C/C++ :: Getting Divide By 0 Error When Using Modulus With Number Below 1
Jan 31, 2013
I am trying to get a remainder of a number with a = 0.9144, rm = ry % a; however, I keep getting a divide by zero error (I believe due to the program rounding 0.9144 down to the integer 0).
i was suppose to write a program that gives a divide error but output is not showing anything like it.I am using orwell devc++ 5.4.2 and all the default built tools(gcc compiler etc.) with default settings.so, what should i do to see this error?
Code:
#include<stdio.h> #include<conio.h> int main() { int a,b,c; float x; }
I am making a random number generator. I have fixed all issues except for one. I am supposed to ask the user for how many digits the user wants the numbers to have. How many numbers does the user want. Then randomly generate numbers according to what the user entered. So if the user said 2 digits and wanted 4 numbers then it would output 4 random numbers in the range of 10 to 99.
My notes from class show this working correctly. And it does work correctly. But I don't understand the math here or how the modulus effects it. I was able to get 1 and 2 digits to work but once I get to 5 it doesn't generate numbers correctly. it will only generate number s
Code: int min =1; int max = 9;int number1 = rand(); cout << number1 % max + min << " "; h
Here is the 3 digit code I have. I also need to figure out how to make it unique so no number generates more then once. I think the issue may be that the numbers are not unique and it is generating the same number and that is somehow effecting the numbers it is outputting. It is either that or my math is wrong.
Code: if (intLength == 5) { for (int i = 0; i<intQuantity; i++) {
how can i use the modulus operator to disect them so i can get the first number of each integer .what i mean is how can i take 1 form 1378 and 2 from 2496 etc...
I have to develop minimalistic implementation of RSA algorithm in C for an embedded device.
I'm doing that for two days but I have run into a problem. The N modulus is the limitation for the maximum message value to be encrypted with RSA.
For example theoretically RSA-1024 can encrypt/decrypt messages 1024 bits long but I still cannot understand how to choose p and q values to produce N == pow(2, 1024).
Is it possible to encrypt/decrypt 1024 bits long messages in practice if the N < pow(2, 1024)?
I liked that the above code does not put the result into a variable and then test the variable which would use more memory, and more lines of code. Is this thinking bad?
How to perform RSA decryption with just the modulus and public exponent. I know that in RSA the private key does the decrypt, but thats not what I need to do. It is actually more of a verify used to recover information for the purposes of integrity (stupid I know) but its the problem I have to solve. Basically I need to do an RSA operation with the modulus and public key. When I try it I receive the error:
System.Security.Cryptography.CryptographicException: Key does not exist. at System.Security.Cryptography.CryptographicException.ThrowCryptographicException(Int32 hr) at System.Security.Cryptography.RSACryptoServiceProvider.DecryptKey(SafeKeyHandle pKeyContext, Byte[] pbEncryptedKey, Int32 cbEncryptedKey, Boolean fOAEP, ObjectHandleOnStack ohRetDecryptedKey) at System.Security.Cryptography.RSACryptoServiceProvider.Decrypt(Byte[] rgb, Boolean fOAEP) at IssuerPublicKeyRecoveryApp.Form1.PerformVerification() in C:devMarvinDevIssuerPublicKeyRecoveryAppIssuerPublicKeyRecoveryAppForm1.cs:line 543.
My Code is as follows: try{ //Create a new instance of RSACryptoServiceProvider. using (RSACryptoServiceProvider RSA = new RSACryptoServiceProvider()) { //Build the RSA key parameters to perform decrypt RSAParameters RSAKeyInfo = new RSAParameters();
[Code] ....
I have verified that my values are good on : [URL] ....
This example is strait off the MSDN website, and I realize it says that you need the private key information on the RSA import. For what I am doing, I do not have that, since I am doing a decrypt(not really). Since I know that the Math of an RSA operation is actually identical weather you are doing encrypt or decrypt, and it is just a convention of length for the exponent, I tried to do an Encrypt operation to decrypt the data, and the error checking caught that it was the wrong length in stead of performing the operation.
I need to write a program that accepts a number from the user, divides that number by two until it reaches one or cannot be divided evenly anymore, then adds all of the quotients from each division and displays them.
So something like this should be displayed:
Please enter a number: 8
8/2=4 4/2=2 2/2=1 4+2+1= 7
I thought about using an array to possibly store the quotients but I just can't see how that would work.
