2020/07/13-Java loop structure exercises (print Yang Hui triangle, print all odd numbers composed of 0-7, there is no repetition of numbers in each digit)

2020/07/13-Java loop structure exercises (print Yang Hui triangle, print all odd numbers composed of 0-7, there is no repetition of numbers in each digit)

1. Print out Yang Hui triangle (10 lines are required to be printed as shown below)

• Program analysis:
• 1
• 1 1
• 1 2 1
• 1 3 3 1
• 1 4 6 4 1
• 1 5 10 10 5 1
• Ideas:
• First of all, the first number in each row is 1
• The law of each line of numbers is: 1; i/1; i(i-1)/1 2; (i-1)(i-2)/2 3;... int i =0
• Important: Analyze the relationship between the characteristics of each row of numbers and i, and also pay attention to the initial value of i is set to 0
• For example, for row 3: 1, 2, 1
• Ordinance v = 1; a = i,
• i=2 represents the third row, v=v*aa–; output: v=v/j, where int j=0, j=0 is defined, which means the first number in the first column, j=1 means one column The second number in

/**
*/
public class test15 {
    
    
    public static void main(String[] args) {
    
    
        int num = 5;
        for (int i = 0; i < num; i++) {
    
    
            int v = 1;
            int a = i;
            for (int j = 0; j <= i; j++) {
    
    
                int b = j;
                if(b==0)
                    System.out.print(1+" ");
                else {
    
    
                    System.out.print((v = v / j) + " ");
                }
                v=v*a;
                a--;

            }
            System.out.println();
        }

    }
}

2. Find the odd number of 0-7

analysis of idea:

• Only odd numbers when the ones place is 1, 3, 5, 7

• Calculate the odd numbers of 2 digits, 3 digits, 4 digits..., for example:

• With 2 digits, there are 6*4 possibilities. Because the highest digit (first digit) cannot be 0, it is 7-1=6 possibilities

• With 3 digits, there are 6 6 4 types, the first digit and the ones digit are a number, and the second digit is 8-2=6 types.

• 4 digits, there are 6 6 5*4 possibilities

• Therefore, it should be written separately for 2 digits, and the digits after 2 can be written according to the circulation law
  */

public class test16 {
    
    
    public static void main(String[] args) {
    
    
        int total = 4;
        for (int i = 2; i <=8 ; i++) {
    
    //外层循环定义位数
            int sum = 4;
            for (int j = 1; j < i; j++) {
    
    
                if (j==1){
    
    
                    sum=sum*6;
                }else {
    
    
                    sum=sum*(8-j);
                }
            }
            total+=sum;


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

3. Find the positive difference between prime numbers and non-prime numbers within 1000

• Ideas:
• 2 is a prime number, 1 is neither a prime number nor a composite number
• Prime number + composite number + 1 = all numbers
• First find the prime numbers, then use the sum-the sum of prime numbers = the sum of non-prime numbers

public class khzy2 {
    
    
    public static void main(String[] args) {
    
    
        int j = 0;
        int sum = 2;
        int num = 0;
        int s = 0;
        for (int i = 1; i <1000 ; i++) {
    
    
            num=num+i;
        }
        System.out.println(num);
        for (int i = 2; i <1000 ; i++) {
    
    
            for ( j = 2; j <i ; j++) {
    
    
                s =i%j;
                if (s==0){
    
    
                    break;
                }
            }
            if (s!=0){
    
    
                sum = sum + i;

            }

        }
        System.out.println("质数:"+sum);
         int n = num - sum;
         int k = n - sum;
        System.out.println(k);
    }
}

4. Randomly generate 6 random numbers between 1 and 33, and a random number between 1 and 15. These 7 numbers cannot be repeated.

• Ideas:
• 1. int n = (int)(Math.random() 10); when generated is a random number between 0-9, so 34 will generate a random number between 0-33
• 2. Since each number cannot be repeated, add a || statement to ensure that every two numbers are not equal to each other.
• 3. If it is not equal, print out 7 numbers. If it is equal, use the contiune statement to exit this loop and continue to the next loop.

public class khzy1 {
    
    
    public static void main(String[] args) {
    
    
        int flag = 0;
        do {
    
    
            int num1 = (int) (Math.random() * 34);
            int num2 = (int) (Math.random() * 34);
            int num3 = (int) (Math.random() * 34);
            int num4 = (int) (Math.random() * 34);
            int num5 = (int) (Math.random() * 34);
            int num6 = (int) (Math.random() * 34);
            int num7 = (int) (Math.random() * 16);
            if ((num1!=num2)||(num1!=num3)||(num1!=num4)||(num1!=num5)||(num1!=num6)
                    ||(num1!=num7)||(num2!=num3)||(num4!=num2)||(num5!=num2)||(num6!=num2)
                    ||(num7!=num2)||(num4!=num3)||(num5!=num3)||(num6!=num3)||(num7!=num3)||
                    (num4!=num5)||(num4!=num6)||(num4!=num7)||(num4!=num2)||(num5!=num6)||(num5!=num7)||(num6!=num7)
            ){
    
    
                flag = 1;
                System.out.println(num1+" "+num2+" "+num3+" "+num4+" "+num5+" "+num6+" "+num7+" ");
            };
        }while (flag == 0);
    }
}