import java.util.Scanner;
public class JavaApplication70 {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
System.out.println("Input a String:");
String txt = scan.nextLine();
System.out.println("Input an integer:");
int num = scan.nextInt();
String newTxt = "";
int w = 0;
for (int i = txt.length()-1; i >= 0; i--){
char c = txt.charAt(i);
while (w < num){
newTxt += c;
w++;
}
w = 0;
}
System.out.println(newTxt);
}
}
I hope this helps!
Answer:
In the presence of packets transfer between client and server in a TCP session, the attack will be successful if the number of the sequence is approximately X+100. Otherwise, the attack will not be successful (i.e. fail).
Explanation:
Generally, it is important to ensure that there is a successful operation in the operation of a TCP session. In the presence of packets transfer between client and server in a TCP session, the attack will be successful if the number of the sequence is approximately X+100. Otherwise, the attack will not be successful (i.e. fail).
Answer:
#include <iostream>
using namespace std;
int main()
{
char fullname[30];
string fname="",lname="";
int i,j;
cout<<"Enter fullname\n";
cin.getline(fullname,30); //so that blank can be read
for(i=0;fullname[i]!=' ';i++)
fname+=fullname[i]; //fistname will be saved
cout<<"\n";
for(j=i;fullname[j]!='\0';j++)
lname+=fullname[j]; //lastname will be saved
cout<<"\nFirstname : "<<fname<<"\nLastname : "<<lname;
return 0;
}
OUTPUT :
Enter fullname
John thomson
Firstname : John
Lastname : thomson
Explanation:
cin.getline() should be used instead of cin in case of strings so that space can be read otherwise after blank string will be ignored.
