How can I print after z again abcd like that?

Question

1.00/5 (1 vote)

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Accept a character C and a positive integer N as input. The program must print N characters starting from C.

Boundary Condition(s):
1 <= N <= 100

Input Format:
The first line contains C and N separated by space(s).

Output Format:
The first line contains N characters.

Example Input/Output 1:
Input:
a 4

Output:
abcd

Example Input/Output 2:
Input:
z 5

Output:
zabcd

What I have tried:

#include<stdio.h>
#include <stdlib.h>

int main()
{
    char a;
    int n,b=0;
    scanf("%c",&a);
    scanf("%d",&n);
    for( char i=a;b<n;i++){
        printf("%c",i);
        b=b+1;
    }
}

Posted 22-May-18 19:51pm

KaranKumar P

Updated 22-May-18 22:23pm

User 7429338

v2

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Comments

[no name] 23-May-18 2:02am

I have changed your question tag to C instead of Java, based on the code you posted.

KaranKumar P 23-May-18 2:14am

yes sorry

3 solutions

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User 7429338 · Answer 1 · 2018-05-22T20:16:00

Solution 1

Since there are 26 letters in the alphabet, you can use % 26 to start from the beginning again after z.
The character 'a' is not 0 however, so you need to subtract it first and then later re-add it.

For example:

for (int i = 0; i < n; i++) {
    printf("%c", (a - 'a' + i) % 26 + 'a');
}

Posted 22-May-18 20:16pm

User 7429338

CPallini · Answer 2 · 2018-05-22T21:28:00

Solution 2

C

for (i=0; i<n; ++i)
{
  putchar(a);
  a = (a == 'z') ? 'a' : (a + 1);
}

Posted 22-May-18 21:28pm

CPallini

Patrice T · Answer 3 · 2018-05-22T22:24:00

Having 2 variables in the loop os basically the idea, but your problem is the you are mixing their usage.
Keep their usage separated, have 1 variable to count printed letters, and the other variable to keep track of the letter to print.

C++

for( int i=0;i<n;i++){ // count here
    printf("%c",a);
    if (a<'z') // set next letter here
        a=a+1;
    else
        a='a';
}