I need help with writing power function. So, I need to write a porogramm, that will output a table from 1 to 10 in a power in a LOOP. NOT USING POW or EXP
Example of output:
0^0 == 1
1^1 == 1
2^2 == 4
3^3 == 27
4^4 == 256
(and so on, up to)
10^10 == 10000000000
NOT USING Cmath (NO POW or EXP)
for example:
e.g. power( 3.0, 5 ) will return 243 because 3*3*3*3*3 is 243
e.g. power( 173, 0 ) will return 1 because any number raised to the power of 0 is 1.
I did this Simple loop, But I have no idea how to insert power formula in it. I was also thinking about while loop
#include <iostream>
#include <string>
using namespace std;
int main(){
int number = 0, tot;
for (int table = 0; table < 10; table++)
{
tot = number * table;
cout << tot << endl;
number++;
}
}
This is a recursive function that can calculate a value raised to an integer power
double power(double base, unsigned int exp)
{
if (exp == 0)
{
return 1.0;
}
else
{
return base * power(base, exp - 1);
}
}
An iterative method to do this would be
double power(double base, unsigned int exp)
{
double product = 1.0;
for (unsigned int i = 0; i < exp; ++i)
{
product *= base;
}
return product;
}
You can test either method with something like
int main()
{
std::cout << power(5, 3);
}
Output
125
I think you already know the answer to your own question by now, but still; some hints:
Exponentiation is a repeated multiplication of the base, the repetition is defined by the exponent.
In C++, or any modern programming language, loops allow repetition of certain blocks of code: when the number of iterations is known beforehand, use the for-loop, otherwise, use the while-loop.
Combining both hints: you'll need to use a loop to repeat a multiplication; the amount of repetition (or iterations) is known beforehand, thus, a for-loop will be best.
int exponentiation(int base, int exponent) {
int result = 1;
for (int i = 0; i < exponent; ++i)
result = result * base;
return result;
}
Note: this will only suffice for integer exponentiation with positive exponents!
You can then call this function in a for-loop to let it compute the values you want:
#include <iostream>
int main(int argc, char** argv) {
for(int i = 0; i <= 10; ++i)
std::cout << exponentiation(i, i) << '\n';
}
Related
Hi I am trying to calculate the results of the Taylor series expansion for sine to the specified number of terms.
I am running into some problems
Your task is to implement makeSineToOrder(k)
This is templated by the type of values used in the calculation.
It must yield a function that takes a value of the specified type and
returns the sine of that value (in the specified type again)
double factorial(double long order){
#include <iostream>
#include <iomanip>
#include <cmath>
double fact = 1;
for(int i = 1; i <= num; i++){
fact *= i;
}
return fact;
}
void makeSineToOrder(long double order,long double precision = 15){
double value = 0;
for(int n = 0; n < precision; n++){
value += pow(-1.0, n) * pow(num, 2*n+1) / factorial(2*n + 1);
}
return value;
int main()
{
using namespace std;
long double pi = 3.14159265358979323846264338327950288419716939937510L;
for(int order = 1;order < 20; order++) {
auto sine = makeSineToOrder<long double>(order);
cout << "order(" << order << ") -> sine(pi) = " << setprecision(15) << sine(pi) << endl;
}
return 0;
}
I tried debugging
here is a version that at least compiles and gives some output
#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
double factorial(double long num) {
double fact = 1;
for (int i = 1; i <= num; i++) {
fact *= i;
}
return fact;
}
double makeSineToOrder(double num, double precision = 15) {
double value = 0;
for (int n = 0; n < precision; n++) {
value += pow(-1.0, n) * pow(num, 2 * n + 1) / factorial(2 * n + 1);
}
return value;
}
int main(){
long double pi = 3.14159265358979323846264338327950288419716939937510L;
for (int order = 1; order < 20; order++) {
auto sine = makeSineToOrder(order);
cout << "order(" << order << ") -> sine(pi) = " << setprecision(15) << sine << endl;
}
return 0;
}
not sure what that odd sine(pi) was supposed to be doing
Apart the obvious syntax errors (the includes should be before your factorial header) in your code:
I see no templates in your code which your assignment clearly states to use
so I would expect template like:
<class T> T mysin(T x,int n=15){ ... }
using pow for generic datatype is not safe
because inbuild pow will use float or double instead of your generic type so you might expect rounding/casting problems or even unresolved function in case of incompatible type.
