yes");
            } else {
                System.out.println("No");
            }
        }
    }
}

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.math.BigInteger;

public class Main {

private boolean isPrime(String number) {

int value = Integer.valueOf(number);

if (value <= 1) return false;

if (value == 2) return true;

if (value % 2 == 0) return false;

int maxLimit = (int) Math.sqrt(value);

for (int i = 3; i <= maxLimit; i += 2) {

if (value % i == 0) {

return false;

}

return true;

}

private String reversePrime(String number, int radix) {

int value = Integer.valueOf(number);

StringBuilder builder = new StringBuilder();

for (; value != 0; value = value / radix) {

int bit = value % radix;

builder.append(bit);

}

return new BigInteger(builder.toString(), radix).toString();

}

public static void main(String[] args) throws IOException {

BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));

Main rp = new Main();

String[] split = reader.readLine().split(" ");

for (; split.length > 1; split = reader.readLine().split(" ")) {

String number = split[0];

int radix = Integer.valueOf(split[1]);

if (rp.isPrime(number) &&

rp.isPrime(rp.reversePrime(number, radix))) {

System.out.println("Yes");

} else {

System.out.println("No");

}