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Brilliant_brown [7]
2 years ago
7

Return 1 if ptr points to an element within the specified intArray, 0 otherwise.

Computers and Technology
1 answer:
Paladinen [302]2 years ago
4 0

Answer:

int withinArray(int * intArray, int size, int * ptr) {

      if(ptr == NULL) // if ptr == NULL return 0

            return 0;

      // if end of intArr is reached

      if(size == 0)

          return 0; // element not found

  else  

          {

            if(*(intArray+size-1) == *(ptr)) // check if (size-1)th element is equal to ptr

                   return 1; // return 1

            return withinArray(intArray, size-1,ptr); // recursively call the function with size-1 elements

         }

}

Explanation:

Above is the completion of the program.

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Answer:

Following are the algorithm to this question:

Explanation:

In the RSA algorithm can be defined as follows:  

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It's more evident as a fix for d provided of DE ≡ 1 (modulo \lambda (N) ).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 \lambda (N) will also be confined as they can be used in measuring D. The Euler totient operates \varphi (N) = (P-1)(Q - 1) 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 \varphi (N), which can always be split into  \lambda (N), and thus any D satisfying DE ≡ 1, it can also satisfy (mod  \lambda (N)). It works because \varphi (N), will always be divided by \varphi (N),. That d issue, in this case, measurement provides a result which is larger than necessary (i.e. D >   \lambda (N) ) 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<   \lambda (N). 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  \lambda (N).

As there are common threads (P− 1) and (Q – 1) which are present throughout the N-1 = PQ-1 = (P -1)(Q - 1)+ (P-1) + (Q- 1)), it's also possible, if there are any, for all the common factors (P -1) \ \ \ and \ \ (Q - 1)to become very small, if necessary.  

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