I am learning C++ using "Programming Principles and Practice using C++" by Bjarne Stroustroup. In the following question of finding median, I have problem with the ceil function. Say, for example, I input 14 entries then value of a should be 6 and value of b should be 7. But the value of b is coming out 6.
#include "std_lib_facilities.h"
using namespace std;
int main()
{ vector<double> temps;
for(double temp;cin>>temp;)
{temps.push_back(temp);
}
sort(temps);
for(double temp : temps)
cout << temp << " " ;
int size = temps.size();
if(size%2 == 0)
{ int a,b,c;
a = temps[floor((size + 1)/2) - 1]; cout << a;
**b = temps[ceil((size + 1)/2) - 1 ]; cout << b;**
cout<<"Median temperature:"<<(a+b)/2<<'\n';
}
else
cout<<"Median temperature:"<<temps[((size + 1)/2) - 1]<<'\n';
return 0;
}
std_lib_facilties.h link
since size is an int, (size+1)/2 will perform an integer division, and will return only the whole part of the division. Applying ceil to it, as you saw, is pointless.
You could work around it by defining size as a double, or casting it inline:
**b = temps[ceil((((double)size) + 1)/2) - 1 ];
Related
Find a sum of series by function s= 1!/y!+y!/3!+5!/y!......n
I don't know how to define fact function and is this code correct??
#include <math.h>
#include <iostream>
using namespace std;
float fact(float, float);
float sum(float, float);
int main() {
float n, x, s;
cout << "enter n and x" << endl;
cin >> n >> x;
s = sum(n, x);
cout << "sum =" << s << endl;
return 0;
}
float sum(float n, float x)
{
float x, s = 0;
for (int i = 0; i <= n; i++)
if (i % 2 == 0)
s = s + float(fact(i + 1) / fact(x));
else
s = s + float(fact(x) / fact(i + 1));
return s;
}
Whilst there are ways of defining the factorial of a floating-point number (using a gamma function), I doubt that is what you want. Similarly, the upper index n shouldn't be a float, either.
Your series as written looks awfully divergent.
As somebody else has said, it is rare to calculate new factorials from scratch: divisors of them tend to cancel, and whole terms of your series are simple multiples of earlier terms.
I am struggling to make this equation equals to each other because of a bad understanding of mathematics.
The problem is that the equation does not equal to each other
here is my code for better understand
#include <iostream>
#include <ccomplex>
using std::cout;
int main() {
int n = 8;
double sum = 0.0;
unsigned long long fact =1;
for (int i = 1; i <= n; i++)
{
fact *= 2*i*(2*i-1);
sum += 1.0 / fact;
}
std::cout << "first equation " << sum << std::endl;
double e = M_E;
double st = 1.0/2.0*(e + (1.0/e));
std::cout <<"second equation " << st << std::endl;
return 0;
}
the output
first equation 0.543081
second equation 1.54308
The result it nearly It must be at least equal before the comma,
You don't account for n = 0, which yields 0! and thus 1. Therefore, you need to add 1 to sum.
I want to calculate cos(nx) = p/q Using recursion and memoization so i tried Chebyshev Method for Recursive Method it works for small inputs of n but i want to do it for larger inputs like 10^18 etc. How should i approach it to calculate cos(nx) in p/q Form?
For larger Values of n i am getting Memory Problem How do i fix that?
cos(x) = a/b Known
#include <bits/stdc++.h>
using namespace std;
map<long, long> numerator;
map<long, long> denominator;
#define MODULO 1000000007
long a, b, n;
int main() {
cin >> a >> b;
cin >> n;
numerator[0] = 1;
numerator[1] = a;
denominator[0] = 1;
denominator[1] = b;
for (long i = 2; i <= n; ++i) {
numerator[i] = (long) (2 * a * numerator[i - 1] - b*numerator[i - 2]) % MODULO;
denominator[i] = (denominator[i - 1] * denominator[i - 2]) % MODULO;
}
cout << "numerator of cos(nx) = " << numerator[n] << " denominator = " << denominator[n] << endl;
}
lol Thats a contests question you're asking for. Shame on you. That's Codechef's February Challenge's Question. How low can you fall for some rating. smh
I am to calculate the arithmetic mean, the geometric mean, and the harmonic mean for five numbers using a single while loop.
