下面程序的功能是输出1~100之间的所有偶数.程序: i=1 DO m=iMOD2 IF①THEN PRINTi ENDIF ② LOOPUNTILi>100 END (1)试将上面的程序补充完整; (2)改写为WHILE型循环结构程序. 相关知识点: 试题来源: 解析 (1)①m=0②i=i+1;(2)见解析 【分析】(1)如果除以的余数为零,则为偶数,故填.每次增加,...
(1)①m=0,②i=i+1.(2)改写为WHILE型循环程序如下:i=1 WHILE i=100 m=i MOD 2 IF m=0 THEN PRINT i END IF i=i+1 WEND END 【解题方法提示】(1)根据条件中的功能要求“偶数”可以写出①中的答案;根据条件输出1~100“所有”的偶数可以判断出②中的答案;(2)注意当型循环和直到型循环的区别...
Print the First 10 Prime Numbers in Python Using a While Loop Here, let me show you two methods to print the first 10 prime numbers using a while loop in Python. Method 1: Basic While Loop with Prime Check Function This method uses a while loop to iterate through numbers and a helper ...
下面的程序是输出1～100间的所有偶数,试补充完整,并改写为WHILE-WEND循环. 程序: i=1 DO IF①THEN PRINT i END IF ② LOOP UNTIL i>100 END 试题答案 在线课程 答案: 解析: 解:①i MOD 2=0或i/2=INT(i/2) ②i=i+1 改写为WHILE循环: ...
In this article, we will learn how to print a sequence of numbers in Java, ranging from 0 to 15. To do this, we'll use recursion rather than using loops like for loop or while loop. Recursion is a programming technique where a method calls itself to perform a sub-operation as ...
解析 B 程序初始化N=1和s=0后进入循环。每次循环中,s先累加当前的n值,随后n自增1。循环终止条件是n>100(即n=101时退出)。循环实际执行时,n的取值从1到100,每个值均被累加到s中。根据等差数列求和公式,1到100的和为100×101/2=5050。选项中5050对应B,故答案为B。
Loop While a>111Print aEnd Sub(A) 100 (B) 120(C) 201(D) 101 相关知识点: 试题来源: 解析D程序执行流程分析:1. 初始化时,`a`被赋值为100。2. 进入`Do...Loop While`循环: - **第1次循环**: - `s = s + a`:此时`s`的值由0变为100(假设`s`初始值为0)。
Question Transcribed Image Text:Q.3): Write a C++ program to print the following numbers using for loop: 1 10 100 1000 10000 2 20 200 2000 20000 3 30 300 3000 30000 4 40 400 4000 40000 5 50 500 5000 50000 Expert Solut...
and dedicated debugging tools might be more appropriate. additionally, for tasks that involve extensive automation or continuous integration, repl may not fit seamlessly into automated workflows. are there any security concerns related to using repl? while repl itself isn't inherently insecure, there ...
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