PAT – 第3页 – liuchuo

[Java] 1011. World Cup Betting (20)-PAT甲级

1011. World Cup Betting (20)
With the 2010 FIFA World Cup running, football fans the world over were becoming increasingly excited as the best players from the best teams doing battles for the World Cup trophy in South Africa. Similarly, football betting fans were putting their money where their mouths were, by laying all manner of World Cup bets.

Chinese Football Lottery provided a “Triple Winning” game. The rule of winning was simple: first select any three of the games. Then for each selected game, bet on one of the three possible results — namely W for win, T for tie, and L for lose. There was an odd assigned to each result. The winner’s odd would be the product of the three odds times 65%.

For example, 3 games’ odds are given as the following:

W T L
1.1 2.5 1.7
1.2 3.0 1.6
4.1 1.2 1.1
To obtain the maximum profit, one must buy W for the 3rd game, T for the 2nd game, and T for the 1st game. If each bet takes 2 yuans, then the maximum profit would be (4.1*3.0*2.5*65%-1)*2 = 37.98 yuans (accurate up to 2 decimal places).

Input

Each input file contains one test case. Each case contains the betting information of 3 games. Each game occupies a line with three distinct odds corresponding to W, T and L.

Output

For each test case, print in one line the best bet of each game, and the maximum profit accurate up to 2 decimal places. The characters and the number must be separated by one space.

Sample Input
1.1 2.5 1.7
1.2 3.0 1.6
4.1 1.2 1.1
Sample Output
T T W 37.98

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 {
    static class Bet {
        static String[] names = new String[]{"W", "T", "L"};
        String name;
        double value;

        Bet(int index, double value) {
            this.name = names[index];
            this.value = value;
        }
    }

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

    public static void main(String[] args) throws IOException {
        StringBuilder builder = new StringBuilder();
        double sum = 1;
        for (int i = 0; i < 3; i++) {
            Bet bet = getBet();
            builder.append(bet.name).append(" ");
            sum = sum * bet.value;
        }
        double profit = 2 * (sum * 0.65 - 1);
        //四舍五入就可以, 可以无视example
        builder.append(String.format("%.2f", profit));
        System.out.println(builder);
    }

    private static Bet getBet() throws IOException {
        String[] split = reader.readLine().split(" ");
        int max_index = 0;
        double max_bet = -1.0;
        for (int i = 0; i < split.length; i++) {
            double v = Double.valueOf(split[i]);
            if (v > max_bet) {
                max_index = i;
                max_bet = v;
            }
        }
        return new Bet(max_index, max_bet);
    }
}

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

public class Main {

static class Bet {

static String[] names = new String[]{"W", "T", "L"};

String name;

double value;

Bet(int index, double value) {

this.name = names[index];

this.value = value;

}

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

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

StringBuilder builder = new StringBuilder();

double sum = 1;

for (int i = 0; i < 3; i++) {

Bet bet = getBet();

builder.append(bet.name).append(" ");

sum = sum * bet.value;

}

double profit = 2 * (sum * 0.65 - 1);

//四舍五入就可以, 可以无视example

builder.append(String.format("%.2f", profit));

System.out.println(builder);

}

private static Bet getBet() throws IOException {

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

int max_index = 0;

double max_bet = -1.0;

for (int i = 0; i < split.length; i++) {

double v = Double.valueOf(split[i]);

if (v > max_bet) {

max_index = i;

max_bet = v;

}

return new Bet(max_index, max_bet);

}

[Java] 1010. Radix (25)-PAT甲级

1010. Radix (25)
Given a pair of positive integers, for example, 6 and 110, can this equation 6 = 110 be true? The answer is “yes”, if 6 is a decimal number and 110 is a binary number.

Now for any pair of positive integers N1 and N2, your task is to find the radix of one number while that of the other is given.

Input Specification:

Each input file contains one test case. Each case occupies a line which contains 4 positive integers:
N1 N2 tag radix
Here N1 and N2 each has no more than 10 digits. A digit is less than its radix and is chosen from the set {0-9, a-z} where 0-9 represent the decimal numbers 0-9, and a-z represent the decimal numbers 10-35. The last number “radix” is the radix of N1 if “tag” is 1, or of N2 if “tag” is 2.

Output Specification:

For each test case, print in one line the radix of the other number so that the equation N1 = N2 is true. If the equation is impossible, print “Impossible”. If the solution is not unique, output the smallest possible radix.