Here is what I have so far:
#define _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <math.h>
float a;
float g;
float h;
sum1 = 0;
sum2 = 0;
sum3 = 0;
n = 5;
int k;
int main()
{
printf("Please Enter Five Integers:\n");
while (k = 0 && k < n && ++k);
{
scanf("%lf", &k);
sum1 = sum1 + k;
sum2 = sum2 * k;
sum3 = sum3 + (1.0 / k);
}
a = sum1 / n;
g = pow(sum2, 1 / n);
h = n / sum3;
printf("Arithmetic mean: %.3f\n", a);
printf("Geometric mean: %.3f\n", g);
printf("Harmonic mean: %.3f\n", h);
return 0;
Your C program has several issues.
You don't declare all the variables you are using, for example, and there's no need for them to be global.
Your initial value for sum2 (0) is wrong, it will never update because you repetedly multiply k times 0.
Then in pow(..., 1 / n) the 1/n is an integer division, so you are elevating to 0.
Your loop and its condition must be modified. Try this, I used double, instead of integers and float, but it depends on your assignment:
#include <stdio.h>
#include <math.h>
#define MAX 80
int main()
{
double a, g, h, k;
double sum = 0, prod = 1, sum_inv = 0;
const int n = 5;
int i = 0;
printf("Please, enter five numbers:\n");
char buffer[MAX];
while ( i < n ) {
fgets(buffer, MAX, stdin);
if ( sscanf(buffer, "%lf", &k) != 1 ) {
printf("Wrong format, floating point number expected\n");
continue;
}
if ( k == 0.0 ) {
printf("You should enter non zero numbers\n");
continue;
}
++i;
sum += k;
prod *= k;
sum_inv += (1.0 / k);
}
a = sum / n;
g = pow(prod, 1.0 / n);
h = n / sum_inv;
printf("Arithmetic mean: %.3f\n", a);
printf("Geometric mean: %.3f\n", g);
printf("Harmonic mean: %.3f\n", h);
return 0;
}
Apologies if this is brutal, but simply saying there are multiple issues and proceeding to correct them without explaining why they are issues or what was done to correct them doesn't make for a very good answer. It makes for homework cut-and-paste.
#define _CRT_SECURE_NO_WARNINGS
This is actually a bad idea. Those security warnings often tell you you're taking unnecessary risks. They are annoying, but often they are right.
#include <stdio.h>
#include <math.h>
These should be <cstdio> and <cmath>. Better still, don't use cstdio. Use the C++ equivalents.
float a;
float g;
float h;
sum1 = 0;
sum2 = 0;
sum3 = 0;
n = 5;
The preceding 4 variables do not have data types. All variables must have a type.
Further initializing sum2 to zero when it will be used to gather a product is a bad idea. 0 will result.
int k;
None of these variables need to be global and all of the variable names are non-descriptive. In a program this size, that's not horrible, but in a large program with dozens or thousands of variables, being able to read from the variable name what it does and what it contains is worth it's weight in gold.
int main()
{
printf("Please Enter Five Integers:\n");
while (k = 0 && k < n && ++k);
The ; is a bad mistake here. ; ends the instruction. It separates the loop from it's body, so you get a while the loops but does nothing else.
But let's look at the loop conditions shall we?
k = 0
this is the same as
k = 0
if (k)
Which is always false since k is 0. This exits the loop right here.
k < n
Which it always is because of k = 0. k is 0. A moot point because this never gets tested.
++k
is always true because at this point k will always be 1.
This screams read the textbook more closely because you missed quite a bit.
{
scanf("%lf", &k);
This line reads a floating point number into an integer. Not a good idea. The results will be bizarre at best.
In addition, the return code from scanf is untested so you have no way to tell whether or not scanf successfully read a value.
And this question is tagged C++. Why use C?
sum1 = sum1 + k;
sum2 = sum2 * k;
sum3 = sum3 + (1.0 / k);
That all looks good to me, other than being really bad, non-descriptive names.
