/* Check numbers of arguments, and read input */
validInput = (argc = 4);
[Code] ....
Is a section of the code (the first section). And as you can probably guess, it goes on to calculate for a > 0 etc...
I dont really understand what the validinput section is saying? And a, b and c are never defined so Xcode is just saying a,b,c,root1,root2 are uninitialized and I also dont know what means. Do I need to define these values?
Code:
int main(int argc, char *argv[]) {
int validInput, solution_type;
double a, b, c, root_1, root_2, q;
The main point of the program is to calculate equations just like a standard calculator but I wanted to do it myself. I don't understand what the problem is right now but I've managed to create a program that asks for both values but somehow it doesn't want to ask for an operator (*, /, + etc). What's wrong with my code that the terminal skips the scanning part for the operator?
Code:
#include <stdio.h> int main() { int value1, value2, answer; char operator;
I'm trying to write a code that will give me two solutions one for x and one for y.
The equation is the following x^2+4=y^3
The solution for y is y=(x^2+4)^1/3
The solution for x is x=(y^3+4)^1/2
The answer needs to printed like that in upper section. I've written a bad code, cause I'm not good at programing. But never the less it's a start, i think.
#include <iostream> using namespace std; int main () { int a; //replaceable int b; //replaceable int ind_a=2; //upper index of a int ind_b=3; //upper index of b
Code: #include <math.h> #include <stdio.h> int main(void) { float a,b,c,root_1,root_2; printf("Please enter value a from the quadratic equation
[Code] ......
And I keep getting this error:
Code: /tmp/ccgtUIun.o: In function `main': assign345.c:(.text+0xc7): undefined reference to `sqrt' assign345.c:(.text+0xef): undefined reference to `sqrt' collect2: ld returned 1 exit status
We are suppose to build a program to do the quadratic formula, which isnt really a issue. my issue is i have a if else loop and my if is being ignored.
note: i know that i can use the i number system but we have been asked to not too and to do this instead.
here is my code
#include <stdio.h> #include <math.h> int main() { float a, b, c, d, e, f, g, h, i; /* i would love to figure out a way to do this without so many variables*/
I need to find out all the possible equations which are same of a given equation. For eg. a+b+c-d and b+c+a-d are same. So if the user inputs a+b+c-d then the output will be all the possible equations that can be formed from the given equation, viz:
a+c+b-d c+a+b-d -d+a+c+b
etc
i.e the program should rearrange the operands and operators in such a way that it should give the same result.
I am creating a program that solves the quadratic equation ax^2 + bx +c.
I have this program almost complete except the output of the equation in the function called display_quadratic. I need the program to display the variables a,b,c in the equation ax^2 + bx + c but I am having 2 problems. My first problem is that I cannot get the right addition and subtraction signs for the equation.
For instance, if the program had the values for a,b,c as 2,2,3
it will display 2x^2 2x 3
How can I get it to display 2x^2 + 2x + 3? or if it was negative like 2x^2 - 2x - 3?
My next question is how do I get to not display the coefficients that are 1?
I had an if-else statement but no matter what I created it would overlap with another statement and print out twice. Here is the code:
This is my code for the quadratic equation. It keeps telling me that my else is illegal since no matching if statement and my else statement is missing a statement
#include <iostream> #include <cmath> #include <string> #include <iomanip> using namespace std; int main() { string Name;
Program which accepts two lines and and determines their intersection point and whether they lie within a circle, also given interactively. I'm racing against time and I've racked my skull to no avail
write codes that could solve a quadratic formula, and my codes are like this:
#include <bjarne/std_lib_facilities.h> int main() { cout << "Enter the coefficients of a quadratic polynomial a*x**2 + b*x +c: "; cout << " a? "; double a; cin >> a; cout << " b? ";
[code]....
Which runs perfectly, but I have 2 questions:
1. How to simplify these code? On the assignment sheet the professor wrote about using void solve_linear(double b, double c); and void solve_ quadratic (double a, double b, double c);which I currently dont understand how these works. He asked us to write a well-encapsulated (as short as possible) program.
2. These are for extra points: the precision problem of floating numbers: professor asked us to find a way to get the precise answer of it, like this: Enter the coefficients of a quadratic polynomial a*x**2 + b*x +c: a? 1 b? -20000 c? 1.5e-9 Trying to solve the quadratic equation 1*x*x + -20000*x + 1.5e-09 == 0 Using classical formula: Two roots, x = 20000 and x = 0 Using stable formula: Two roots, x = 20000 and x = 7.5e-14
My guess is that A. Using code like if(x1*x2=a/c) to check if numbers were approximated. B. Somehow determine the larger one in x1 and x2. C. Somehow use that larger one to do something
So, I successfully made a program that will perform the quadratic equation on three numbers, imaginary or real. however, i am now trying to simplify the result, as to get rid of the "/2a" on the bottom. Hence the simplify() function. I just started to create the simplification function, and am attempting to divide the imaginary part of the solution as well as the real part of the solution by 2a. Somehow, it gives the error, "error:invalid operands of types 'int' and 'double *' to binary 'operator*'" on lines 105 and 106. I suspect it has to do with the pointers and references that i am passing as parameters. Also, just an aside, I have never actually seen "/=" be used. It can be, right? I know "+=" can be.
