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
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 maximum number, appears several times * advantage:
/***//**
* 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
=
2
, F, Sum
=
0
;Scanf (
# 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
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 ("" + I + "); If (! Maxstep) {if (I % 2 = 1) {s
Content on the machine: accumulate data using cyclic statements. Objective: To learn how to use cyclic statements. /** Copyright (c) 2012, Emy of computer science, Yantai University * All rights reserved. * Prepared by: Li Yang * completion date: January 1, November 01, 2012 * version No.: v1.0 * enter the description: none. * Problem description: the result of 1/3-3/5
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,
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
Package COM. WZS; // Add difficulty to the first question. Use the numbers 1, 2, 3, 4, and 5 to write a main function in Java and print out all the different orders, // For example, 51234, 12345, etc., the requirement: "4" cannot be In the third place, "3" and "5" cannot be connected. Public class test3 {public static
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], so a plus 1 is actually a + sizeof (int)
Exercise 3-3 the last three bits of the productEnter several words, enter a number of integers (which can be positive, negative, or 0), and output the last three bits of their product. These integers are mixed with a string of uppercase letters, and your program should ignore them. Tip: Try to enter a string when executing scanf ("%d").#include Summary: 1 Note overflow2 for a break in the bad, the next code
Chapter 5 data-centric-Data Access (3), Chapter 5 Access5.1.3 free operation-serialization and deserialization
In Android development, apart from common text and XML files, data is often transmitted or accessed through serialization and deserialization.
Android serialization objects can be implemented in two ways: Serializable interface or Parcelable interface. T
PHP 5 provides sample code for managing any database tables in MySQL 5 (3). Read the sample code for managing any database tables in MySQL 5 (3). continue: click edit to jump to edit. php // edit. phphtmlheadtitleEditinganentryfromthedatabase/titlemetahttp-equ
Continued: cli
Problem description: there are now n ordered arrays in M groups, such as {1, 2, 3, 4}, {2, 3, 6}, {1, 3, 5, 7 }, select the data smaller than K in these arrays and return this value.
Idea: Compare the minimum data selected each time by referring to the process of merging two Arrays
1. Define the selection position arra
); path[] paths =NewPath[args.length]; for(inti =0; i NewPath (Args[i]); } filestatus[] status = Fs.liststatus (paths); path[] Listedpaths = fileutil.stat2paths (status); for(Path p:listedpaths) {System. out. println (P); } }}Z) rather than have to enumerate each file and directory to specify the input, it's convenient to use wildcard characte RS to match multiple files with a single expression, an operation, which is known as globbing. Hadoop provides FileSystem methods for processing
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