Here is a code I have written, although there is no double variable in it, g++ throws this warning: "invalid operands of types ‘double’ and ‘int’ to binary ‘operator%’
c = pow(m,e)%n;
" How is that?
#include <iostream>
#include <boost/dynamic_bitset.hpp>
#include <math.h>
using namespace std;
int main()
{
boost::dynamic_bitset <> seq(5);
int key = 0;
int p = 0, q = 0;
int n = 0;
int f = 0;
int d = 0;
int e = 0;
int c = 0;
int m = 0;
int t = 0;
cout << "Enter a sequence of bits: ";
cin >> seq;
for (unsigned int i = 0; i < seq.size(); i++)
{
if ( seq[i]==1)
{
d = d + pow(2,i);
cout << d << "\n";
}
}
//cout << key << "\n";
cout << "Enter p: ";
cin >> p;
cout << "Enter q: ";
cin >> q;
cout << "Enter m: ";
cin >> m;
n = p*q;
f = (p-1)*(q-1);
for ( int k = 0; t < 1; k++)
{
if ((1+k*f)%d==0)
{
t = 2;
e = (1+k*f)/d;
}
}
cout << "E is: " << e << "\n";
c = pow(m,e)%n;
cout << "C is: " << c << "\n";
cin.get();
return 0;
}
Function pow is declared as
double pow (double base, double exponent)
that is a double.
Related
I am trying to get the height of these slashes to be a certain length based on input. So far, I have:
#include <iostream>
using namespace std;
int main() {
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
}
I need it to then reverse and go back.
It prints:
//
////
//////
If the user entered 3.
It should print:
//
////
//////
////
//
Can anyone lead me in the right direction? I am new to cpp.
You can use a different kind of loop and add a bool variable to track when the program have reached "n". Then, after the program reaches "n", it sets the bool variable to true and starts to substract until i equals 0
Code below, read comments and ask if you have any further questions:
#include <iostream>
using namespace std;
int main()
{
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You have entered: " << n << "\n";
int i = 1;
bool reachedN = false; // tells if [i] has reached [n]
while (i != 0)
{
// Print required slashes
for (int j = 1; j <= i; j++)
{
cout << "//";
}
cout << '\n'; // new line
// Add until i == n, then substract
if (i == n)
{
reachedN = true;
}
if (reachedN)
{
--i;
}
else
{
++i;
}
}
}
If you enter 3, the output is the following:
This is one way to achieve that:
#include <iostream>
using namespace std;
int main() {
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
for (int i = n - 1; i > 0; i--) {
for (int j = 1; j <= i; j++)
cout << '/' << '/';
cout << "\n";
}
}
This is a shorter solution with only two for-loops.
#include <iostream>
using namespace std;
int main()
{
int n = 0;
cout << "Enter value: ";
cin >> n;
cout << "You entered: " << n << "\n";
n = n * 2 - 1;
int r = 0;
for (int j = 0; j < n; j++)
{
if (j > n / 2) r--;
else r++;
for (int i = 0; i < r; i++)
{
cout << '/' << '/';
}
cout << "\n";
}
return 0;
}
The program should read n resistances and a voltage from the keyboard and then calculate the equivalent resistance and the current.
My problem is that it calculates based only on the last entered resistance.
Is it possible to declare a method inside a function? or should I give up this completely unpractical approach
#include "stdafx.h"
#include<iostream>
#include<conio.h>
using namespace std;
class rez {
float r;
public:
void set(int n);
float val() { return r; }
};
void rez :: set(int n) { //n is the number of resistances
int i;
for (i = 1; i <= n; i++) {
cout << "R" << i << "=";
cin >> r;
}
}
float serie(rez r1,int n)
{
float s=0;
int i;
for (i = 1; i <= n; i++)
{
s = s+ r1.val();
}
return s;
}
float para(rez r1, int n)
{
float s = 0;
int i;
for (i = 1; i <= n; i++)
{
s = s + (1/r1.val());
}
return 1/s;
}
int main()
{
char c, k = 'y'; // 'c' selects series or para
rez r1;
int n;
cout << "number of resis:";
cin >> n;
cout << endl;
while (k != 'q')
{
r1.set(n);
float i, u;
cout << "\n Vdc= ";
cin >> u;
cout << endl;
cout << "series or para(s/p)?"<<endl;
cin >> c;
switch (c)
{
case('s'):cout <<"\n equiv resistance = "<< serie(r1,n)<<endl;
i = u / serie(r1, n);
cout << "curr i = " << i << " amp";
break;
case('p'):cout << "\n equiv res = " << para(r1, n)<<endl;
i = u / para(r1, n);
cout << "cur i = " << i << " amp";
break;
}
cout <<endl<< "\n another set?(y/q)?"<<endl;
cin >> k;
}
return 0;
}
It is because when you read in the resistances you are setting the value of the total resistance each time not adding to the total resistance.