Sample Input 1:
6 110 1 10
Sample Output 1:
2
Sample Input 2:
1 ab 1 2
Sample Output 2:
Impossible

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 static String binarySearch(final String value, final BigInteger expect) {
        //重点在于最大值和最小值的界定 下界大于 所有位中最大值 考虑2 2 1 10 这个case
        BigInteger start = BigInteger.valueOf(maxBitValue(value) + 1);
        //上界不小于下界,不大于expect 考虑200000 1 1 10 这个case
        BigInteger end = expect.compareTo(start) < 0 ? start : expect;

        BigInteger result_radix = null;
        while (end.compareTo(start) >= 0) {
            BigInteger middle = start.add(end).divide(BigInteger.valueOf(2));
            BigInteger ten_value = convert2ten(value, middle);
            int compare = ten_value.compareTo(expect);
            if (compare >= 0) {
                if (compare == 0 && (result_radix == null || middle.compareTo(result_radix) < 0)) {
                    result_radix = middle;
                }
                end = middle.subtract(BigInteger.valueOf(1));
            } else {
                start = middle.add(BigInteger.valueOf(1));
            }
        }
        return result_radix == null ? "Impossible" : result_radix + "";
    }

    private static BigInteger convert2ten(String value, BigInteger radix) {
        char[] chars = value.toCharArray();
        BigInteger sum = BigInteger.ZERO;
        for (char aChar : chars) {
            int v = char2int(aChar);
            if (radix.compareTo(BigInteger.valueOf(v)) < 0) {
                return BigInteger.valueOf(Long.MAX_VALUE);
            }
            sum = sum.multiply(radix).add(BigInteger.valueOf(v));
        }
        return sum;
    }

    private static int maxBitValue(String value) {
        int max = 0;
        for (char c : value.toCharArray()) {
            int i = char2int(c);
            if (i > max) max = i;
        }
        return max;
    }

    private static int char2int(char c) {
        int ret = c - 48;
        return ret < 10 ? ret : ret - 39;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        String[] split = reader.readLine().split(" ");
        String v1 = split[0], v2 = split[1];
        String know_v = v1, unknow_v = v2;
        BigInteger know_radix = BigInteger.valueOf(Integer.valueOf(split[3]));
        if ("2".equals(split[2])) {
            know_v = v2;
            unknow_v = v1;
        }
        //期望值转化为十进制 相当于把输入转化为 N1 N2 1 10
        BigInteger expect = convert2ten(know_v, know_radix);
        String s = binarySearch(unknow_v, expect);
        System.out.println(s);
    }
}

import java.io.BufferedReader;

import java.io.IOException;

import java.io.InputStreamReader;

import java.math.BigInteger;

public class Main {

private static String binarySearch(final String value, final BigInteger expect) {

//重点在于最大值和最小值的界定下界大于所有位中最大值考虑2 2 1 10 这个case

BigInteger start = BigInteger.valueOf(maxBitValue(value) + 1);

//上界不小于下界,不大于expect 考虑200000 1 1 10 这个case

BigInteger end = expect.compareTo(start) < 0 ? start : expect;

BigInteger result_radix = null;

while (end.compareTo(start) >= 0) {

BigInteger middle = start.add(end).divide(BigInteger.valueOf(2));

BigInteger ten_value = convert2ten(value, middle);

int compare = ten_value.compareTo(expect);

if (compare >= 0) {

if (compare == 0 && (result_radix == null || middle.compareTo(result_radix) < 0)) {

result_radix = middle;

}

end = middle.subtract(BigInteger.valueOf(1));

} else {

start = middle.add(BigInteger.valueOf(1));

}

return result_radix == null ? "Impossible" : result_radix + "";

}

private static BigInteger convert2ten(String value, BigInteger radix) {

char[] chars = value.toCharArray();

BigInteger sum = BigInteger.ZERO;

for (char aChar : chars) {

int v = char2int(aChar);

if (radix.compareTo(BigInteger.valueOf(v)) < 0) {

return BigInteger.valueOf(Long.MAX_VALUE);

}

sum = sum.multiply(radix).add(BigInteger.valueOf(v));

}

return sum;

}

private static int maxBitValue(String value) {

int max = 0;

for (char c : value.toCharArray()) {

int i = char2int(c);

if (i > max) max = i;

}

return max;

}

private static int char2int(char c) {

int ret = c - 48;

return ret < 10 ? ret : ret - 39;

}

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

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

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

String v1 = split[0], v2 = split[1];

String know_v = v1, unknow_v = v2;

BigInteger know_radix = BigInteger.valueOf(Integer.valueOf(split[3]));

if ("2".equals(split[2])) {

know_v = v2;

unknow_v = v1;

}

//期望值转化为十进制相当于把输入转化为 N1 N2 1 10

BigInteger expect = convert2ten(know_v, know_radix);

String s = binarySearch(unknow_v, expect);

System.out.println(s);

}

[Java] 1007. Maximum Subsequence Sum (25)-PAT甲级

1007. Maximum Subsequence Sum (25)
Given a sequence of K integers { N1, N2, …, NK }. A continuous subsequence is defined to be { Ni, Ni+1, …, Nj } where 1 <= i <= j <= K. The Maximum Subsequence is the continuous subsequence which has the largest sum of its elements. For example, given sequence { -2, 11, -4, 13, -5, -2 }, its maximum subsequence is { 11, -4, 13 } with the largest sum being 20.

