How can I write an "X to the power of k" procedure in C++? (k is a positive integer)
I did the same thing in python, and it was a breeze, but in C++, I don't even know where to begin.
How can I write an "X to the power of k" procedure in C++? (k is a positive integer)
Write a short loop in a function like
int pow(int X, int k) {
int result = 1;
for(int i = 0; i < k; ++i) result *= X;
return result;
}
It's easy to express this in a lambda as well:
auto pow = [] (int X, int k) {
int result = 1;
for(int i = 0; i < k; ++i) result *= X;
return result;
};
cout << pow(5,3);
See a working sample please.
Ummm, maby try this:
#include <iostream>
#include<cmath> //adds math functions: power, square root etc.
using namespace std;
int main(){
int x;
int k;
cin >> x;
cin >> k;
x = pow(x, k);
cout << "\nX to the power of k: " << x << endl << endl;
return 0;
}
Related
The formula is listed in the following article: https://en.wikipedia.org/wiki/Formula_for_primes. I am trying to implement it but to no success, for whatever reason the code is producing number which seem to be nth power of two + 1, which is obviously not what I want to achieve.
#include <iostream>
#include <cmath>
using namespace std;
int nth_prime(int n) {
double s = 1;
for (int i = 1; i <= pow(2, n); i++) {
double c = 0;
for (int j = 1; j <= i; j++) {
double f = (tgamma(j)+1)/j;
c+=floor(pow(cos(M_PI*f), 2));
}
s+=floor(pow(n/c, 1/n));
}
return s;
}
int main() {
int n;
while (cin >> n) {
cout << nth_prime(n) << endl;
}
return 0;
}
I am doing a code in c++ where I am supposed to be finding the series and I build the function for the series myself yet and I call the function I don't find my answer
here is my code
#include <iostream>
#include <cmath>
using namespace std;
double harmonicSeries(int n);
int main() {
int n;
cout << "Enter n" << endl;
cin >> n;
harmonicSeries(n);
}
double harmonicSeries(int n) {
for (int i = 1; i <= n; i++) {
float s;
float sum = 0.0;
s = 1 / n;
sum += s;
return sum;
}
}
I will be thankful for any help
See I have made the changes in your code,this works fine in this finding numbers and adding to get their sum.You should use return outside the function and basically harmonic series is of form 1/n which can be any float number or double number so I use s as double and i has float(which by this).
s=1/i(double=1/float,gets converted to double)
#include <iostream>
#include <cmath>
using namespace std;
double harmonicSeries(int n);
int main() {
int n;
cout << "Enter n" << endl;
cin >> n;
cout<<harmonicSeries(n);
}
double harmonicSeries(int n) {
double sum=0.00;
double s;
for (float i = 1; i <= n; i++) {
s = 1 / i;
sum += s;
}
return sum;
}
If you find anything wrong do ask for sure:)
Screenshot of my code
Hey, I have just started learning C++ and I am trying to get it to sum the series:
K+N−1∑n=K [-1^(n)/(n+1)2]
I have managed to get it to tell me the nth term previously, but now I would like to for each term in the series, starting with the kth and going in sequence to the last (k+n-1st), add this term to the running sum.
I need to use a function direct_up() which uses the function term(). I defined initially and test it in the main.
I know I am missing something and am a bit confused about how to use the functions and that I may have made a few mistakes. So I would be very grateful for some advice or help. I have attached a picture of what I have done so far, as well as typed it below.
using namespace std;
double term(int n) {
double y;
if(n%2==1) {
y = -1.0/((n+1.0)*(n+1.0));
} else {
y = 1.0/((n+1.0)*(n+1.0));
}
return y;
}
double direct_up(int k, int n) {
int startingnumber = k;
for (int i = startingnumber; i <= k+n; i++) {
cout << n << term(n) << endl;
}
return n;
}
int main() {
double n;
int k;
cout.precision(16);
cout << "enter number of terms";
cin >> n;
cout << "enter a value for k";
cin >> k;
cout << "here is the value" << direct_up(k,n);
return 0;
}
This is what you want to do:
double direct_up(int k, int n)
{
int startingnumber = k;
double sum = 0;
for (int i = startingnumber; i <= k+n; i++) {
sum += term(i);
}
return sum;
}
Here's how to do it without keeping your own running sum as you asked in your comment:
#include <vector>
#include <numeric>
double direct_up(int k, int n)
{
int startingnumber = k;
std::vector<double> terms;
for (int i = startingnumber; i <= k+n; i++) {
terms.push_back(term(i));
}
return accumulate(terms.begin(), terms.end(), 0.0);
}
Here is the question:
Comparing two numbers written in index form like 2^11 and 3^7 is not difficult, as any calculator would confirm that 2^11=2048<3^7=2187.
However, confirming that 632382^518061>519432^525806 would be much more difficult, as both numbers contain over three million digits.
You are given N base exponent pairs, each forming a large number you have to find the Kth smallest number of them. K is 1−indexed.
Input Format
First line containts an integer N, number of base exponent pairs. Followed by N lines each have two space separated integers B and E, representing base and exponent.
Last line contains an integer K, where K<=N
Constraints
1≤N≤105
1≤K≤N
1≤B≤109
1≤E≤109
No two numbers are equal.