Say I wanted to overload the modulus operator to return the remainder of a division between two floating point numbers. Why isn't a custom double operator%(double, double) allowed even though that function isn't available in the standard anyway?
I am new to C programming and I am just wondering how to multiply / divide two different variables which the user type in as the promt is asking like this:
Code: void inmatning3 (double *a, double *b) { printf("Mata in tv217 stycken flyttal: "); /* asks you to type in 2 numbers */ scanf("%lf %lf", a, b); }
When you've enterd the two numbers I need to eather multiply or divide the two variables "a" & "b" .....
I need to make a program that takes in a user’s number and keep dividing it by 2 until it is 1.I need to display the division steps ...
I do not know how to keep making it divide by 2 until 1. I tired this but it obviously did not work..
Code: #include <iostream> using namespace std; int main() { //User Input int input,total; cout <<"Enter A Number To Be Divided By 2 Until 1" << endl << endl;
I've been stuck on a divide a conquer algorithm problem for about an hour now, and I'm not sure how to solve it. I need to use divide-and-conquer to implement an algorithm that finds the dominant element of an array of positive integers and returns -1 if the array does not have a dominant element (a dominant element is one that occurs in more than half the elements of a given array).
No sorting may be used, and the only comparison that may be used is a test for equality.
I understand the general process I need to follow to solve this problem, but I'm not sure exactly how to convert my thoughts to code. I know that if a number x is the dominant element of an array A, the x must be the dominant element in either the first half of A, the second half of A, or both.
Here is what I have so far.
#include <iostream> #include <string> #include <vector> using namespace std; int find_dominant(int *A, int p, int r) {
I'm practicing so I wrote this simple program that suppose to add and divide two numbers. It does that but the result comes out with a 0 at the front and don't know why.
#include<iostream> using namespace std; int main() { int a ,b; int result; a = 0; b = 0;
The first line indicates how many paths exist in my design. The second and third lines are those lines(the first number indicates 2 pairs, so 10,1 and 2,7 are 2 pairs) The first element in the third line indicates 3 pairs and the pairs are 8,3 and 7,7 and 10,7.
I have been able to read the text file and store these into ONE array. However, I need to divide them up and have each line in a seperate array.
So for example
array alpha to contain 10,1 and 2,7 array beta to contain 8,3 and 7,7, and 10,7
For example, to calculate GPA, we add all the grade point of a course and divide by the number of courses we are taking. How do I create a program in such a way that when the user enters grade point for two courses, it will add the grade points for the two course and divide by 2. And if another user enters grade points for six courses, it adds the grade points and divides by 6.
The problem is that you have a set of numbers and you need to divide that set into two subsets where the difference between the sums of the subset is minimal.
Example: a set of numbers {1,5,9,3,8}, now the solution is two subsets, one subset with elements {9,3} and the other {8,5,1} the sum of the first one is 13 and the sum of the second is 13 so the difference between the sums is 0. The result shows the difference between the sums.
Another example: a set of numbers where the difference between the subsets cannot be zero, {9 51 308 107 27 91 62 176 28 6}, the minimal difference between the two subsets is 2.
I want to know how the function finds the two subsets, it works great because I've tested it for up to 300 inputs which sum adds up to 100,000.
Code: #include <iostream> #include <stdio.h> #include <math.h> #include <stdlib.h> #include <limits.h> using namespace std; int BalancedPartition ( int a[] , int n ) {
Create a program that adds, subtracts, multiplies, or divides two integers. The program will need to get a letter (A for addition, S for subtractions, M for multiplication, or D for division) and two integers from the user. If the user enters an invalid letter, the program should display an appropriate error message before the program ends. If the letter is A (or a), the program should calculate and display the sum of both integers. If the letter is S (or s), the program should display the difference between both integers. When calculating the difference, always subtract the smaller number from the larger one. If the letter is M (or m), the program should display the product of both integers. If the letter is D (or d), the program should divide both integers, always dividing the larger number by the smaller one."
And here is the test data. I am posting the results from my desk-check table.
operation first integer second integer answer A 10 20 30 a 45 15 60 S 65 50 15 s 7 13 6 G -1 M 10 20 200 d 45 15 3 d 50 100 2
Then, I transferred my program into a source file. Here it is:
//Exercise16.cpp - display answer of two integers
#include <iostream> using namespace std; int main() { //declare variables int firstInteger = 0;
[Code] ....
After putting in the data, everything worked fine, except the last two operations, which are M (multiplication) and D (division). All the answers for the last two operations essentially give me a 0.