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I am currently testing some questions to practice my programming skills. (I didn’t take it at school or something else, I taught it myself). I ran into this problem that required me to read the number from a given txt file. This number will be N. Now I have to find an N-dimensional number for N <= 10,000. After I find it, I have to print it in another txt file. Now for most parts of the question, I can understand and develop a method for getting N. The problem is that I use an array to store previously found primes, so I can use them to check for future numbers. Even when my array was 100 in size, as long as the integer was approximately <15, the program would crash.

#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <fstream>
using namespace std;

int main() {
    ifstream trial;
    trial.open("C:\\Users\\User\\Documents\\trial.txt");
    int prime;
    trial >> prime;
    ofstream write;
    write.open("C:\\Users\\User\\Documents\\answer.txt");
    int num[100], b, c, e;
    bool check;
    b = 0;
    switch (prime) {
        case 1:
        {
            write << 2 << endl;
            break;
        }
        case 2:
        {
            write << 3 << endl;
            break;
        }
        case 3:
        {
            write << 5 << endl;
            break;
        }
        case 4:
        {
            write << 7 << endl;
            break;
        }
        default:
        {
            for (int a = 10; a <= 1000000; a++) {
                check = false;
                if (((a % 2) != 0) && ((a % 3) != 0) && ((a % 5) != 0) && ((a % 7) != 0)) // first filter
                {
                    for (int d = 0; d <= b; d++) {
                        c = num[d];
                        if ((a % c) == 0) {
                            check = true; // second filter based on previous recorded primes in array
                            break;
                        }

                    }
                    if (!check) {
                        e = a;
                        if (b <= 100) {
                            num[b] = a;
                        }

                        b = b + 1;
                    }
                }
                if ((b) == (prime - 4)) {
                    write << e << endl;
                    break;
                }
            }
        }
    }
    trial.close();
    write.close();
    return 0;
}

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14
#include <cstdio>
#include <iostream>
#include <cstdlib>
#include <fstream>

using namespace std;

int main()
{
ifstream trial;
trial.open("C:\\Users\\User\\Documents\\trial.txt");
int prime, e;
trial>>prime;
ofstream write; 
write.open("C:\\Users\\User\\Documents\\answer.txt");
int num[10000], currentPrime, c, primePrint;
bool check;
currentPrime=0;
num[currentPrime] = 2;
currentPrime=1;

for(int currentInt=2; currentInt<=1000000; currentInt++) 
{check = false; 
for( int arrayPrime=0; arrayPrime<currentPrime; arrayPrime++) 
        { c=num[arrayPrime];
            if ((currentInt%c)==0) { check = true;// second filter based on previous recorded primes in array
                    break;}
               }


 if (!check)
    { e=currentInt;  
    if( currentInt!= 2 ) { 
    num[currentPrime]= currentInt;}
    currentPrime = currentPrime+1;}
    if(currentPrime==prime)
    {
    write<<e<<endl;
    break;}
    }
trial.close();
write.close();
return 0;
}

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int someArray[100];
someArray[150] = 10;

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#include <vector> // std::vector

...

const int MAX_ITEMS = 100;

std::vector<int> intVector;

intVector.reserve(MAX_ITEMS); // allocates all memory up-front

// add items
for (int i = 0; i < MAX_ITEMS; i++)
{
  intVector.push_back(i);  // this is how you add a value to a vector;
}

// print them
for (int i = 0; i < intVector.size(); i++)
{
  int elem = intVector[i]; // this access the item at index 'i'
  printf("element %d is %d\n", i, elem);
}
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for( int d=0; d<b; d++)

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#include <iostream>
#include <list>
#include <math.h>
#include <functional>
#include <algorithm>

using namespace std;

class is_multiple : public binary_function<int, int, bool>
{
    public:
        bool operator()(int value, int test) const
        {
            if(value == test) // do not remove the first value
                return false;
            else
                return (value % test) == 0;
        }
};

int main() 
{
    list<int> numbersToTest;
    int input = 500;

