Create a thread corresponding to the type, three threads are processed separately.Call three different threads, cycle through 5 numbers, wake up other dormant threads after printing, and hibernate yourself.1 PackageThread;2 /**3 * Three threads in turn print 1-75, one thread prints 5 numbers at a time4 * @authorAdmin

Sword refers to the Java Implementation of offer programming questions-interview question 5 prints the linked list from start to end and offer from end to endTopic Description * sword refers to offer interview question 5: print the linked list from the end to the endEnter the head node of a linked list and print the value of each node from the end to the end.
Sol

Package com.day3.one;public class PrimeNumber1 {/*** @param args* Print a prime number between 101-200 and count, and one line per 5 output*/public static void Main (string[] args) {int count=0;for (int m=101;m{Boolean a=true;for (int i=2;i{if (m%i==0){A=false;Break}}if (a==true){System.out.print (m+ "");count++;}if (count%5==0)System.out.println ();}System.out.p

1 ImportJava.util.Scanner;2 3 Public classdayin_100 {4 Public Static voidMain (string[] args) {5System.out.println ("Enter the first number of integers you want to print:");6Scanner San1 =NewScanner (system.in);7System.out.println ("Please enter the mantissa of the integer to be printed:");8Scanner San =NewScanner (system.in);9 intNUM1 =San1. Nextint ();Ten intNum =San. Nextint (); One Dayin (num1,num); A } - Public Stat

It is not very difficult to see an algorithm question on the Internet. There are also solutions for searching, but there are usually several layers of for loops. I tried to write it down.
/*** Give you a set of strings such as {5, 2, 3, 2, 4, 5,}, so that you can output the maximum number of occurrences and the maximu

Package com; public class test {public static void main (string [] ARGs) {system. out. println (getsteps1 (); system. out. println (">>>>>>>>>>>>>>>>>>>>>>>>>>>>"); system. out. println (test. getsteps (); system. out. println (">>>>>>>>>>>>>>>>>>>>>>>>>>>> ");} public static int getsteps () {// use the minimum public multiple to reduce the number of traversal times. Int I = 1; int step = 2; Boolean maxstep = false; while (true) {system. Out. Print ("

There is a fractional sequence of 2/1 + 3/2 + 5/3 + 8/5 + 13/8 +... Find the sum of the first 20 items of the series.Program:
# Include
Output result: 32.660261 Press any key to continue

C language: There is a score sequence: 2/1 + 3/2 + 5/3 + 8/5 + 13/8 +... Find the sum of the first 20 items in this seriesProgram:
# Include
Output result: 32.660261 Press any key to continue

Silicon Valley Mall version 2 5-personal center module, Silicon Valley Mall version 2 5 --
Fragment_user.xml
1. The title header uses GradationTitleBar to implement the gradient effect of the title bar of the imitation QQ space;

Int a [5] = {1, 2, 3, 4, 5}; printf ("% d \ n", * (int *) ( a + 1)-2 );, printf % d
What is the result of a certain convincing pen question in a certain year? The answer is 4. Why?
My understanding (do not know if it is correct ):
A is an array pointer of the int type [5

2-5-single-chain table for merge chain storage-linear table-Chapter 2nd-source code of the data structure textbook-yan Weimin Wu Weimin edition, 2-5-Data Structure
Textbook source code
Chapter 2 linear table-merge a single-chain table (Chain Storage)
-- Data Structure-yan We

/***//**
* Fractionserial. Java
* There is a fractional sequence: 2/1, 3/2, 5/3, 8/5, 13/8, 21/13...
* Calculate the sum of the first 20 items of the series.
* @ Author Deng Chao (codingmouse)
* @ Version 0.2
* Development/test environment: jdk1.6 + eclipse SDK 3.3.2
*/
Public class fractionserial ...{
Public static

Calculate the sum of the First n items of the Fibonacci fractional sequence (N is a constant, and the Fibonacci fractional sequence is 2/1, 3/2, 5/3, 8/5 ,...).
# Include
Stdio. h
>
# Include
Conio. h
>
Void
Main (){
Int
I, N;
Float
F1
=
1
, F2
=

# Include
}/* The numerator behind the score is equal to the numerator plus the denominator of the previous score, and the denominator of the subsequent score is equal to the numerator with the previous score */
There is a fractional sequence: 2/1, 3/2, 5/3, 8/5, 13/8, 21/13... find the sum of the first 20 items of

/**5-1* Define interface printable, which includes a method Printitmyway (),* This method has no formal parameters and the return value is null**/Interface Printable{void Printitmyway ();}/**5-2* Rewrite the rectangle class in experiment 3 to implement the printable interface,* Use the Printitmyway () method to relate information about the rectangle (length, widt

Starting 3 threads, thread 1 printing 1 to 5, thread 2 printing 5 to 10, thread 3 printing 11 to 15, then thread 1 printing 16 to 20, and so on ...
Print until 30 public class Mainthread {private static int num;//current record number private static final int threadnum =3;//open Task thread count private static final int loopnum = number of

In order to omit the space and make it visible to the operator who manually fills in the paper, the volume number on the card of the materials shelf is determined to be classified and sorted, as shown in
A-3, A-4, A-5, A-8 forming A-3 ~ 5, 8, etc.
The following code uses a few auxiliary list
/// /// Similar to 1, 2, 3, 5

View code
//// Main. M // money /// enter a specified amount (in Yuan, for example, 345.78) from the keyboard, and then display the number of different denominations that pay the amount, required to display 100 yuan, 50 yuan, 10 yuan, 5 yuan, 2 yuan, 1 yuan, 5 cents, 1 cent, 5 points, 1 cent each how many sheets // cr

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