So this program will print perfect numbers, but one of them, 2096128, is being printed for some reason? Would really appreciate some help figuring out what is happening! Thank you! I can't figure out why one non perfect number is finding it way into the sequence!
#include <iostream>
#include <string>
#include <math.h>
#include <iomanip>
bool isPerfect(int n);
using namespace std;
int main() {
long long perfect = 0;
int first = 0;
first = (pow(2, 2 - 1))*(pow(2, 2) - 1);
cout << first << endl;
for (int i = 3, j = 1; j < 5; i += 2) {
if (isPerfect(i)) {
perfect = (pow(2, i - 1)*(pow(2, i) - 1));
cout << perfect << endl;
j++;
}
}
// pause and exit
getchar();
getchar();
return 0;
}
bool isPerfect(int n)
{
if (n < 2) {
return false;
}
else if (n == 2) {
return true;
}
else if (n % 2 == 0) {
return false;
}
else {
bool prime = true;
for (int i = 3; i < n; i += 2) {
if (n%i == 0) {
prime = false;
break;
}
}
return prime;
}
}
You're pretty much complicating this task.
Here's what I came up with:
#include <iostream>
using namespace std;
bool isPerfect(long long n);
int main()
{
int count = 5;
long long sum = 1;
for (int i = 3; count >= 0; i += 2)
{
sum += i * i * i;
if (isPerfect(sum))
{
cout << sum << endl;
count--;
}
}
system("pause");
return 0;
}
bool isPerfect(long long n)
{
int sum = 0;
for (int i = 1; i < n; i++)
{
if (n % i == 0)
sum += i;
}
return sum == n;
}
It sure isn't perfect, but will do for 5 numbers. Consider that it'll be very slow for more than 5 numbers.
Related
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int n = 5;
int nums[] = {0 ,1};
if (n == 0)
{
return 0;
} else if (n == 1)
{
return 1;
} else{
for (int i = 2; i < n + 1; i++)
{
cout << i << endl;
nums[i] = nums[i-2] + nums[i-1];
}
return nums[n]
}
return 0;
}
It is just a simple fibonacci array, but my code only gives stdout one time and it is two. FYI: this code may not be correct to compute the nth term of fibonacci array, but i am strugglling with this for loop.
I think the condition of i < n+1 is not meet, but why this for loop ends
nums[i] has size 2. You can't access nums[i] for i >= 2. An array doesn't grow. You can't change the size of an array. Use a std::vector. You've already included the header. A return statement finishes a function. In case of the main function it causes the program to stop. The return value of the main function by convention describes if the program was successful or errors occurred. You are returning
return nums[n];
That's probably not your intention. I assume you want to print the vector.
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int n = 5;
std::vector<int> nums {0 ,1};
if (n == 0)
{
return 0;
} else if (n == 1) {
return 1;
} else {
nums.reserve(n + 1);
for (int i = 2; i < n + 1; i++)
{
cout << i << '\n';
nums.emplace_back(nums[i-2] + nums[i-1]);
}
for (const auto num : nums)
{
cout << num << '\n';
}
}
return 0;
}
it's because you can't access to nums[2] so the correct version of your code is :
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int n = 5;
int nums[5] = {0 ,1};
if (n == 0)
{
return 0;
} else if (n == 1)
{
return 1;
} else{
for (int i = 2; i < n + 1; i++)
{
cout << i << endl;
nums[i] = nums[i-2] + nums[i-1];
}
return nums[n]
}
return 0;
}
you must provide estimated size to array or you will create vector or some dynamic array..
Your approach always give Segmentation fault;
#include <iostream>
#include <vector>
using namespace std;
int main()
{
int n = 5;
vector<int> nums(n+1);
nums[0]=0;
nums[1]=1;
if (n == 0)
{
return 0;
} else if (n == 1)
{
return 1;
} else{
for (int i = 2; i < n + 1; i++)
{
// cout << i << endl;
nums[i] = nums[i-2] + nums[i-1];
}
for(int i=0;i<n;i++)
{
cout<<nums[i]<<endl; }
}
return 0;
}
*/
a beginner at coding here.
