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 Main
To get this topic, we will first think of using loops to complete.But not every operator is a "+" sign.Therefore, we are here to use (-1) of the I-side to do "+" "-" number control.The loop variable i is then treated as the denominator.Here we have the idea of the loop body is basically OK.It is important to note that the calculation results here are expressed in decimals, so it is not possible to define variables with int integers.The code is as foll
Note: When calculating 1 to use a double type that is 1.0 .
Odd even numbers are calculated separately and then merged.
#include
Label control +1,-1 with flag.
#include
Use the Function Pow Pow ( -1,i+1) equivalent ( -
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 ti
#include Be careful to define its type, divide it into two parts, and define it as "I" to see if the denominator is an odd or even number, and the sum is summed. C language: Calculate 1/1-1/2+1/3-1/4+
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
The authoritative guide for HTML5 and CSS3 (2nd edition) has become a benchmark in the field of HTML 5 and CSS 3 books, dubbed "one of the best guide books on system learning HTML 5 and CSS 3" and "one of the Web front-end Engineers ' desk books". The 2nd edition first updated and supplemented the content from the tech
There are many formulas for calculating pi pai in history, in which Gregory and Leibniz found the following formula:
Pai = 4* (1-1/3+1/5-1/7 ...)
The formula is simple and graceful, but in a bad way, it converges too slowly.
I
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 ("
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
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
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
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
*/
Pub
The content source of this page is from Internet, which doesn't represent Alibaba Cloud's opinion;
products and services mentioned on that page don't have any relationship with Alibaba Cloud. If the
content of the page makes you feel confusing, please write us an email, we will handle the problem
within 5 days after receiving your email.
If you find any instances of plagiarism from the community, please send an email to:
info-contact@alibabacloud.com
and provide relevant evidence. A staff member will contact you within 5 working days.