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  • java程序设计基础篇 复习笔记 第四单元

    1 
    think before coding
    code incrementally
    2
    sentinel value
    sentinel-controlled loop
    3 
    输入输出重定向 > <
    input redirection
    output redirection
    4
    pretest loop
    posttest loop
    5
    从小到大添加浮点数比从大到小精确
    6
    Integer.toBinaryString(int)
    Integer.toHexString(int)
    7
    PIE =4* (1 - 1 / 3 + 1 / 5 - 1 / 7 ... - 1 / (2 * i - 1) + 1 / (2 * i + 1))
    
    Keyword
    break statement
    continue statement
    do-while loop
    for loop
    loop control statement
    infinite loop
    input redirection
    iteration
    labeled continue statement
    loop
    loop-condition condition :循环控制条件
    loop body
    nested body
    off-by-one error
    output redirection
    sentinel value
    while loop
    
    4.1 
    always true
    4.2
    无限循环
    4.3
    无限次;9次;9次
    11111111111111111
    2,4,6,8
    3,5,7,9
    4.4
    循环控制条件和循环体的执行顺序
    int sum = 0;
    int number = 0;
    do{
    	sum += number;
    	number = input.nextInt();
    }while(number != 0)
    4.5
    相同
    4.6
    循环初始化,循环控制条件,每次迭代后操作
    for(int i = 0;i < 100;i++)System.out.println(i+1);
    4.7
    max is 5
    number is 0
    4.8
    sum is 14
    number is 0
    4.9
    max is 5
    number is 0
    4.10
    无限循环
    4.11
    不能
    4.12
    可以,但是for循环更直接,明晰
    4.13
    long sum =0;
    int i = 0;
    while(i <= 1000){
    	sum += i++;
    }
    /*
    do{
    	sum+=i++;
    }while(i<=1000)
    */
    4.14
    cant because of integer division
    4.15
    跳出当前循环;跳出当前迭代;能,1;不能
    4.16
    int sum  = 0;
    int i = 1;
    for(;sum < 10000;i++)sum += i;
    4.17
    if(i % 3 == 0){i++;continue;}
    4.18
    .........
    4.19
    System.out.println(i);1;
    4.20
    System.out.println(i);1;
    4.21
    2 public static void
    3:int i 在循环体外声明
    5:声明int j 并初始化
    7:分号
    11:去掉分号
    4.22
    未初始化;多加了分号;
    4.23
    0 0 1 0 1 2 3 0 1 2 3 4;
    compile Error;
    1xxx2xxx4xxx8xxx16xxx
    1xxx2xxx4xxx8xxx
    1xxx2xxx4xxx
    1xxx2xxx
    1xxx
    ;
    1G
    1G2G
    1G2G4G
    1G2G4G8G
    1G2G4G8G16G
    ;
    4.24
    no output result;because the loop never end
    4.25
    n
    n-5
    n+1
    (n-3)/3
    
    编程题
    4.23
    //从左向右
    double sum = 0;
    for(int i =1;i<=50000;i++){
    	sum += 1d/i;
    }
    System.out.println(sum);
    //11.397003949278504
    //从右向左
    double sum = 0;
    for(int i = 50000;i >= 1;i--){
    	sum += 1d/i;	
    }
    System.out.println(sum);
    //11.397003949278519
    

      

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  • 原文地址:https://www.cnblogs.com/xuesu/p/4188231.html
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