I am new to coding and my task is to make the variable 'sum' greater than (not equal to) m by summing up 1/n for an increasing 'n'.
I need to solve the same problem twice (one using a for-loop and once using a while-loop). But both ways end in an infinte loop.
My code is working fine when I replace the "<=" with "<". But that
Can someone help me?
#include <iostream>
using namespace std;
int main () {
unsigned int m = 1;
double sum = 0;
long n;
n = 1;
while (sum <= m) //THIS LINE
{
double sumsum = 1/n;
sum += sumsum;
n++;
}
cout << "n = " << n << endl ;
cout << "sum = " << sum << endl ;
return 0;
}
Epression 1/n with n being of type long and having a value > 1 will always yield 0, since you operate with integral types. Hence, sum will be assigned 1 in the first run, but will never come to a value > 1, as forth on always 0 is added.
Change your code to
double sumsum = 1.0/n;
and it should work. Note that 1.0 enforces to operate with floating points.
Related
I'm writing a method to check if a number is palindrome or not. For example 12321 is palindrome and 98765 is not.
In my program I've used a recursive function to create exactly opposite of given number, like 56789 for 98765 and then checking if two numbers are equal or not. But I'm not getting exact opposite of 98765 which is 56789 instead I'm getting 56787.
Here's my code-
#include<iostream>
#include<bits/stdc++.h>
using namespace std;
long int oppositeNum(int n){
if(n<10 && n>=0) return n;
if(n<0) return 0;
static int m=0;
int x = n%10;
long int num = oppositeNum(n/10);
cout << num << "\n";
return (num+ (x*pow(10,++m)));
}
int main(){
int n = 98765;
int oppNum = oppositeNum(n);
cout << oppNum;
if(oppNum==n){
cout << "Number is palindrome";
}else{
cout << "Number is not palindrome";
}
return 0;
}
I'm not getting the exact opposite of my original nnumber. the last digit is getting decremented by 1 every time is what I've observed.
Can anyone help?
I can not reproduce the result you are getting. Maybe it is a consequence of using the function pow
But in any case your function may not be called the second time for a different number because the static variable m is not reinitialized to 0. m continues to keep the value after the previous call of the function for a different number.
You can write the function without using the function pow.
Take into account that a reversed number can be too large to be stored an object of the type long because in some systems the type long has the same width as the type int. So I adjust instead of the type long to use the type long long as the return type.
Here you are.
#include <iostream>
long long int oppositeNum( int n )
{
static long long int multiplier = 1;
const int Base = 10;
int digit = n % Base;
return ( n /= Base ) == 0
? ( multiplier = 1, digit )
: ( n = oppositeNum( n ) , multiplier *= Base, digit * multiplier + n );
}
int main()
{
int n = 12321;
std::cout << n << " -> " << oppositeNum( n ) << '\n';
n = - 12321;
std::cout << n << " -> " << oppositeNum( n ) << '\n';
n = 98765;
std::cout << n << " -> " << oppositeNum( n ) << '\n';
n = -98765;
std::cout << n << " -> " << oppositeNum( n ) << '\n';
return 0;
}
The program output is
12321 -> 12321
-12321 -> -12321
98765 -> 56789
-98765 -> -56789
I changed only as much as it seemed neccessary. It still could use some tail recursion but this would probably required rewriting the whole algorithm.
long int oppositeNum(int n, int &m){
if(n<10) return n;
int x = n%10;
long int num = oppositeNum(n/10, m);
cout << num << "\n";
m *= 10;
return num + x * (long int)m;
}
long int oppositeNum(int n){
if(n<0) return 0;
int m = 1;
return oppositeNum(n, m);
}
What did I change:
Removed the static modifier from m and instead passing it by reference. This allows to use the function more than one time during the program execution. (Also it would allow to use the function by multiple threads at once but I guess this is not a concern here.)
Removed the floating point function pow and instead just multiplying the variable m by ten each iteration.
Added a wrapper for the recursive function so it still can be called with just one argument. Additionally this allow to check for negative numbers only once.
The main source of problem was the function pow. Because it works on floating point numbers, it may not give exact results. It depends on the compiler and the processor architecture but generally you shouldn't expect it to give an exact result. Rounding it to integer additionally increases the difference.
return (num + float(x*pow(10,++m)));
just do this and it will work. the ans is wrong cos the pow func is giving out 4999 and 5999 that is why it changes the units place.
Try this:
return round(num+ (x*pow(10,++m)));
Sometimes the pow function will return the approximated result. e.g. pow(10, 5) could be 99999.9999999, if you round it, you'll get 100000 else it takes the floor value (I think).