Here is my code:
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int N,i = 0,k,j=0,x,m;
long long int *arr,*arr2,*arr3;
cin >> N;
arr = (long long int *)malloc(sizeof(long long int)*2*N);
arr2 = (long long int *)calloc(N,sizeof(long long int));
arr3 = (long long int *)calloc(N,sizeof(long long int));
x = 2*N;
while(x>0)
{
cin >> arr[i];
i++;
x--;
}
cin >> k;
for(i=0;i<2*N;i+=2)
{
arr2[j] = pow(arr[i],arr[i+1]);
j++;
}
arr3 = arr2;
sort(arr2,arr2+N);
for(i=0;i<N;i++)
{
if(arr3[i] == arr2[k-1])
{
m = i;
break;
}
}
cout << arr[2*m] << " " << arr[2*m + 1];
return 0;
}
The program works for small numbers only, can't make it work for large numbers. what to do?
Maybe you are generating an overflow on the big numbers. You could consider using a multiprecision arithmetic library such as https://gmplib.org/. I haven't used this library myself.
Have a look at this post How to detect integer overflow? on how to detect integer overflow.
From your choosing of long long int type I guess you calculated the a^b of the numbers in order to sort them, which leads to very big numbers and may lead to overflow.
Note that in order to sort the numbers there is no need for this calculation, for knowing if a^b > d^c it is sufficient to check log(a^b) > log(c^d) and therefore b*log(a) > d*log(c).
And it's better to use a struct or class to create a data structure for this big numbers.
This is the code for it:
#include <iostream>
#include <algorithm>
#include <math.h>
using namespace std;
struct BigNumber{
int base;
int exponent;
};
int Compare(BigNumber x, BigNumber y);
void Sort(BigNumber* arr, int N);
int main() {
int N,i = 0,k;
BigNumber *numbers;
cout<<"\nEnter N:";
cin >> N;
numbers = (BigNumber *)calloc(N,sizeof(BigNumber));
for(i=0; i<N; i++)
{
cout<<"\nEnter base and exponent for number "<<i<<":";
cin >> numbers[i].base>>numbers[i].exponent;
}
cout<<"\nEnter K:";
cin >> k;
Sort(numbers,N);
cout << "Kth number is :" << numbers[k].base << "^" << numbers[k].exponent;
return 0;
}
void Sort(BigNumber* arr, int N){
for(int i=0; i< N; i++ ){
for(int j=0; j< N; j++){
if(Compare(arr[i], arr[j])<0){
BigNumber temp = arr[j];
arr[j] = arr[i];
arr[i] = arr[j];
}
}
}
}
int Compare(BigNumber x, BigNumber y){
double X = x.exponent * log10(x.base);
double Y = y.exponent * log10(x.base);
return X == Y? 0: X > Y ? 1: -1;
}
I changed the code a little. Only problem I was having was I was calculating the exponent rather than comparing log of exponent.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
int main() {
int N,i = 0,k,j=0,x,m;
int *arr;
double *arr2,*arr3;
cin >> N;
arr = (int *)malloc(sizeof(int)*2*N);
arr2 = (double *)calloc(N,sizeof(double));
arr3 = (double *)calloc(N,sizeof(double));
x = 2*N;
while(x>0)
{
cin >> arr[i];
i++;
x--;
}
cin >> k;
for(i=0;i<2*N;i+=2)
{
arr2[j] = arr[i+1]*log10(arr[i]);
j++;
}
for (i = 0; i < N; i++) {
arr3[i] = arr2[i];
}
sort(arr2,arr2+N);
for(i=0;i<N;i++)
{
if(arr3[i] == arr2[k-1])
{
m = i;
break;
}
}
cout << arr[2*m] << " " << arr[2*m + 1];
return 0;
}
Can someone please help me understand why this code isn't working properly? I know it's very close and I think that I'm just overlooking something. Any help is appreciated. Here is what I have so far:
#include <iostream>
#define TEST_ARRAY_SIZE 4
long int factorial(int num);
long int factorial(int num){
for(unsigned int i = 1; i <= num; i++) {
num *= i;
}
return num;
}
int main() {
int test[TEST_ARRAY_SIZE] = {1, 2, 5, 7};
for (unsigned int i = 0; i < TEST_ARRAY_SIZE; i++) {
std::cout << "Factorial of " << test[i] << " is " << factorial(test[i]) << std::endl;
}
return 0;
}
You should move return num; outside the loop. As it is now, control always return after the first number is multiplied.
-- the correct code --
long int factorial(int num){
long int res = 1;
for(unsigned int i = 1; i <= num; i++) {
res *= i;
}
return res;
}
The body of the factorial function is not correct. You can use a recursive method to compute the factorial of a specific number. The corrected program of your version is below:
#include <iostream>
#define TEST_ARRAY_SIZE 4
long int factorial(int num);
int main() {
int test[TEST_ARRAY_SIZE] = {1, 2, 5, 7};
for (unsigned int i = 0; i < TEST_ARRAY_SIZE; i++) {
std::cout << "Factorial of " << test[i] << " is " << factorial(test[i]) << std::endl;
}
return 0;
}
long int factorial(int num){
if (num == 1)
return num;
else
return num * factorial(num-1);
}
That's only part of the problem; it's a case in which proper indentation would make the bug apparent in an instant.
In addition to that, why are you using num inside the loop? You should leave it as-is and declare a new variable to compound and return the result.
long int factorial(unsigned int num) {
int x = 1;
for(unsigned int i = 1; i <= num; i++) {
x *= i;
}
return x;
}
Other things to note:
You shouldn't be using a #define; you should declare a const instead.
You're comparing an unsigned against a signed in the factorial() for loop.