What is wrong with line 1?
1 answer:
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Answer:
To check if the year comes under each 100th year, lets check if the remainder when dividing with 100 is 0 or not.
Similarly check for 400th year and multiple 0f 4. The following C program describes the function.
#include<stdio.h>
#include<stdbool.h>
bool is_leap_year(int year);
void main()
{
int y;
bool b;
printf("Enter the year in yyyy format: e.g. 1999 \n");
scanf("%d", &y); // taking the input year in yyyy format.
b= is_leap_year(y); //calling the function and returning the output to b
if(b==true)
{
printf("Thae given year is a leap year \n");
}
else
{
printf("The given year is not a leap year \n");
}
}
bool is_leap_year(int year)
{
if(year%100==0) //every 100th year
{
if(year%400==0) //every 400th year
{
return true;
}
else
{
return false;
}
}
if(year%4==0) //is a multiple of 4
{
return true;
}
else
{
return false;
}
}
Explanation:
Output is given as image
Answer:
Following are the algorithm to this question:
Explanation:
In the RSA algorithm can be defined as follows:
In this algorithm, we select two separate prime numbers that are the "P and Q", To protection purposes, both p and q combines are supposed to become dynamically chosen but must be similar in scale but 'unique in length' so render it easier to influence. Its value can be found by the main analysis effectively.
Computing N = PQ.
In this, N can be used for key pair, that is public and private together as the unit and the Length was its key length, normally is spoken bits. Measure,
where
is the total function of Carmichaels. It is a privately held value. Selecting the integer E to be relatively prime from
and
that is
. D was its complex number equivalent to E (modulo
); that is d was its design multiplicative equivalent of E-1.
It's more evident as a fix for d provided of DE ≡ 1 (modulo ).E with an automatic warning latitude or little mass of bigging contribute most frequently to 216 + 1 = 65,537 more qualified encrypted data.
In some situations it's was shown that far lower E values (such as 3) are less stable.
E is eligible as a supporter of the public key.
D is retained as the personal supporter of its key.
Its digital signature was its module N and the assistance for the community (or authentication). Its secret key includes that modulus N and coded (or decoding) sponsor D, that must be kept private. P, Q, and will also be confined as they can be used in measuring D. The Euler totient operates
however, could even, as mentioned throughout the initial RSA paper, have been used to compute the private exponent D rather than λ(N).
It applies because , which can always be split into
, and thus any D satisfying DE ≡ 1, it can also satisfy (mod
). It works because
, will always be divided by
,. That d issue, in this case, measurement provides a result which is larger than necessary (i.e. D >
) for time - to - time). Many RSA frameworks assume notation are generated either by methodology, however, some concepts like fips, 186-4, may demand that D<
. if they use a private follower D, rather than by streamlined decoding method mostly based on a china rest theorem. Every sensitive "over-sized" exponential which does not cooperate may always be reduced to a shorter corresponding exponential by modulo
.
As there are common threads (P− 1) and (Q – 1) which are present throughout the , it's also possible, if there are any, for all the common factors
to become very small, if necessary.
Indication: Its original writers of RSA articles conduct their main age range by choosing E as a modular D-reverse (module ) multiplying. Because a low value (e.g. 65,537) is beneficial for E to improve the testing purpose, existing RSA implementation, such as PKCS#1, rather use E and compute D.