    // add all numbers to list
    for(int x = 1; x < input; x++)
        numbersToTest.push_back(x);

    // starting at 2 go through the list and remove all multiples until you reach the squareroot
    // of the last element in the list
    for(list<int>::iterator itr = ++numbersToTest.begin(); *itr < sqrt((float) input); itr++)
    {
        int tmp = *itr;
        numbersToTest.remove_if(bind2nd(is_multiple(), *itr));  
        itr = find(numbersToTest.begin(), numbersToTest.end(), tmp); //remove_if invalidates iterator 
                                                                 // so find it again. kind of ugly
    }

    // output primes
    for(list<int>::iterator itr = numbersToTest.begin(); itr != --numbersToTest.end(); itr++)
        cout << *itr << "\t";

    system("PAUSE");

    return 0;
}

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+1

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#include <iostream>
#include <fstream>
#include <cmath>
using namespace std;

int main()
{
    int m;
    int n=0;
    char ch;
    fstream fp;
    cout<<"What prime numbers do you want get within?  ";
    if((cin>>m)==0)
    {
                 cout<<"Bad input! Please try again!\n";
                 return 1;
    }
    if(m<2)
    {
        cout<<"There are no prime numbers within "<<m<<endl;
        return 0;
    }
    else if(m==2)
    {
        fp.open("prime.txt",ios::in|ios::out|ios::trunc);//create a file can be writen and read. If the file exist, it will be overwriten.
        fp<<"There are only 1 prime number within 2.\n";
        fp<<"2\n";
        fp.close();
        cout<<"Congratulations! It has worked out!\n";
        return 0;
    }
    else
    {
        int j;
        int sq;
        fp.open("prime.txt",ios::in|ios::out|ios::trunc);
        fp<<"2\t\t";
        n++;
        for(int i=3;i<=m;i+=2)
        {
            sq=static_cast<int>(sqrt(i))+1;
            fp.seekg(0,ios::beg);
            fp>>j;
            for(;j<sq;)
            {
                if(i%j==0)
                {
                    break;
                }

                else
                {
                    if((fp>>j)==NULL)
                    {
                        j=3;
                    }
                }
            }
            if(j>=sq)
            {
                fp.seekg(0,ios::end);
                fp<<i<<"\t\t";
                n++;
                if(n%4==0)
                    fp<<'\n';
            }
        }
        fp.seekg(0,ios::end);
        fp<<"\nThere are "<<n<<" prime number within "<<m<<".\n";
        fp.close();
        cout<<"Congratulations! It has worked out!\n";
        return 0;
    }
}
+1

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                for (int d = 0; d <= b; d++) {
                    c = num[d];

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for(int currentInt=2; currentInt<=1000000; currentInt++) 

{check = false;  // Basically the idea for this for loop is to run checks against integers. This is the main for loop in this program. I re initialize check to false ( check is a bool declared above this. )

for( int arrayPrime=0; arrayPrime<currentPrime; arrayPrime++) // This for loop is used for checking the currentInt against previously found primes which are stored in the num array.

{ c=num[arrayPrime];
        if ((currentInt%c)==0) { check = true;// second filter based on previous recorded primes in array
                break;}  // this is the check. I check the number against every stored value in the num array. If it divisible by any of them, then bool check is set to true.


if ( currentInt == 2)
{ check = false; } // since i preset num[0] = 2 i make an exception for the number 2.

if (!check)
{
e=a;
if(currentPrime <= 100){
num[currentPrime]= currentInt;} // This if uses check to see if the currentInt is a prime. 

currentPrime = currentPrime+1;} // increases the value of currentPrime ( previously b ) by one if !check.

if(currentPrime==prime)
{
write<<e<<endl;
break;}           // if currentPrime == prime then write the currentInt into a txt file and break loop, ending the program.

polythinker =)

0

Since you will need more prime numbers for later questions , I suggest you follow the tips in trivia and make a sieve. This is a very useful arrow for your quiver.

0
source

Source: https://habr.com/ru/post/1709444/

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