I was practising loops(c++) when I stumbled upon this problem:-
Write a program in C++ to find the perfect numbers between 1 and 500. (6,28 and 496)
Perfect number: It is a positive integer that is equal to the sum of its proper divisors. The smallest perfect number is 6, which is the sum of 1, 2, and 3.
I wrote the following code:-
#include <iostream>
using namespace std;
int main() {
int n=2; //test numbers from 2 to 500.
int div=1; //divisor for n.
int sum=0; //sum of divisors which divide n.
while (n<=500) {
while (div<n){ //if div divides n, then it will added to sum and incremented, else only incremented.
if (n%div==0){
sum=sum+div;
div++;
} else{
div++;
}
}
if (sum==n){
cout<<n<<" is a perfect number."<<endl;
n++;
} else{
n++;
}
}
return 0;
}
The code is supposed to print that 6, 28 and 496 are perfect numbers.
But instead, it's not printing anything. Haven't been able to find the error yet after checking for 30+ minutes.
Could anyone point out the error?
You forget to re-initialize some variables in your loop.
for seems more appropriate than while here.
Create sub function also help to "identify" scope.
#include <iostream>
bool isPerfectNumber(int n)
{
int sum = 0;
for (int div = 1; div != n; ++div) {
if (n % div == 0) {
sum += div;
}
}
return sum == n && n > 0;
}
int main()
{
for (int i = 2; i != 501; ++i) {
if (isPerfectNumber(i)) {
std::cout << n << " is a perfect number." << std::endl;
}
}
return 0;
}
#include<iostream>
using namespace std;
bool perfect_num(int x);
int main() {
int m, n, x;
cout << "input the range m, n: " << "\n";
cin >> m >> n;
for (x = m; x <= n; ++x) {
if (perfect_num(x)) {
cout << x << " ";
}
}
return 0;
}
bool perfect_num(int x) {
bool flag = false;
//initialize
int sum = 0, i;
//loop 1 to x
for (i = 1; i < x; ++i) {
//judge whether is the factor
if (x % i == 0) {
sum += i;
}
}
//update flag
flag = (sum == x);
return flag;
}
#include<iostream>
using namespace std;
//judge function
bool isPerfectNum(int num){
int tmp = 0;
for (int i = 1; i < num; ++i) {
if (num % i == 0) {
tmp += i;
}
}
return tmp == num;
}
int main(){
cout << "Perfect Number contains: ";
for (int i = 1; i <= 500; ++i){
if (isPerfectNum(i)) {
cout << i << " ";
}
}
cout << "\n";
return 0;
}
at the end of your first loop, you should bring back div and sum to their default value.
int main() {
int n=2; //test numbers from 2 to 500.
int div=1; //divisor for n.
int sum=0; //sum of divisors which divide n.
while (n<=500) {
while (div<n){ //if div divides n, then it will added to sum and incremented, else only incremented.
if (n%div==0){
sum=sum+div;
div++;
} else{
div++;
}
}
if (sum==n){
cout<<n<<" is a perfect number."<<endl;
n++;
} else{
n++;
}
div = 1; // you should bring them back here.
sum = 0;
}
return 0;
}
I wrote this code for project euler problem #7(Find the 10001st prime number) but it isn't working and is throwing out all sorts of obviously wrong answers(such as even numbers)
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
long long numprime = 1;
long long counter = 3;
long long arrofprimes[10001];
for(int i = 0; i < 10001; i++)
{
arrofprimes[i] = 2;
}
while(true)
{
if(numprime == 10001)
{
break;
}
bool isprime = true;
for(int i = 0; i < numprime; i++)
{
if((counter % arrofprimes[i]) == 0)
{
isprime = false;
break;
}
}
if(isprime)
{
numprime++;
arrofprimes[numprime - 1] = counter;
}
counter++;
}
cout << counter << endl;
}
You have incremented the counter after finding the 10001st prime, giving you an off-by-one error.