void rez :: set(int n) { //n is the number of resistances
int i;
for (i = 1; i <= n; i++) {
cout << "R" << i << "=";
cin >> r; // <- this sets the value of r, it does not add to it
}
}
To fix this you should create a temporary variable to store the input resistance and then add it to the total resistance
void rez :: set(int n)
{
int i;
for (i = 1; i <= n; i++)
{
float input;
cout << "R" << i << "=";
cin >> input;
r += input;
}
}
I would like to analyze the complexity of my code algorithm.Therefore,i must have 2 different programs giving the same functions to allow me to start off.
Currently this is my own code.
I'm not sure if it is allowed that i would like to have someone that could volunteer his own way code to compute summation of factorial for me as the 2nd program code.
Preferrably a nested loop.
#include <iostream>
using namespace std;
int main()
{
int val;
int i;
int a = 0;
int c = 1;
cout << "Please enter a number: ";
cin >> val;
cout << endl;
for (i = 1; i <= val; i++)
{
c = c * i;
a = a + c;
}
cout << "The sum of the factorials is " << a << endl;
system("pause");
return 0;
}
#include <iostream>
using namespace std;
int main()
{
int val;
cout << "Please enter a number: ";
cin >> val;
cout << endl;
static const int results[] = {
0, 1, 3, 9, 33, 153, 873, 5913, 46233, 409113,
4037913, 43954713, 522956313
};
cout << "The sum of the factorials is " << results[val < 0 ? 0 : val] << endl;
system("pause");
return 0;
}
Note that I replicated the defect in the original program which causes it to return the incorrect value if the user enters 0.
This alternate version assumes 32-bit integers because it takes advantage of overflow behavior. Extending to 64-bit integers is left as an exercise.
I do not understand what you do with another nested way but i hope this can help...
#include <iostream>
using namespace std;
int main()
{
int val;
int i;
int a = 0;
int c = 1;
cout << "Please enter a number: ";
cin >> val;
cout << endl;
for (i = 1; i <= val; i++){
c *= i;
a += c;
}
int c2=1;
for (i = val; i > 1; i--){
c2*=i;
c2++;
}
cout << "The sum of the factorials is " << a << endl;
cout << "The sum of the factorials is " << c2 << endl;
system("pause");
return 0;
}
#include <iostream>
using namespace std;
int main()
{
int suma = 0;
int n = 0;
cout << "Sum of factorials\n";
cout << "-------------------------------\n";
cout << "Insert number of n: ";
cin >> n;
int i = 1;
while (i <= n)
{
int factorial = 1;
for(int j=1; j<=i; j++)
{
factorial = factorial * j;
}
suma += factorial;
i++;
}
cout << "Sum of factorials is: " << suma;
system("pause");
return 0;
}
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Improve this question
#include <iostream>
using namespace std;
int main()
{
int a;
int b;
int c;
int d;
int e;
int f;
int aa = 0;
int bb = 0;
int cc = 0;
int dd = 0;
int ee = 0;
int ff = 0;
const string odd = "ODD";
const string even = "EVEN";
cout << "enter 6 numbers " << endl;
cin >> a;
cin >> b;
cin >> c;
cin >> d;
cin >> e;
cin >> f;
aa = a % 2;
bb = b % 2;
cc = c % 2;
dd = d % 2;
ee = e % 2;
ff = f % 2;
if(aa == 0){
cout << even << endl;
}else{
cout << odd << endl;
}
if(bb == 0){
cout << even << endl;
}else{
cout << odd << endl;
}
if(cc == 0){
cout << even << endl;
}else{
cout << odd << endl;
}
if(dd == 0){
cout << even << endl;
}else{
cout << odd << endl;
}
if(ee == 0){
cout << even << endl;
}else{
cout << odd << endl;
}
if(ff == 0){
cout << even << endl;
}else{
cout << odd << endl;
}
return 0;
}
for example is there a way to make it do the same thing but with less code, anything I should have included?
is there an easier way than having to write 6 if/else statements - is there a way to do all 6 in one statement or loop?
how could i improve its efficiency?