I've been playing with finding divisors of numbers recently and decided to try and make a program, which prints all ways to express a number N as a product of two integers. I wrote one which works on positive numbers and only considers positive numbers to make up the product.
#include <iostream>
#include <cmath>
int main()
{
int n;
std::cin >> n;
int root = std::sqrt(n);
for(int i = 1; i <= root; i++)
{
if(n % i != 0)
continue;
int j = n / i;
std::cout << i << ", " << j << std::endl;
}
return 0;
}
The code above just finds all divisors of N and prints them as pairs. It works fine, but I wanted to try and make it find all possible ways to get to N, not only with positive numbers.
For example, if I input 10 in the program above, the results will be (1, 10), (2, 5); These are correct, but there are other ways to multiply two numbers and get to 10. It involves negative numbers: (-1, -10), (-2, -5) are also solutions, since when you multiply two negative numbers, you end up with a positive one.
If I wanted the program to only work on positive N values but also find negative multiples, I could just print the negative versions of i and j, since you can only get to a positive number by either multiplying two positive or two negative together.
That works, but now I want to get this code to work on negative N values. For example, an expected output for N = -10 would be: (-1, 10), (1, -10), (2, -5), (-2, 5);
The problem is, the algorithm above can only find positive divisors for positive numbers, since it involves square root, which is only defined for positive numbers, and the loop starts at a positive and ends at a positive.
I noticed that I can just calculate the square root of the absolute value of N, then make the loop start at -root and end at root to go over the negative divisors of N as well. I had to make sure to skip 0, though, because division with 0 isn't defined and that made it crash. The code I ended up with looks like this:
#include <iostream>
#include <cmath>
int main()
{
int n;
std::cin >> n;
int root_n = std::sqrt(std::abs(n));
for(int i = -root_n; i <= root_n; i++)
{
if(i == 0 || n % i != 0)
continue;
int j = n / i;
std::cout << i << ", " << j << std::endl;
}
return 0;
}
It worked properly for all the tests I came up with, but I am not sure if it's the best way to write it. Is there anything that I can improve?
Thanks in advance!
EDIT: Tried using std::div as suggested by Caleth (also used ReSharper addon in VS to give me refactoring suggestions):
#include <iostream>
#include <cstdlib>
int main()
{
int n;
std::cin >> n;
const int sqrt_n = std::sqrt(std::abs(n));
for(auto i = -sqrt_n; i <= sqrt_n; i++)
{
if (i == 0)
continue;
const auto div_res = std::div(n, i);
if (div_res.rem)
continue;
std::cout << i << ", " << div_res.quot << std::endl;
}
return 0;
}
Instead of calculating the remainder, then calculating the quotient, I can just do a single call to std::div, which returns a struct, containing both values.
A major observation is that the negative divisors and the positive divisors of a number are the same, except for sign. You could save half of the time if you looked only at positive numbers and handled signs on your own. For example,
int main()
{
int n;
std::cin >> n;
int root_n = std::sqrt(std::abs(n));
for(int i = 1; i <= root_n; i++) \\ don't loop over negative numbers now
{
if(n % i != 0) \\ not necessary to check for 0 anymore
continue;
int j = n / i;
std::cout << i << ", " << j << "\n";
std::cout << -i << ", " << -j << "\n"; \\ corresponding negative
}
return 0;
}
You might notice that I removed std::endl --- you don't really need to flush the stream that often. It's faster to let output buffer as it wants.
If you are looking for other ways to modify your program, you might try finding the prime factorization and then computing lists of divisors from it. For large, highly composite inputs, this will be faster. But it's also quite different.
I need help figuring out where my code went wrong. I want to reset the values for the loop so that it isn't compiling because my output right now is using past input values in current calculation whereas I want the output to be different every time as though it is the firs time running the code. The code works fine when I don't use the while loop, but then I have to rerun the program each time. I want the output to prompt a new input every time, but not use past inputs in the new calculations. I know I'm not explaining it very well, but I'm just sort of lost. Anything helps!
This is my problem:
An approximate value of pi can be calculated using the series given
below:
pi = 4 · [ 1 – 1/3 + 1/5 – 1/7 + 1/9 ... + (–1 ^ n)/(2n + 1) ]
Write a C++ program to calculate the approximate value of pi using
this series. The program takes an input n that determines the number
of terms in the approximation of the value of pi and outputs the
approximation. Include a loop that allows the user to repeat this
calculation for new values n until the user says she or he wants to
end the program.