To remedy that you can rewrite the code to not use pow as its just consequent multiplication in loop so why computing pow again and again?
using factorial function is waste
you can compute it similar to pow in the same loop no need to compute the already computed multiplications again and again. Also not using template for your factorial makes the same problems as using pow
so putting all together using this formula:
along with templates and exchanging pow,factorial functions with consequent iteration I got this:
template <class T> T mysin(T x,int n=15)
{
int i;
T y=0; // result
T x2=x*x; // x^2
T xi=x; // x^i
T ii=1; // i!
if (n>0) for(i=1;;)
{
y+=xi/ii; xi*=x2; i++; ii*=i; i++; ii*=i; n--; if (!n) break;
y-=xi/ii; xi*=x2; i++; ii*=i; i++; ii*=i; n--; if (!n) break;
}
return y;
}
so factorial ii is multiplied by i+1 and i+2 every iteration and power xi is multiplied by x^2 every iteration ... the sign change is hard coded so for loop does 2 iterations per one run (that is the reason for the break;)
As you can see this does not use anything funny so you do not need any includes for this not even math ...
You might want to add x=fmod(x,6.283185307179586476925286766559) at the start of mysin in order to use more than just first period however in that case you have to ensure fmod implementation uses T or compatible type to it ... Also the 2*pi constant should be in target precision or higher
beware too big n will overflow both int and generic type T (so you might want to limit n based on used type somehow or just use it wisely).
Also note on 32bit floats you can not get better than 5 decimal places no matter what n is with this kind of computation.
Btw. there are faster and more accurate methods of computing goniometrics like Chebyshev and CORDIC
I have a vector of numbers (floats), representing everything after the second of some time stamp. They have varying lengths. It looks something like this:
4456
485926
346
...
Representing 0.4456, 0.485926, and 0.346 seconds, respectively. I need to convert each of these to milliseconds, however I can’t simply multiply each by some constant since they’re all of different lengths. I’m fine with loosing accuracy, I just need the first 3 digits (the millisecond bit). How can this be done?
Try this:
#include <iostream>
#include <string>
using namespace std;
int getFirstThreeDigits(int number){
return stoi(to_string(number).substr(0,3));
}
int main()
{
float values[] = {4456, 485926, 346};
int arrLength = (sizeof(values)/sizeof(*values));
for( int i = 0 ; i < arrLength ; i++){
cout << getFirstThreeDigits(values[i]) << endl;
}
}
I'm assuming here that the integral portion of the float represents a subsecond value, so that 1234f is actually 0.1234 seconds. That seems to be what your question states.
If that's the case, it seems to me you can just continuously divide the value by ten until you get something less than one. Then multiply it by one thousand and round. That would go something like:
#include <iostream>
int millis(float value) {
if (value < 0) return -millis(-value);
//while (value >= 1000f) value /= 1000f;
while (value >= 1.0f) value /= 10.f;
return static_cast<int>(value * 1000 + .5f);
}
int main(int argc, char *argv[]) {
for (int i = 1; i < argc; ++i) {
float f= atof(argv[i]);
std::cout << " " << f << " -> " << millis(f) << "\n";
}
}
I've also put in a special case to handle negative number and a (commented-out, optional) optimisation to more quickly get down to sub-one for larger numbers.
A transcript follows with your example values:
pax> ./testprog 4456 485926 346
4456 -> 446
485926 -> 486
346 -> 346
If instead the values are already sub-second floats and you just want the number of milli-seconds, you do the same thing but without the initial divisions:
int millis(float value) {
if (value < 0) return -millis(-value);
return static_cast<int>(value * 1000 + .5f);
}
Using "(int)log10 + 1" is an easy way to get the number of integer digits
auto x = 485926;
auto len = (int)std::log10(x) + 1;
https://godbolt.org/z/v1jz7P
So..
Here is the code:
#include <iostream>
#include <limits>
#include <math.h>
using namespace std;
int main()
{
unsigned long long i,y,n,x=45;
unsigned long long factorial = 1;
for(n = 0; n <= 5; n++)
{
y = (pow(-1,n)*pow(x,2*n)) / factorial;
cout << "COS IS " << y << endl;
}
for(int i = 1; i <=n; i++)
{
factorial *= 2*i;
}
}
I get an overflow but I really don't know why. I use unsigned long long just to make sure that I on't get but.. I still get it. Even limited to small numbers. I tried to implement this:
https://en.wikibooks.org/wiki/Trigonometry/Power_Series_for_Cosine_and_Sine
But I really can't do it because of the overflow. Do you have any ideea on what can I do ? I am newbie in programming so, take it easy on me :D
There are many issues.
you use integer types when you should use floating point types
you use unsigned types for signed calculations
you don't use radians but degrees (45° ≈ 0.78539 radians)
you don't calculate the factorial in the loop, it is always 1, you only calculate it at the end of the loop but then it's too late, and your calculation of the factorial is wrong anyway.
the algorithm is wrong, it just doesn't do what Maclaurin's therorem says, you need to sum up the terms, but you just print the terms.
