#include <stdio.h>
int main (int argc, char **argv) {
printf ("1 1 2 1 1 2 1 3 1 2 1 1 2 1 3 1 2 1 4 1 2 1 3 1 2 1\n");
}
a program to generate the sequence 1 4 9 16 25 is:
for (int i=1; i<=5; ++i) std::cout << i*i << ' ';
1, 12, 123, 1234, 12345, 123456, 1234567, 12345678, 123456789.
It keeps adding, but I'm not sure about the last 1 2 3 4 5 6 7.
It is not a program, you seem to have misunderstood something.
for (int i=1; i<=5; ++i) std::cout << i*i << ' ';
1, 10, 3, 9, 5, 8, 7, ,7 ,9, 6 What is next weries? 11, 5 10 5 10 4 11 6 Which one?
Source Code :: import java.io.*; class automorphic { protected static void main()throws IOException { BufferedReader in=new BufferedReader(new InputStreamReader(System.in)); System.out.print("Enter the number: "); int a=Integer.parseInt(in.readLine()),b=a,c=0,e=a*a; while(b>0) { c++; b/=10; } double d=Math.pow(10,c-1); if(a==e%d) System.out.println("Automorphic number!!"); else System.out.println("Not an Automorphic number!!"); }}
Perform encryption on the following PT using RSA and find the CT p = 3; q = 11; M = 5
You can use the symbol % to see the remainder of the division problem. eg. document.write(11 % 3 + "") will display the number 2 since 3 goes into 11 three times with a remainder of 2. So here's some sample code: function oddEven(num){ if(num % 2 == 0){ return "even" } else{ return "odd" } }
Decimal 30 = binary 11110. The decimal binary code (BCD), however, is 11 0000.
1,2,3,4,5,6,7,8,9,0,10,11,12,13,14,15
10 because it's an even number !
11 out of 10 11 10
Those two networks only program until 10pm. After 10, local affiliates program the schedule.
11 = 11
11
For 8 9 9 9 10 11 11 12: σ=1.3562
In retrospect, no.
This sort code belongs to Halifax. This code is assigned to the Solihull High Street branch.
This sort code belongs to Halifax. This code is assigned to the Bromley High Street branch.
11
what is the mode of the following data 18, 17, 12, 14, 8, 21, 10, 11, 19, 20, 10, 5, 17, 12, 10, 20