#include <iostream> #include <string> #include <cmath> #include <cstdlib>//simplify the answer using namespace std; int count=0; //prototyping double ans_1(double,double,double);
I wrote a program that solves an equation of two numbers, but in addition to that, I want it to be able to continue to solve longer equations. Ex: ( solves 2 * 4, or 2 * 4 - 5).I want to put this part of the program's result into a variable and go from there. How do I place the result of the calculation into a variable, and where would it go?
Code:
#include <stdio.h> #include <string.h> int main(void) { int a; scanf("%d", &a); char s[2]; scanf("%s", s); int b; scanf("%d", &b); }
I have to make a numerical integration program, how I can write my code so that the user is able to write their own function that they want to integrate?
E.g. they would see the message: 'please enter your function' and would be able to write whatever they wanted e.g. 'x +5' then this would then be integrated by the program.
I have already written a program that can integrate a known function but would prefer that the user could choose their own.
I am trying to write a code that solves a system of linear equations such as A*B=C. My system has a dimension equal to 1600. The matrix A cab be separated into 4 sub matrices and each can be handled by a different thread. I tried to solve this using the following code:
int main() { int count = 0; //Inputing matrix A ifstream matrix; matrix.open("example.txt");
[Code] ....
Although the above code gives the correct answer, the time needs to find the solution is bigger than that needed without using threads.
Calculated by the explicit scheme. Produces some very large numbers.
task:
[math] U_t = 3 (1,1-0,5 x) U_ {xx} + e ^ t-1 [/ math] [math] U (0, t) = 0 [/ math] [math] U (1, t) = 0 [/ math] [math] U (x, 0) = 0.01 (1-x) x [/ math]
Need to find a solution with accuracy [math] 0.0001 [/ math] on the interval [math] T = 1 / a ^ *, where a ^ * = max a (x, t) [/ math] Plot graphs of functions [math] u (x ^ *, t), u (x, jt ^ *) [/ math] where [math] x ^ * = 0.6, t ^ * = T/10, j = 1,2,4 [/ math]
I have a problem: solve the system of equations by the Gauss-Jordan methods with pthreads. I have a big matrix A (for example 2000x2000), so i must solve Ax = b. Before I devide matrix to number of threads and each thread must work only with his peace of matrix.
Code: #include <iostream> #include "synchronize.h" #include <pthread.h> using namespace std; typedef struct _ARGS { int thread_count; int thread_number;
[Code] .....
I write it on Ubuntu, and when I compile [g++ main.cpp -o main -lpthread -lm] the it works good(for example, if in 1 thread I get time_of_working = 10 sec, on 2 threads time_of_working = 5.4, i.e. about 2 times faster ), if I compile like this [g++ main.cpp -o main -lpthread -lm -O3] it is only 1.2-1.3 times faster.
I'm trying to write a program to calculate an approximate value of e using the formula e = 1 + 1/1! + 1/2! + ....... 1/n!...However, I'm always getting nearly 2.291481 as answer.Here is my code :
Code:
//Approximate the value of e #include <stdio.h> int main(void) { int n, j; printf("Please enter value of n : "); scanf("%d", &n);
[code]....
I know I'm not supposed to use system("PAUSE"), but this is self-study.
I'm trying to successfully run a program that calculates your BMI but, although it runs, it is not giving the the correct math. Instead, it gives me the number i've submitted as my weight as an answer. I'm using Visual Studio 2008 and the formula: weight / (height/100)*2
Here is my code
#include <iostream> #include <cmath> using namespace std; int main() { int weight; int height; double BMI;
I want calculate sum of every digit in my zip code so I wrote implementation function like that but, it shows "0" as answers, then I realize it should be call first, I add correctionDigitOf() in to constructor "Zipcode::Zipcode" then My zipcode are all "0' after I do this .....
#include <iostream> #include <stdio.h> #include <stdlib.h> #include <ctime> #include <cmath> using namespace std; class Zipcode {
I was given an assignment for class to calculate the area of a circle using only the radius as a user input and not using Pi in the code. I need to do this by calculating the areas of a series of rectangles under the curve and adding them together. Using nested loops to continuously reduce the size of these rectangles until the approximated area value is within a margin of error less than 0.1%.
Code: #include<iostream> #include<cmath> using namespace std; int main ()