You probably want this:
#include <iostream>
#include <cmath>
using namespace std;
long factorial(int n)
{
long result = 1;
for (int i = 1; i <= n; i++)
result *= i;
return result;
}
int main()
{
double x = 0.785398163397448309616; //PI/4 expectd result COS(PI/4) = 0.7071067
double mycosinus = 0;
for (int n = 0; n <= 5; n++)
{
mycosinus += (pow(-1, n) * pow(x, 2 * n)) / factorial(2*n);
cout << "COS IS " << mycosinus << endl;
}
}
This is your wrong algorithm for calculating the factorial of 5:
int main()
{
int n = 5;
int factorial = 1;
for (int i = 1; i <= n; i++)
{
factorial *= 2 * i;
}
cout << "factorial 5 = " << factorial << endl;
}
The calculated value is 3840 instead of 120. I let you find out what's wrong yourself.
For performing this sort of maths you need to use a floating point like float or double not integral types like long, int or long long, given that sin and cos can both return negative numbers you shouldn't be using unsigned either.
the point of this exercise is to multiply a digit of a number with its current position and then add it with the others. Example: 1234 = 1x4 + 2x3 + 3x2 + 4x1 .I did this code successfully using 2 parameters and now i'm trying to do it with 1. My idea was to use - return num + mult(a/10) * (a%10) and get the answer, , because from return num + mult(a/10) i get the values 1,2,3,4- (1 is for mult(1), 2 for mult(12), etc.) for num, but i noticed that this is only correct for mult(1) and then the recursion gets wrong values for mult(12), mult(123), mult(1234). My idea is to independently multiply the values from 'num' with a%10 . Sorry if i can't explain myself that well, but i'm still really new to programming.
#include <iostream>
using namespace std;
int mult(int a){
int num = 1;
if (a==0){
return 1;
}
return ((num + mult(a/10)) * (a%10));
}
int main()
{
int a = 1234;
cout << mult(a);
return 0;
}
I find this easier and more logically to do, Hope this helps lad.
int k=1;
int a=1234;
int sum=0;
while(a>0){
sum=sum+k*(a%10);
a=a/10;
k++;
}
If the goal is to do it with recursion and only one argument, you may achieve it with two functions. This is not optimal in terms of number of operations performed, though. Also, it's more of a math exercise than a programming one:
#include <iostream>
using namespace std;
int mult1(int a) {
if(a == 0) return 0;
return a % 10 + mult1(a / 10);
}
int mult(int a) {
if(a == 0) return 0;
return mult1(a) + mult(a / 10);
}
int main() {
int a = 1234;
cout << mult(a) << '\n';
return 0;
}
I am trying to solve a problem, a part of which requires me to calculate (2^n)%1000000007 , where n<=10^9. But my following code gives me output "0" even for input like n=99.
Is there anyway other than having a loop which multilplies the output by 2 every time and finding the modulo every time (this is not I am looking for as this will be very slow for large numbers).
#include<stdio.h>
#include<math.h>
#include<iostream>
using namespace std;
int main()
{
unsigned long long gaps,total;
while(1)
{
cin>>gaps;
total=(unsigned long long)powf(2,gaps)%1000000007;
cout<<total<<endl;
}
}
You need a "big num" library, it is not clear what platform you are on, but start here:
http://gmplib.org/
this is not I am looking for as this will be very slow for large numbers
Using a bigint library will be considerably slower pretty much any other solution.
Don't take the modulo every pass through the loop: rather, only take it when the output grows bigger than the modulus, as follows:
#include <iostream>
int main() {
int modulus = 1000000007;
int n = 88888888;
long res = 1;
for(long i=0; i < n; ++i) {
res *= 2;
if(res > modulus)
res %= modulus;
}
std::cout << res << std::endl;
}
This is actually pretty quick:
$ time ./t
./t 1.19s user 0.00s system 99% cpu 1.197 total
I should mention that the reason this works is that if a and b are equivalent mod m (that is, a % m = b % m), then this equality holds multiple k of a and b (that is, the foregoing equality implies (a*k)%m = (b*k)%m).
