if 5 3 2

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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,

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 (">>>>>>>>>>>>>>>>>>>>>

> 1. For example, if n is 60, the output is 2 2 3 5. Add the missing parts.

/* Factorization is a very basic mathematical operation and is widely used. The followingProgramReturns the factorization of integer n (n> 1. For example, if n is 60, the output is 2 2 3 5. Add the missing parts. */Public class factorization {public

There are now n ordered arrays in the M group, such as {1, 2, 3, 3}, {2, 3, 4, 6}, {1, 3, 5, 7}. In these arrays, select the data smaller than K, then return this value

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

(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

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,},

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 +

Question 5: f = 1! -2! + 3! -4! +... + N! (N is a large number. If n is too large, it will overflow)

/*************************************** ************************Accumulated (C language)AUTHOR: liuyongshuiDATE :************************************************ ***********************//*Question 5: f = 1! -2! + 3! -4! +... + N! (N is a large

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 ): & A is an

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,

Multiples of 2 3 5 7

Topic Five, 2 3 5 7multiples ofGiven a number n, ask 1 to N, how many are not multiples of 2 3 5 7. For example, n = 10, only 1 is not a multiple of 2 3 5 7.InputEnter 1 number n (1 OutputThe output is not the number of multiples of 2 3 5 7.Input

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