Greatest string.

Problem:-

You are given a string S and an integer Q. You are allowed to perform at most Q operations on the string. In one operation, you can change any vowel to it's next character (e.g., 'a'->'b', 'e'->'f', 'i'->'j', 'o'->'p', 'u'->'v'). Generate the lexicographically greatest string by performing at most Q operations on string S.
Note- Vowels in English alphabet are- 'a','e','i','o','u'.
Input Format:
First line contains an integer T denoting the number of test cases .
For each test case,in first line you will be given the string S and in second line an integer Q (maximum number of operations allowed).
Output Format:
For each test case , print the lexicographically greatest string that can be formed after applying at most Q operations on the given string.
Answer for each test case should come in a new line.
Constraints:
$1\le T\le 10$
$1\le |S|\le {10}^5$
$0\le Q\le {10}^5$
String will consist of only lowercase English alphabets.

SAMPLE INPUT

 

2
abcde
3
xyzwu
0

SAMPLE OUTPUT

 

bbcdf
xyzwu

Explanation

For case 1:
We have string "abcde" and we are allowed to perform at max 3 operations, we can form lexicographically greatest string by applying the operation on first and last character of string by changing the string to "bbcdf",which is lexicographically greatest.

For Case 2:
We are not allowed to do any operations, so the answer will be the string itself.
 

Time Limit:1.0 sec(s) for each input file.

Memory Limit:256 MB

Source Limit:

solution:-

#include<stdio.h>

void main()

{

     int t,q,k=1;

     char s[100000];

     scanf("%d",&t);

     for(int i=1;i<=t;i++)

     {

        scanf("%s",s);

        scanf("%d",&q);

        for(int j=0;s[j];j++)

        {

            if(k>q)

             break;

            else

            {

                int x;

                x=s[j];

                if(s[j]=='a'||s[j]=='e'||s[j]=='i'||s[j]=='o'||s[j]=='u')

                {

                x=x+1;

                s[j]=x;

                k++;

            }}

        }

        printf("%s\n",s);

        k=1;

     }

}