12345678910#include <stdio.h> int main() { int i = 1, j = 5, result = 0; while (i <= j && result <= 40) { result += i * (j - i + 1); printf("%d ", result); i++; } return result; }
Question
12345678910#include <stdio.h> int main() { int i = 1, j = 5, result = 0; while (i <= j && result <= 40) { result += i * (j - i + 1); printf("%d ", result); i++; } return result; }
Solution
The code you've posted is written in C language. It's a simple program that uses a while loop to perform a calculation and print the result. Here's a step-by-step explanation:
-
The program starts with the inclusion of the standard input-output library
stdio.h. -
The
main()function is declared, which is the entry point of the program. -
Inside the
main()function, three integer variablesi,j, andresultare declared and initialized.iis set to 1,jis set to 5, andresultis set to 0. -
A while loop is started, which will continue as long as
iis less than or equal tojandresultis less than or equal to 40. -
Inside the loop,
resultis incremented by the product ofiand(j - i + 1). -
The value of
resultis then printed to the console. -
The variable
iis incremented by 1. -
The loop continues until
iis greater thanjorresultis greater than 40. -
The function then returns the value of
result. -
The program ends.
The output of this program will be the values of result at each iteration of the loop, printed on the console.
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