#include <cstdio>
#include <iostream>
using namespace std;
int main()
{
int n;
double sum=0;
cout << "Enter the number of terms to approximate (or zero to quit):\n";
cin >> n;
if (n == 0)
{
return 0;
}
while (n != 0)
{
{ for(int i=0;i<n;i++)
{
if (i%2==0)
{
sum += 1.0/(2*i+1);
}
else
{
sum += -1.0/(2*i+1);
}
}
cout.setf(ios::showpoint);
cout.precision(3);
cout << "The approximation is " << sum*4 << " using " << n << " terms." << endl;
}
cout << "Enter the number of terms to approximate (or zero to quit):\n";
cin >> n;
}
return 0;
}
This is my output:
This is what the output should be:
You do not reset sum before entering the for loop. Just add
sum=0;
before the for line.
Notice that 2.67 = 6.67 - 4.00.
You want your program to compute a sum for several values of n.
The sum must be initialized to 0 at the beginning of each calculation, inside the while loop.
Actually, it should even be declared there.
C++ doesn't require you to define variables at the start of a function, so it is perfectly legal to write:
while (n != 0)
{
double sum = 0.0;
this would solve your problem. Alternatively, if you want to keep the declaration of sum at the top of the function, just change your code to
while (n != 0)
{
sum = 0.0;
I have this program to write that I have a array of 11 numbers entered from me. Then I need to find the avarage sum of those numbers, and then im asked to find the closest number of this array to the avarage sum, and then the most distant element of the array to the avarage sum again. SO far I manage to write a program to create this array and find the avarage sum. I asssume there is something to do with abs function of cmath libary , but so far I only fail to make it.
#include <iostream>
using namespace std;
int main() {
unsigned const int size = 11;
float number[size];
for (unsigned i = 0; i<size; i++) {
cout << "Please enter value for number "
<< i + 1 << ":";
cin >> number[i];
}
for (unsigned i = 0; i<size; i++) {
cout << "Number " << i + 1 << " is : "
<< number[i] << endl;
}
unsigned int sum = 0;
for (unsigned i = 0; i<size; i++) {
sum += number[i];
}
What is the problem? You are not asking a question, just making a statement... It does seem that you have not posted the whole code..
In c++ usually to use "abs" you should use fabs from the "math.h" library!
You will be okay with the compare operators.
Just traverse your array in a loop and calculate the difference between your compare value and the current value on your array. Initiate a temporary variable that keeps the array entry that created the smallest difference.
Every time a difference that is smaller than the current one comes up replace the value in your temporary variable.
So you replace under the following condition: If |number[i] - average_value| < |tmp_closest_val -average_val| Then tmp_closest_val = number[i] EndIf.
I hope you get the concept from that rough draft.
This is my code for finding the sum of a harmonic series of 1/n. I want it to stop when the sum is greater than or equal to 15, but the code cannot run. Can anyone let me know what I'm doing wrong? It seems to follow the correct while loop structure. Thanks!
#include <iostream>
using namespace std;
int main ()
{
int divisor = 1;
int sum = 1;
while ((sum <= 15) && (divisor >=1))
{
sum = sum + (1/divisor);
divisor++;
}
cout << "You need " << divisor << " terms to get a sum <= 15" << endl;
return 0;
}
Your loop is actually running. However, your sum variable is of type int, and so is divisor.
1 (an int) / divisor (also an int) will return 1 or 0. This is because you are doing integer division. 1/1 == 1. However, 1/2 == 0, 1/3 == 0, etc... To solve this, cast divisor to double:
(1 / (double)divisor)
So that solves the issue of that segment returning only 1 or 0. However, you will still gain a sum of 1 as sum is of type int. Attempting to assign a double to an int variable will result in a truncation, or floor rounding. Sum will add the first 1, but it will remain 1 indefinitely after that. In order to solve this, change the type of sum to double.
Your assignment of sum = 1; is a logical error. Your result will be 1 higher than it should be. Your output statement is also mistaken... It should be...
cout << "You need " << divisor << " terms to get a sum > 15" << endl;
In addition, the condition of divisor >= 1 is needless... It is always greater than or equal to one because you assign it as 1 and are incrementing... If you do want a sum that is >= 15, change the while condition to...
while (sum < 15)
Your code should look like this...
#include <iostream>
using namespace std;
int main()
{
int divisor = 1;
double sum = 0; //Changed the type to double and assigned 0 rather than 1
while (sum <= 15) //While condition shortened...
{
sum = sum + (1 / (double)divisor); //Added type cast to divisor
divisor++;
}
//cout statement adjusted...
cout << "You need " << divisor << " terms to get a sum > 15" << endl;
return 0;
}