Chris proposed GMP, but if you need just that and want to do things The C++ Way, not The C Way, and without unnecessary complexity, you may just want to check this out - it generates few warnings when compiling, but is quite simple and Just Works™.
You can split your 2^n into chunks of 2^m. You need to find: `
2^m * 2^m * ... 2^(less than m)
Number m should be 31 is for 32-bit CPU. Then your answer is:
chunk1 % k * chunk2 * k ... where k=1000000007
You are still O(N). But then you can utilize the fact that all chunk % k are equal except last one and you can make it O(1)
I wrote this function. It is very inefficient but it works with very large numbers. It uses my self-made algorithm to store big numbers in arrays using a decimal like system.
mpfr2.cpp
#include "mpfr2.h"
void mpfr2::mpfr::setNumber(std::string a) {
for (int i = a.length() - 1, j = 0; i >= 0; ++j, --i) {
_a[j] = a[i] - '0';
}
res_size = a.length();
}
int mpfr2::mpfr::multiply(mpfr& a, mpfr b)
{
mpfr ans = mpfr();
// One by one multiply n with individual digits of res[]
int i = 0;
for (i = 0; i < b.res_size; ++i)
{
for (int j = 0; j < a.res_size; ++j) {
ans._a[i + j] += b._a[i] * a._a[j];
}
}
for (i = 0; i < a.res_size + b.res_size; i++)
{
int tmp = ans._a[i] / 10;
ans._a[i] = ans._a[i] % 10;
ans._a[i + 1] = ans._a[i + 1] + tmp;
}
for (i = a.res_size + b.res_size; i >= 0; i--)
{
if (ans._a[i] > 0) break;
}
ans.res_size = i+1;
a = ans;
return a.res_size;
}
mpfr2::mpfr mpfr2::mpfr::pow(mpfr a, mpfr b) {
mpfr t = a;
std::string bStr = "";
for (int i = b.res_size - 1; i >= 0; --i) {
bStr += std::to_string(b._a[i]);
}
int i = 1;
while (!0) {
if (bStr == std::to_string(i)) break;
a.res_size = multiply(a, t);
// Debugging
std::cout << "\npow() iteration " << i << std::endl;
++i;
}
return a;
}
mpfr2.h
#pragma once
//#infdef MPFR2_H
//#define MPFR2_H
// C standard includes
#include <iostream>
#include <string>
#define MAX 0x7fffffff/32/4 // 2147483647
namespace mpfr2 {
class mpfr
{
public:
int _a[MAX];
int res_size;
void setNumber(std::string);
static int multiply(mpfr&, mpfr);
static mpfr pow(mpfr, mpfr);
};
}
//#endif
main.cpp
#include <iostream>
#include <fstream>
// Local headers
#include "mpfr2.h" // Defines local mpfr algorithm library
// Namespaces
namespace m = mpfr2; // Reduce the typing a bit later...
m::mpfr tetration(m::mpfr, int);
int main() {
// Hardcoded tests
int x = 7;
std::ofstream f("out.txt");
m::mpfr t;
for(int b=1; b<x;b++) {
std::cout << "2^^" << b << std::endl; // Hardcoded message
t.setNumber("2");
m::mpfr res = tetration(t, b);
for (int i = res.res_size - 1; i >= 0; i--) {
std::cout << res._a[i];
f << res._a[i];
}
f << std::endl << std::endl;
std::cout << std::endl << std::endl;
}
char c; std::cin.ignore(); std::cin >> c;
return 0;
}
m::mpfr tetration(m::mpfr a, int b)
{
m::mpfr tmp = a;
if (b <= 0) return m::mpfr();
for (; b > 1; b--) tmp = m::mpfr::pow(a, tmp);
return tmp;
}
I created this for tetration and eventually hyperoperations. When the numbers get really big it can take ages to calculate and a lot of memory. The #define MAX 0x7fffffff/32/4 is the number of decimals one number can have. I might make another algorithm later to combine multiple of these arrays into one number. On my system the max array length is 0x7fffffff aka 2147486347 aka 2^31-1 aka int32_max (which is usually the standard int size) so I had to divide int32_max by 32 to make the creation of this array possible. I also divided it by 4 to reduce memory usage in the multiply() function.
- Jubiman