In this HackerRank For loop in c programming problem solution, In this challenge, you will learn the usage of the for loop, which is a programming language statement which allows code to be executed until a terminal condition is met. They can even repeat forever if the terminal condition is never met.
The syntax for the for loop is:
for ( <expression_1> ; <expression_2> ; <expression_3> )
<statement>
expression_1 is used for intializing variables which are generally used for controlling the terminating flag for the loop.
expression_2 is used to check for the terminating condition. If this evaluates to false, then the loop is terminated.
expression_3 is generally used to update the flags/variables.
The following loop initializes i to 0, tests that i is less than 10, and increments i at every iteration. It will execute 10 times.
for(int i = 0; i < 10; i++) {
...
}
Task
For each integer [a,b] in the interval (given as input) :
- If 1 <= n <= 9, then print the English representation of it in lowercase. That is "one" for 1, "two" for 2, and so on.
- Else if n > 9 and it is an even number, then print "even".
- Else if n > 9 and it is an odd number, then print "odd".
HackerRank For loop in c programming problem solution.
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int a, b; scanf("%d\n%d", &a, &b); char labels[11][6] = {"one", "two", "three", "four", "five", "six", "seven", "eight", "nine", "even", "odd"}; int labels_index; for (int i=a; i<=b; i++) { labels_index = i <= 9 ? i - 1 : 9 + i % 2; printf("%s\n", labels[labels_index]); } return 0; }
Second solution
#include <stdio.h> #include <string.h> #include <math.h> #include <stdlib.h> int main() { int a, b; scanf("%d\n%d", &a, &b); // Complete the code. for (int i=a; i<b+1; i++) { switch(i) { case 1: printf("one\n"); break; case 2: printf("two\n"); break; case 3: printf("three\n"); break; case 4: printf("four\n"); break; case 5: printf("five\n"); break; case 6: printf("six\n"); break; case 7: printf("seven\n"); break; case 8: printf("eight\n"); break; case 9: printf("nine\n"); break; default: if (i % 2) printf("odd\n"); else printf("even\n"); } } return 0; }
3 Comments
anyone can explain this problem for me guys pls pls
ReplyDeletewhat if the b value is smaller than value a ? the loop won't run isn't it?
ReplyDeleteIn the second solution
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