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(Two new ideas about an algorithm question) give you a set of strings, such as {5, 2, 3, 2, 4, 5, 1, 5}, so that you can output the one with the most occurrences and the largest number, appears several times

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

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

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 reco

Using regular expressions to implement the Operation Express = ' 1-2* ((60-30 + ( -40/5) * (9-2*5/3 +7/3*99/4*2998 +10 *568/14))-( -4*3)/(16-3*2)) '

#!/usr/bin/env python# Coding:utf-8Import Redef Dealwith (Express): Express.replace ('+-','-') Express.replace ('--','+') returnexpressdef Col_suanshu (exp):if '/' inchexp:a,b= Exp.split ('/') returnStrfloat(a)/float(b))if '*' inchexp:a,b= Exp.split ('*') returnStrfloat(a) *float(b) def get_no_barcate (Express): Express=express.strip ('()') Print ('>>>', Express) whileTrue:ret= Re.search ("-?\d+\.? \d*[*/]-?\d+\.? \d*", Express)ifRet:res=Col_suanshu (Ret.group ()) Express= Ex

Int a [5] = {1, 2, 3, 4, 5}; printf (& quot; % d \ n & quot;, * (int *) (& amp; a + 1)-2);, printf % d

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

C language computing 1/1-1/2 + 1/3-1/4 + 1/5-... + 1/99-1/100

C language computing 1/1-1/2 + 1/3-1/4 + 1/5-... + 1/99-1

Implemented in C: Calculate 1/1-1/2+1/3-1/4+1/5 ... + 1/99-1/100 value.

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

The "C language" calculates the value of 1/1-1/2+1/3-1/4+1/5 ... + 1/99-1/100.

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 ( -

Calculate 1/1-1/2+1/3-1/4+1/5 ... + 1/99-1/100 value

#include Calculate 1/1-1/2+1/3-1/4+1/5 ... + 1/99-1

Those years, learn together Java 5-1 5-2

/**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

Thunder programming questions: programming: Find a number in addition to 2 + 1 In addition to 3 + 2 in addition to 4 + 3 in addition to 5 + 4 in addition to 6 + 5 in addition to 7 + 0

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

Use the for and while loops to calculate the value of e [e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + 1/5! +... + 1/n!], While

Use the for and while loops to calculate the value of e [e = 1 + 1/1! + 1/2! + 1/3! + 1/4! + 1/5

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.

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 series

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

1, 2, 3, 5, 7, 8, 10, 11, 12, 13, 14, 15, 16, 21, 22-"1 ~ 3, 5, 7 ~ 8, 10 ~ 16,21 ~ 22

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

Enter a specified amount (in Yuan, such as 345.78) on the keyboard, and then display the number of different denominations paid for this amount, required to display 100 yuan, 50 yuan, 10 yuan, 5 yuan, 2 yuan, 1 yuan, 5 cents, 1 cent each

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

C language: Calculate 1/1-1/2+1/3-1/4+1/5 ... + 1/99-1/100 value

#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+

Obtain the fractional sequence: 2/1, 3/2, 5/3, 8/5, 13/8, 21/13... The sum of the first 20 items

/***//** * 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

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 ,...)

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 =

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 this series

# 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

Use 1, 2, 3, 4, and 5 to form a five-digit repeat. 4 cannot be In the third place, and 3 cannot be connected to 5 (the simplest way)

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

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