Answer:
#include<stdio.h>
#include<string.h>
int main(){
char str[20];
int i=0;
printf("Enter a name: ");
gets(str);
printf("%c",*str);
while(str[i]!='\0'){
if(str[i]==' '){
i++;
printf("%c",*(str+i));
}
i++;
}
return 0;
}
Explanation:
d would be your answer....................................
Answer:
very
Explanation:
people pay more attention and learn better in one on one conversations than in groups . I believe we would have more understanding persons in the world and also the world would be a better place
Answer:
The solution code is written in Java
- public static void checkCommonValues(int arr1[], int arr2[]){
- if(arr1.length < arr2.length){
- for(int i = 0; i < arr1.length; i++){
- for(int j = 0; j < arr2.length; j++){
- if(arr1[i] == arr2[j]){
- System.out.print(arr1[i] + " ");
- }
- }
- }
- }
- else{
- for(int i = 0; i < arr2.length; i++){
- for(int j = 0; j < arr1.length; j++){
- if(arr2[i] == arr1[j]){
- System.out.print(arr2[i] + " ");
- }
- }
- }
- }
- }
Explanation:
The key idea of this method is to repeated get a value from the shorter array to check against the all the values from a longer array. If any comparison result in True, the program shall display the integer.
Based on this idea, an if-else condition is defined (Line 2). Outer loop will traverse through the shorter array (Line 3, 12) and the inner loop will traverse the longer array (Line 4, 13). Within the inner loop, there is another if condition to check if the current value is equal to any value in the longer array, if so, print the common value (Line 5-7, 14-16).