PAT基础编程题目集----2

最新推荐文章于 2022-11-19 14:55:43 发布

Suzuki Mirai

最新推荐文章于 2022-11-19 14:55:43 发布

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CC 4.0 BY-SA版权

分类专栏：基础原理-数据结构文章标签： PAT JAVA

本文链接：https://blog.youkuaiyun.com/github_39312212/article/details/82226664

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13 篇文章

订阅专栏

本文提供了多个Java编程示例，包括分段计算水费、简单计算器、日K蜡烛图绘制、求整数段和等，通过实际案例展示了Java语言的基础用法。

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分段计算居民水费

import java.util.Scanner;
public class Main {
    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);
        double s = scan.nextDouble();
        double y;
        if (s <= 15){
            y = 1.0*4*s/3;
        }else {
            y = 2.5 * s - 17.5;
        }
        System.out.printf("%.2f",y);
    }
}

简单题……

两个数的简单计算器

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int i = scan.nextInt();
        String next = scan.next();
        int i1 = scan.nextInt();
        switch (next) {
            case "+":
                i += i1;
                break;
            case "-":
                i -= i1;
                break;
            case "*":
                i *= i1;
                break;
            case "/":
                i /= i1;
                break;
            case "%":
                i %= i1;
                break;
            default:
                next = "ERROR";
        }
        if ("ERROR".equals(next))
            System.out.print("ERROR");
        else
            System.out.print(i);
    }
}

简单题

日K蜡烛图

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        double open = scan.nextDouble();
        double high = scan.nextDouble();
        double low = scan.nextDouble();
        double close = scan.nextDouble();

        String candle = "";
        if (close < open) {
            candle = "BW-Solid";
        } else if ((close > open)) {
            candle = "R-Hollow";
        } else {
            candle = "R-Cross";
        }
        if (low < open && low < close && high > open && high > close) {
            System.out.print(candle + " with Lower Shadow and Upper Shadow");
            return;
        } else if (high > open && high > close) {
            System.out.print(candle + " with Upper Shadow");
            return;
        } else if (low < open && low < close) {
            System.out.print(candle + " with Lower Shadow");
            return;
        } else {
            System.out.print(candle);
        }
    }

}

简单题

求整数段和

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int A = scan.nextInt();
        int B = scan.nextInt();
        int sum = 0;
        for (int i = A ;i<=B;i+=5){
            for (int j = i;j<i+5&&j<=B;j++){
                sum += j;
                System.out.printf("%5d",j);
            }
            System.out.println();
        }
        System.out.print("Sum = "+sum);

    }
}

简单题

计算圆周率

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        double threshold = scan.nextDouble();
        double pi = 1;
        double Numerator=1,denominator=3;
        int numeratorTem=1,denominatorTem=3;
        while (true) {
            double numn = Numerator / denominator;
            pi += numn;
            if (numn < threshold)
                break;
            Numerator *= (++numeratorTem);
            denominatorTem += 2;
            denominator *= denominatorTem;
        }
        System.out.printf("%.6f", pi * 2);
    }
}

这道题有巨坑，分子分母必须一开始就是double，否则数据类型转换会出问题，过不了第二个测试点，具体哪里出问题我也不清楚，总之看了比人的博客知乎改成double就好了。

求符合给定条件的整数集

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int A = scan.nextInt();
        for (int i = A; i < A + 4; i++) {
            int flag = 1;
            for (int j = A; j < A + 4; j++) {
                if (j==i)
                    continue;
                for (int k = A ;k<A+4;k++){
                    if (k ==i||k==j)
                        continue;
                    System.out.print(i*100+j*10+k);
                    flag++;
                    if(flag!=7)
                        System.out.print(" ");
                }

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

还行吧，算简单的

爬动的蠕虫

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int N = scan.nextInt();
        int U = scan.nextInt();
        int D = scan.nextInt();
        int length = 0;
        int mins = 1;
        while (true){
            if (mins%2 !=0)
                length += U;
            else {
                length -= D;
            }
            if (length>=N)
                break;
            mins++;
        }
        System.out.println(mins);
    }
}

蠕虫长度是糊弄人的，只需要关注头部即可。

二分法求多项式单根

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        double a3 = scan.nextDouble();
        double a2 = scan.nextDouble();
        double a1 = scan.nextDouble();
        double a0 = scan.nextDouble();
        double a = scan.nextDouble();
        double b = scan.nextDouble();
        double aResult = calculateResult(a3, a2, a1, a0, a);
        double bResult = calculateResult(a3, a2, a1, a0, b);
        double abResult = calculateResult(a3, a2, a1, a0, (a + b) / 2.0);
        while (b - a >= 0.001) {

            if (aResult == 0) {
                System.out.printf("%.2f", a);
                return;
            } else if (bResult == 0) {
                System.out.printf("%.2f", b);
                return;
            } else if (abResult == 0) {
                System.out.printf("%.2f", (a + b) / 2.0);
                return;
            } else if (aResult * abResult < 0) {
                b = (a + b) / 2.0;
                bResult = calculateResult(a3, a2, a1, a0, b);
                abResult = calculateResult(a3, a2, a1, a0, (a + b) / 2.0);
            } else {
                a = (a + b) / 2.0;
                aResult = calculateResult(a3, a2, a1, a0, a);
                abResult = calculateResult(a3, a2, a1, a0, (a + b) / 2.0);
            }
        }
        System.out.printf("%.2f", (a + b) / 2.0);
    }

    private static double calculateResult(double a3, double a2, double a1, double a0, double x) {
        return a0 + x * (a1 + x * (a2 + x * a3));
    }
}

有点坑，精确到小数点后两位是什么意思呢？首先区间长度内的值都算它的解，这说明区间长度就是选出的这个解的误差范围，所以要保证小数点后两位精确，只需要把区间长度设置为0.001（或许0.0001才真正能保证），那这个解就可以表示为（X正负0.001，就是表示误差的那个意思），这是一个。还有一个是我自己是因为while循环内的输出格式忘了控制了导致有一个检测点出问题，妈的找了半天。

支票面额

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        double f = 0;
        int y=0;
        int cont = 0;
        while (y < 100 && f < 100) {
            f = (n + 199*y)/98.0;
            if (f-(int)f == 0){
                System.out.println(y+"."+(int)f);
                cont++;
            }
            y++;
        }
        if (cont == 0){
            System.out.println("No Solution");
        }
    }
}

思路就是得到公式之后一个一个试，两个值都是从0~99。

打印九九口诀表

import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int n = scan.nextInt();
        for (int i=1;i<=n;i++){
            for (int j =1;j<=i;j++){
                System.out.printf("%d*%d=%-4d",j,i,i*j);
            }
            System.out.println();
        }
    }
}

简单题