}
a = sum1 / n;
Syntactically and logically sound.
g = pow(sum2, 1 / n);
1 / n will be performed entirely in integer arithmetic and certainly result in a fraction. Integers can't do fractions, so this will result in 0. Any number to the power of 0 is one.
h = n / sum3;
Looks good.
printf("Arithmetic mean: %.3f\n", a);
printf("Geometric mean: %.3f\n", g);
printf("Harmonic mean: %.3f\n", h);
Again, using C in C++. printf has it's uses, even in C++, and frankly this is one of those cases where I might use it (but with caution because there is a performance hit) because the C++ equivalent std::cout << "Arithmetic mean: " << std::fixed << std::setprecision(3) << a << '\n'; is brutally verbose.
return 0;
}
Revising this for C++
#include <cmath>
#include <iostream>
#include <iomanip>
#include <limits>
int main()
{
// discarded a, g, and h. Renamed the rest for easier reading
float sum = 0;
float product = 1;
float invSum = 0;
constexpr int MAX = 5;
int input;
std::cout <<"Please Enter Five Integers:" << std::endl;
int count = 0;
while (count < MAX)
{
if (std::cin >> input)
{ // read a good, or at least not horrible, number
// this will not handle the problem of "123abc" as input. "123" will be
// accepted and "abc" will be seen as a second token and rejected.
// proper handling of this is a question unto itself and has been asked
// hundreds of times.
sum += input;
product *= input;
invSum += (1.0 / input);
count++;
}
else
{ // clean up and ask for new input
std::cin.clear();
std::cin.ignore(std::numeric_limits<std::streamsize>::max(), '\n');
std::cout <<"Bogus integer. Input again: " << std::endl;
}
}
std::cout << "Arithmetic mean: " << std::fixed << std::setprecision(3) << sum / MAX << '\n';
std::cout << "Geometric mean: " << std::fixed << std::setprecision(3) << pow(product, (1.0 / MAX)) << '\n';
std::cout << "Harmonic mean: " << std::fixed << std::setprecision(3) << MAX / invSum << '\n';
return 0;
}
I am having the hardest time figuring out what is wrong here:
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
double fact(double);
double sinTaylor(double);
double cosTaylor(double);
int main()
{
double number, sineOfnumber, cosineOfnumber;
cout << "Enter a number, then I will calculate the sine and cosine of this number" << endl;
cin >> number;
sineOfnumber = sinTaylor(number);
cosineOfnumber = cosTaylor(number);
cout << fixed << endl;
cout << cosineOfnumber << endl;
cout << sineOfnumber << endl;
return 0;
}
double fact(double n)
{
double product = 1;
while(n > 1)
product *= n--;
return product;
}
double sinTaylor(double x)
{
double currentIteration, sumSine;
for(double n = 0; n < 5; n++)
{
currentIteration = pow(-1, n)*pow(x, 2*n+1) / fact(2*n+1);
sumSine += currentIteration;
}
return sumSine;
}
double cosTaylor(double y)
{
double currentIteration, sumCosine;
for(double n = 0; n < 5; n++)
{
double currentIteration = pow(-1, n)*pow(y, 2*n) / fact(2*n);
sumCosine += currentIteration;
}
return sumCosine;
}
Ok, so here's my code. I'm pretty content with it. Except for one thing:
sineOfnumber and cosOfnumber, after the calling of sinTaylor and cosTaylor, will add each other in the following cout line that will print each other.
In other words, if number is equal to lets say, .7853, 1.14 will be printed in the line that is intended to print cosineOfnumber, and sineOfnumber will print the result normally.
Can anyone help me identify why this is? Thank you so much!
Are you ever initializing the variables sumSine and sumCosine in your functions? They're not guaranteed to start at zero, so when you call += inside your loop you could be adding computed values to garbage.
Try initializing those two variables to zero and see what happens, as other than that the code seems okay.
The series for the sine is (sorry for the LaTeX):
sin(x) = \sum_{n \ge 0} \frac{x^{2 n + 1}}{(2 n + 1)!}
If you look, given term t_{2 n + 1} you can compute term t_{2 n + 3} as
t_{2 n + 3} = t_{2 n + 1} * \frac{x^2}{(2 n + 2)(2 n + 3)}
So, given a term you can compute the next one easily. If you look at the series for the cosine, it is similar. The resulting program is more efficient (no recomputing factorials) and might be more precise. When adding up floating point numbers, it is more precise to add them from smallest to largest, but I doubt that will make a difference here.