Now you are supposed to find the largest sum, together with the first and the last numbers of the maximum subsequence.

Input Specification:

Each input file contains one test case. Each case occupies two lines. The first line contains a positive integer K (<= 10000). The second line contains K numbers, separated by a space.

Output Specification:

For each test case, output in one line the largest sum, together with the first and the last numbers of the maximum subsequence. The numbers must be separated by one space, but there must be no extra space at the end of a line. In case that the maximum subsequence is not unique, output the one with the smallest indices i and j (as shown by the sample case). If all the K numbers are negative, then its maximum sum is defined to be 0, and you are supposed to output the first and the last numbers of the whole sequence.

Sample Input:
10
-10 1 2 3 4 -5 -23 3 7 -21
Sample Output:
10 1 4
题目大意：求最大连续子序列和，输出最大的和以及这个子序列的开始值和结束值。如果所有数都小于0，那么认为最大的和为0，并且输出首尾元素。

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));
        reader.readLine();
        String[] split = reader.readLine().split(" ");
        int final_max_sum = Integer.MIN_VALUE;
        int sub_sum = 0, start_index = 0, final_start = 0, final_end = split.length - 1;
        for (int i = 0; i < split.length; i++) {
            int value = Integer.valueOf(split[i]);
            sub_sum = sub_sum + value;
            if (sub_sum >= 0) {
                if (sub_sum > final_max_sum) {
                    final_max_sum = sub_sum;

                    if (start_index != final_start) {
                        final_start = start_index;
                    }
                    final_end = i;
                }
            } else {
                sub_sum = 0;
                start_index = i + 1;
            }
        }
        if (final_max_sum < 0) {
            final_max_sum = 0;
        }
        System.out.print(final_max_sum + " ");
        System.out.print(Integer.valueOf(split[final_start]) + " ");
        System.out.println(Integer.valueOf(split[final_end]));
    }
}

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));

reader.readLine();

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

int final_max_sum = Integer.MIN_VALUE;

int sub_sum = 0, start_index = 0, final_start = 0, final_end = split.length - 1;

for (int i = 0; i < split.length; i++) {

int value = Integer.valueOf(split[i]);

sub_sum = sub_sum + value;

if (sub_sum >= 0) {

if (sub_sum > final_max_sum) {

final_max_sum = sub_sum;

if (start_index != final_start) {

final_start = start_index;

}

final_end = i;

}

} else {

sub_sum = 0;

start_index = i + 1;

}

if (final_max_sum < 0) {

final_max_sum = 0;

}

System.out.print(final_max_sum + " ");

System.out.print(Integer.valueOf(split[final_start]) + " ");

System.out.println(Integer.valueOf(split[final_end]));

}

[Java] 1008. Elevator (20)-PAT甲级

1008. Elevator (20)
The highest building in our city has only one elevator. A request list is made up with N positive numbers. The numbers denote at which floors the elevator will stop, in specified order. It costs 6 seconds to move the elevator up one floor, and 4 seconds to move down one floor. The elevator will stay for 5 seconds at each stop.
For a given request list, you are to compute the total time spent to fulfill the requests on the list. The elevator is on the 0th floor at the beginning and does not have to return to the ground floor when the requests are fulfilled.
Input Specification:
Each input file contains one test case. Each case contains a positive integer N, followed by N positive numbers. All the numbers in the input are less than 100.
Output Specification:
For each test case, print the total time on a single line.
Sample Input:
3 2 3 1
Sample Output:
41

题目：电梯从0层开始向上，给出该电梯依次按顺序停的楼层数，并且已知上升需要6秒/层，下降需要4秒/层，停下来的话需要停5秒，问走完所有需要停的楼层后总共花了多少时间～

分析：累加计算输出～

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));
        String[] split = reader.readLine().split(" ");
        int sum = 0;
        int lastFloor = 0;
        for (int i = 1; i < split.length; i++) {
            int value = Integer.valueOf(split[i]);
            int diff = value - lastFloor;
            if (diff > 0) {
                sum += diff * 6 + 5;
            } else if (diff < 0) {
                sum += -diff * 4 + 5;
            } else {
                sum += 5;
            }
            lastFloor = value;
        }
        System.out.println(sum);
    }
}

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));

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

int sum = 0;

int lastFloor = 0;

for (int i = 1; i < split.length; i++) {

int value = Integer.valueOf(split[i]);

int diff = value - lastFloor;

if (diff > 0) {

sum += diff * 6 + 5;

} else if (diff < 0) {

sum += -diff * 4 + 5;

} else {

sum += 5;

}

lastFloor = value;

}

System.out.println(sum);

}

[Java] 1006. Sign In and Sign Out (25)-PAT甲级

1006. Sign In and Sign Out (25)
At the beginning of every day, the first person who signs in the computer room will unlock the door, and the last one who signs out will lock the door. Given the records of signing in’s and out’s, you are supposed to find the ones who have unlocked and locked the door on that day.

