JAVAJ Classic Program Programming 50 questions __ Programming

 

Java BASIC Programming exercises 50 questions

In this paper, 50 Classic Java program to explain the detailed, for beginners can skip a number of logic too strong topics, such as the first problem with the method of recursion, beginners may not understand, it is best to see those who have if, for, while can be a simple solution to the problem. However, for more in-depth study of the students, or hope to be able to read at one breath, this is a more comprehensive thinking exercise.

"Program 1"

Title: Classical question: There are a pair of rabbits, from the 3rd month after the birth of a pair of rabbits each month, the rabbit long to the third month after the birth of a pair of rabbits each month, if the rabbit is not dead, ask the rabbit log every month for how much.

Procedure analysis: Rabbit's rule for series 1,1,2,3,5,8,13,21 ....

The best way to do this is to find the rules, like the high School series.

There are: a[n]=a[n-1]+a[n-1], and the first second is known, the following value can be obtained

Public class Programme1 { Public static void main (string[] args) { System. out. Print ("Please enter the month of the number of rabbits you want to know:"); Scanner scanner=New Scanner (System. in); int n=scanner.nextint ();//Get input integer System. out. println ("The first" +n+ "the total number of rabbits" +fun (n)); Scanner.close (); } The number of rabbits for the required month, and the number of rabbits to return the value. private static int fun (intn) { if (N==1 | | n==2) return 1; Else return Fun (n-1) +fun (n-2); } }

"Program 2"

Title: Determine how many primes there are between 101-200, and output all primes.

Program Analysis:

* Prime number is: can only be divisible by 1 or itself, such as: 3,5,7,11,131 ...

* Method of determining prime numbers: Remove 2 to sqrt (this number) with a number,

* Actually use this number to remove 2 to his own less than 1 of the number can also be, but the calculation time increased

* If divisible, it indicates that this number is not prime, whereas it is prime.

Public class Programme2 { Public static void main (string[] args) { int sum=0; for (inti = m < 200;i++) { if (Isrightnum (i)) {//Judge whether this number is prime System. out. Print (i+ ""); sum++; if (sum%10==0) {//10 one line System. out. println (); } } } System. out. println ("Integer of Prime number:" +sum); } The specific code that determines whether this number is prime private Static Boolean Isrightnum (inti) { for (intj = 2; J < math.sqrt (i); j + +) { if (i%j==0) {//If divisible, it means not prime, can be interrupted immediately, continue to judge the next number return false; } } return true; } }

"Program 3"

Title: Print out all the "Narcissus number", the so-called "Narcissus number" refers to a three-digit number, its members of the digital cubic and equal to the number itself. For example: 153 is a "narcissus number", because the 153=1 three times side +5 of three times +3 of the three times side.

Program Analysis: Use for loop control 100-999 number, each number decomposition out of single-digit, 10, hundred.

Public class Programme3 { Public static void main (string[] args) { The total number of sum=0;//daffodils in int for (inti = m < 1000;i++) { intbite=i%10; Get a digit intten=i/10%10; Get 10 bits inthundred=i/100;//Get Hundred If the number of Narcissus conditions is printed out if (i== (Bite*bite*bite) + (Ten*ten*ten) + (hundred*hundred*hundred)) { System. out. Print (i+ ""); sum++; } } System. out. println ("Total number of daffodils:" +sum); } }

"Program 4"

Title: Decompose a positive integer into decomposition. For example: Enter 90, print out 90=2*3*3*5.

Program analysis: N to decompose decomposition, you should find a minimum prime number k, and then complete the following steps:

(1) If this prime number is equal to n, then the process of decomposing decomposition is finished and printed out.

(2) if n<>k, but n can be divisible by k, you should print out the value of K, and n divided by the quotient of K, as a new positive integer n, repeat the first step.

(3) If n cannot be divisible by K, then repeat the first step using k+1 as the value of K.

Public class Programme4 { Public static void main (string[] args) { System. out. Print ("Please enter a positive integer you want to decompose:"); Scanner scanner=New Scanner (System. in); int input=scanner.nextint ()//Get the number entered System. out. println (); System. out. Print (input+ "="); for (inti = 2; i < input+1; i++) { while (input%i==0&&input!=i) { input=input/i; System. out. Print (i+ "*"); } if (input==i) {//None of the above is divisible, which means it is a prime number System. out. Print (i); break; } } Scanner.close (); } }

"Program 5"

Topic: Use the nesting of the conditional operator to complete this problem: the students of the academic achievement >=90 score with a said, 60-89 points between the use of B, 60 points below the C-means.

Program Analysis: (a>b)? A:b This is a basic example of a conditional operator.

Public class Programme5 { Public static void main (string[] args) { System. out. println ("Please enter your score:"); Scanner scanner=New Scanner (System. in); int input=scanner.nextint ()//Get input Grade Judgment String belong=input>=90? " A ":(input>=60?" B ":" C "); System. out. println (input+ "Cent belongs to:" +belong); Scanner.close (); } }

"Program 6"

Title: Enter two positive integers m and n, and ask for their gcd and LCM.

Program Analysis: The use of rolling division.

* Here's a little bit of knowledge to remember, GCD and LCM.

* 1, first seek GCD bigdivisor

* 2, can be very convenient access to LCM Multiple=input1*input2/bigdivisor

* The most important thing here is to ask for GCD: The method is as follows

* (1) with a large number to the small number of the remainder

* (2) Assigning a small number to a large number, assigning the result of the remainder to a small number,

* (3) loop the previous operation until the result of the remainder is zero

* (4) The number of the last step is the gcd we want, if you don't believe it, you can try it.

Public class Programme6 { Public static void main (string[] args) { int bigdivisor=0;//definition gcd int multiple=0;//definition LCM System. out. println ("Please enter two integers:"); Scanner Scanner = new Scanner (System. in); int input1 = Scanner.nextint ();//Get First number int input2 = Scanner.nextint ();//Get second number multiple=input1*input2;//this value to save, after the convention number, to facilitate the LCM Intermediate number of int Temp =1;//Interchange if (Input2 >input1) {//Ensure that the first number is not less than the second number TEMP=INPUT1; Input1=input2; Input2=temp; } while (temp!=0) {//The remainder result is not equal to zero, it loops temp=input1%input2;//The result of finding the remainder input1=input2;//The big number is useless, with a small number to replace input2=temp;//assigns the result of the remainder to a small number } bigdivisor=input1;//the last time the result is zero, the number of the remainder Multiple=multiple/bigdivisor; System. out. println ("GCD is:" +bigdivisor); System. out. println ("LCM is:" +multiple); Scanner.close (); } }

"Program 7"

Title: Enter a line of characters to count their English letters, spaces, numbers, and other characters.

Program Analysis:

Here's the need for knowledge points:

1, get a line of strings, nextline ()

2. Assignments the value of each character of the string into a number

3, compare each number in the Ask code range, you can determine the category of its symbols

4, char character ask code range

(1) Number 0 to 9:48~57

(2) Letter A to z:65 to z:97 to 122

(3) The space is 32

Public class Programme7 { Public static void main (string[] args) { int num=0;//number of digits Number of int letter=0;//Letters int space=0;//number of spaces int others=0;//number of other System. out. println (&qu