Write this function:
void outputEvenness(int n)
{
static const string odd = "ODD";
static const string even = "EVEN";
if(n % 2){
cout << odd<< endl;
} else {
cout << even << endl;
}
}
then call it using outputEvenness(a); outputEvenness(b); etc.
First of all you should include header <string>if you use class std::string.
Also there is no sense to define these strings when they are used as string literals. Also instead of different variables it would be better to define only one array. The auxiliary variables are also unnecessary.
If to assume that you may not use arrays then I would write the program the following way
#include <iostream>
#include <initializer_list>
int main()
{
const size_t N = 6;
int a = 0;
int b = 0;
int c = 0;
int d = 0;
int e = 0;
int f = 0;
const char *odd = "ODD";
const char *even = "EVEN";
std::cout << "enter " << N << " numbers: ";
std::cin >> a >> b >> c >> d >> e >> f;
for ( int x : { a, b, c, d, e, f } )
{
if ( x % 2 == 0 )
{
std::cout << x << " is " << even << std::endl;
}
else
{
std::cout << x << " is " << odd << std::endl;
}
}
return 0;
}
If you are allowed to use arrays then the program could look as
#include <iostream>
int main()
{
const size_t N = 6;
int a[N] = {};
const char *odd = "ODD";
const char *even = "EVEN";
std::cout << "enter " << N << " numbers: ";
for ( int &x : a ) std::cin >> x;
for ( int x : a )
{
if ( x % 2 == 0 )
{
std::cout << x << " is " << even << std::endl;
}
else
{
std::cout << x << " is " << odd << std::endl;
}
}
return 0;
}
For this simple program there is no sense to define a separate function that will check whether a number is even or odd because it is this program that is such a function.:)
Use arrays and loops:
int a[6]; // Array of 6 ints
cout << "enter 6 numbers" << endl;
// Input the 6 numbers
for (int i = 0; i < 6; i++)
{
cin >> a[i];
}
// Output the results
for (int i = 0; i < 6; i++)
{
cout << a[i] << " is " << (a[i] & 1 ? "ODD" : "EVEN") << endl;
}
int value = 0;
string response = "";
cout << "enter 6 numbers " << endl;
for(int i=0; i<6; i++)
{
cin >> value;
value % 2 == 0 ? response+="even\n" : response+="odd\n";
}
cout << response;
I am currently doing a task in a book which asks me to calculate the mathematical constant e using the while loop. I managed that fairly easily, however I am having troubles calculating e^x, whereas the user inputs x and the degree of accuracy. The code I used for computing e is:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int degreeOfAccuracy, x = 1;
long double e = 1;
cout << "Enter degree of accuracy of mathimatical constant e: ";
cin >> degreeOfAccuracy;
while (x <= degreeOfAccuracy)
{
int conter = x;
int intial = x;
long double number = x;
int counter = 1;
while (conter > 1)
{
number = number*(intial-counter);
counter++;
conter--;
}
e += (1/number);
x++;
}
cout << endl << "The mathematical constantr e is: "
<< setprecision(degreeOfAccuracy) << fixed << e << endl;
system("pause");
return 0;
}
However, when I tried e^x the following code returned a completely wrong value:
#include <iostream>
#include <iomanip>
using namespace std;
int main()
{
int degreeOfAccuracy, x = 1, exponent;
long double e = 1;
cout << "Enter degree of accuracy of mathimatical constant e: ";
cin >> degreeOfAccuracy;
cout << "Enter the number of which you wish to raise to e: ";
cin >> exponent;
int temp = exponent;
while (x <= degreeOfAccuracy)
{
exponent = temp;
int conter = x;
int intial = x;
long double number = x;
int counter = 1;
while (conter > 1)
{
number = number*(intial-counter);
counter++;
conter--;
}
int counterr = 1;
while (counterr < x)
{
exponent *= exponent;
counterr++;
}
e += (exponent/number);
x++;
}
cout << endl << "The mathematical constantr e is: " << setprecision(degreeOfAccuracy) << fixed << e << endl;
system("pause");
return 0;
}
Any ideas where the calculations went wrong?
This line:
exponent *= exponent;
is wrong. It should be:
exponent *= temp;