Input Specification:

Each input file contains one test case. Each case contains the records for one day. The case starts with a positive integer M, which is the total number of records, followed by M lines, each in the format:

ID_number Sign_in_time Sign_out_time
where times are given in the format HH:MM:SS, and ID number is a string with no more than 15 characters.

Output Specification:

For each test case, output in one line the ID numbers of the persons who have unlocked and locked the door on that day. The two ID numbers must be separated by one space.

Note: It is guaranteed that the records are consistent. That is, the sign in time must be earlier than the sign out time for each person, and there are no two persons sign in or out at the same moment.

Sample Input:
3
CS301111 15:30:28 17:00:10
SC3021234 08:00:00 11:25:25
CS301133 21:45:00 21:58:40
Sample Output:
SC3021234 CS301133

题目大意：给出n个人的id、sign in时间、sign out时间，求最早进来的人和最早出去的人的ID～

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

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.text.DateFormat;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Date;

public class Main {
    static class Record {
        static DateFormat format = new SimpleDateFormat("HH:mm:ss");
        String id;
        Date start;
        Date end;

        Record(String id, String start, String end) throws ParseException {
            this.id = id;
            this.start = format.parse(start);
            this.end = format.parse(end);
        }
    }

    public static void main(String[] args) throws Exception {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        int record_counts = Integer.valueOf(reader.readLine());


        String[] split = reader.readLine().split(" ");
        Record record = new Record(split[0], split[1], split[2]);
        Record earliest = record, latest = record;
        for (int i = 1; i < record_counts; i++) {
            split = reader.readLine().split(" ");
            Record newRecord = new Record(split[0], split[1], split[2]);
            if (newRecord.start.compareTo(earliest.start) < 0) {
                earliest = newRecord;
            }
            if (newRecord.end.compareTo(latest.end) > 0) {
                latest = newRecord;
            }
        }
        System.out.println(earliest.id + " " + latest.id);
    }
}

import java.io.BufferedReader;

import java.io.InputStreamReader;

import java.text.DateFormat;

import java.text.ParseException;

import java.text.SimpleDateFormat;

import java.util.Date;

public class Main {

static class Record {

static DateFormat format = new SimpleDateFormat("HH:mm:ss");

String id;

Date start;

Date end;

Record(String id, String start, String end) throws ParseException {

this.id = id;

this.start = format.parse(start);

this.end = format.parse(end);

}

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

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

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

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

Record record = new Record(split[0], split[1], split[2]);

Record earliest = record, latest = record;

for (int i = 1; i < record_counts; i++) {

split = reader.readLine().split(" ");

Record newRecord = new Record(split[0], split[1], split[2]);

if (newRecord.start.compareTo(earliest.start) < 0) {

earliest = newRecord;

}

if (newRecord.end.compareTo(latest.end) > 0) {

latest = newRecord;

}

System.out.println(earliest.id + " " + latest.id);

}

[Java] 1005. Spell It Right (20)-PAT甲级

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 (<= 10100).

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

题目大意：给一个非负正数N，计算N的每一位相加的和，然后输出和的每一位的英文读音～

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

import java.io.IOException;
import java.io.InputStream;

public class Main {
    static String[] words_table = new String[]{"zero",
            "one", "two", "three", "four", "five",
            "six", "seven", "eight", "nine"};

    public static void main(String[] args) throws IOException {
        InputStream reader = System.in;
        int input, sum = 0;
        while ((input = reader.read()) != '\n') {
            sum += input - 48;
        }
        char[] chars = (sum + "").toCharArray();
        for (int i = 0; i < chars.length; i++) {
            System.out.print(words_table[chars[i] - 48]);
            if (i == chars.length - 1) {
                System.out.println();
            } else {
                System.out.print(" ");
            }
        }
    }
}

import java.io.IOException;

import java.io.InputStream;

public class Main {

static String[] words_table = new String[]{"zero",

"one", "two", "three", "four", "five",

"six", "seven", "eight", "nine"};

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

InputStream reader = System.in;

int input, sum = 0;

while ((input = reader.read()) != '\n') {

sum += input - 48;

}

char[] chars = (sum + "").toCharArray();

for (int i = 0; i < chars.length; i++) {

System.out.print(words_table[chars[i] - 48]);

if (i == chars.length - 1) {

System.out.println();

} else {

System.out.print(" ");

}