I'm coding a recursive algorithm to take a user input N and make a N x N grid where the same number does not appear twice on either a row or a column. Almost everything's working, and duplicates don't appear in columns, but I'm having trouble getting rows working.
My code for checking duplicates in rows is the function noRowDuplicates. Duplicates are still appearing, and occasionally it'll throw a segmentation fault, but I'm not sure why.
Thanks in advance for the help!
// Author: Eric Benjamin
// This problem was solved using recursion. fill() is the recursive function.
#include <iostream>
#include <cstdlib>
#include <time.h>
using namespace std;
void fillOptions();
void fill(int arrayPosition);
int inputNum;
int gridSize;
int *grid;
int allOptionsSize = 0;
int *allOptions;
int main() {
cout << "Please enter a number!" << endl;
cin >> inputNum;
gridSize = inputNum * inputNum;
grid = new int[gridSize];
allOptions = new int[inputNum];
for (int i = 0; i < inputNum; i++) {
allOptions[i] = i + 1;
allOptionsSize++;
}
srand((unsigned)time(0));
fill(0);
delete[] grid;
delete[] allOptions;
return 0;
}
bool noColumnDuplicates(int arrPosition, int valueToCheck) {
for (int i = 1; i < inputNum; i++) {
if (arrPosition - (inputNum * i) >= 0) {
if (grid[arrPosition - (inputNum * i)] == valueToCheck) {
return false;
}
}
}
return true;
}
bool noRowDuplicates(int arrPosition, int valueToCheck) {
int rowPosition = arrPosition % inputNum; // 0 to num - 1
if (rowPosition > 0) {
for (int p = 1; p < rowPosition; p++) {
if (grid[arrPosition - p] == valueToCheck) {
return false;
}
}
}
return true;
}
void fill(int arrayPosition) {
if (arrayPosition < gridSize) {
int randomPosition = rand() % allOptionsSize;
grid[arrayPosition] = allOptions[randomPosition];
if (noColumnDuplicates(arrayPosition, grid[arrayPosition])) {
if (noRowDuplicates(arrayPosition, grid[arrayPosition])) {
if (arrayPosition % inputNum == 0) {
cout << endl;
}
cout << grid[arrayPosition] << " ";
fill(arrayPosition + 1);
} else {
fill (arrayPosition);
}
} else {
fill(arrayPosition);
}
}
}
noRowDuplicates never tests the first element of a row, which makes sense when you are trying to fill the first element of a row, but not any other time.
What is the sum of all the primes below 2000000?
Example of sum below 10 is 2+3+5+7 = 17
I wrote this code, but still getting the wrong answers:
I tested for numbers lower than a few hundreds, and it has shown the correct answers.
#include <iostream>
#include <math.h>
using namespace std;
bool isPrime(long n)
{
if (n < 2)
return false;
if (n == 2)
return true;
if (n == 3)
return true;
int k = 3;
int z = (int)(sqrt(n) + 1); // square root the n, because one of the product must be lower than 6, if squared root of 36
if (n % 2 == 0)
return false;
while (n % k != 0)
{
k += 2;
if (k >= z)
return true;
}
return false;
}
long primeSumBelow(long x)
{
long long total = 0;
for (int i = 0; i < x; i++) // looping for times of prime appearing
{
if (isPrime(i) == true)
total += i;
if (isPrime(i) == false)
total += 0;
}
cout << "fd" << endl;
return total;
}
int main()
{
cout << primeSumBelow(20) << endl;
cout << primeSumBelow(2000000) << endl;
system("pause");
return 0;
}
The total counter's type is correctly long long. Unfortunately the function primeSumBelow returns only long so, depending on the platform, the correctly calculated result is truncated when it's returned from this function.