I'm doing question 4.11 in Bjarne Stroustrup Programming-Principles and Practice Using C++.
Create a program to find all prime numbers in the range from 1 to max using a vector of primes in order(prime[2,3,5,...]). Here is my solution:
#include <iostream>
#include <string>
#include <vector>
#include <bits/stdc++.h>
using namespace std;
bool check_prime(vector<int> &prime, int n) {
int count = 0;
for (int i = 0; prime[i] <= n || i <= prime.size() - 1; ++i) {
if (n % prime[i] == 0) {
count++;
break;
}
}
bool result = 0;
if (count == 0)
result = 1;
else
result = 0;
return result;
}
int main() {
vector<int> prime{2};
int max;
cout << "Please enter a max value:";
cin >> max;
for (int i = 2; i <= max; ++i) {
if (check_prime(prime, i))
prime.push_back(i);
}
for (int i = 0; i <= prime.size() - 1; ++i) {
cout << prime[i];
if (i <= prime.size() - 2)
cout << ',';
}
}
My code is working for numbers smaller than 23 but fail to work for anything bigger. If I open the program in Windows 10 the largest working number increase to 47, anything bigger than that fail to work.
This condition
prime[i]<=n||i<=prime.size()-1
makes the loop continue as long as at least one of them is true, and you're accessing prime[i] without checking the value of i.
This will cause undefined behaviour as soon as i == prime.size().
This means that anything can happen, and that you're experiencing that any specific values are working is just an unfortunate coincidence.
You need to check the boundary first, and you should only continue for as long as both conditions are true:
i <= prime.size() - 1 && prime[i] <= n
which is more idiomatically written
i < prime.size() && prime[i] <= n
(It's never too soon to get comfortable with the conventional half-open intervals.)
You check prime[i]<=n before i<=prime.size()-1. Then, if it's true (even if i>prime.size()-1, which is random behaviour), you work on it, generating wrong results.
Related
#include <iostream>
#include<ctime>
#include<cstdlib>
#include<string>
#include<cmath>
using namespace std;
int main()
{
bool cont = false;
string str;
int num, num2;
cin >> str >> num;
int arr[10];
int a = pow(10, num);
int b = pow(10, (num - 1));
srand(static_cast<int>(time(NULL)));
do {
num2 = rand() % (a - b) + b;
int r;
int i = 0;
int cpy = num2;
while (cpy != 0) {
r = cpy % 10;
arr[i] = r;
i++;
cpy = cpy / 10;
}
for (int m = 0; m < num; m++)
{
for (int j = 0; j < m; j++) {
if (m != j) {
if (arr[m] == arr[j]) {
break;
}
else {
cont = true;
}
}
}
}
cout << num2 << endl;
} while (!cont);
return 0;
}
I want to take a number from the user and produce such a random number.
For example, if the user entered 8, an 8-digit random number.This number must be unique, so each number must be different from each other,for example:
user enter 5
random number=11225(invalid so take new number)
random number =12345(valid so output)
To do this, I divided the number into its digits and threw it into the array and checked whether it was unique. The Program takes random numbers from the user and throws them into the array.It's all right until this part.But my function to check if this number is unique using the for loop does not work.
Because you need your digits to be unique, it's easier to guarantee the uniqueness up front and then mix it around. The problem-solving principle at play here is to start where you are the most constrained. For you, it's repeating digits, so we ensure that will never happen. It's a lot easier than verifying if we did or not.
This code example will print the unique number to the screen. If you need to actually store it in an int, then there's extra work to be done.
#include <algorithm>
#include <iostream>
#include <numeric>
#include <random>
#include <vector>
int main() {
std::vector<int> digits(10);
std::iota(digits.begin(), digits.end(), 0);
std::shuffle(digits.begin(), digits.end(), std::mt19937(std::random_device{}()));
int x;
std::cout << "Number: ";
std::cin >> x;
for (auto it = digits.begin(); it != digits.begin() + x; ++it) {
std::cout << *it;
}
std::cout << '\n';
}
A few sample runs:
Number: 7
6253079
Number: 3
893
Number: 6
170352
The vector digits holds the digits 0-9, each only appearing once. I then shuffle them around. And based on the number that's input by the user, I then print the first x single digits.
The one downside to this code is that it's possible for 0 to be the first digit, and that may or may not fit in with your rules. If it doesn't, you'd be restricted to a 9-digit number, and the starting value in std::iota would be 1.
First I'm going to recommend you make better choices in naming your variables. You do this:
bool cont = false;
string str;
int num, num2;
cin >> str >> num;
What are num and num2? Give them better names. Why are you cin >> str? I can't even see how you're using it later. But I presume that num is the number of digits you want.
It's also not at all clear what you're using a and b for. Now, I presume this next bit of code is an attempt to create a number. If you're going to blindly try and then when done, see if it's okay, why are you making this so complicated. Instead of this:
num2 = rand() % (a - b) + b;
int r;
int i = 0;
int cpy = num2;
while (cpy != 0) {
r = cpy % 10;
arr[i] = r;
i++;
cpy = cpy / 10;
}
You can do this:
for(int index = 0; index < numberOfDesiredDigits; ++index) {
arr[index] = rand() % 10;
}
I'm not sure why you went for so much more complicated.
I think this is your code where you validate:
// So you iterate the entire array
for (int m = 0; m < num; m++)
{
// And then you check all the values less than the current spot.
for (int j = 0; j < m; j++) {
// This if not needed as j is always less than m.
if (m != j) {
// This if-else is flawed
if (arr[m] == arr[j]) {
break;
}
else {
cont = true;
}
}
}
}
You're trying to make sure you have no duplicates. You're setting cont == true if the first and second digit are different, and you're breaking as soon as you find a dup. I think you need to rethink that.
bool areAllUnique = true;
for (int m = 1; allAreUnique && m < num; m++) {
for (int j = 0; allAreUnique && j < m; ++j) {
allAreUnique = arr[m] != arr[j];
}
}
As soon as we encounter a duplicate, allAreUnique becomes false and we break out of both for-loops.
Then you can check it.
Note that I also start the first loop at 1 instead of 0. There's no reason to start the outer loop at 0, because then the inner loop becomes a no-op.
A better way is to keep a set of valid digits -- initialized with 1 to 10. Then grab a random number within the size of the set and grabbing the n'th digit from the set and remove it from the set. You'll get a valid result the first time.
I wrote a C++ program that prints all prime numbers lower than n, but the program keeps crashing while executing.
#include <iostream>
using namespace std;
bool premier(int x) {
int i = 2;
while (i < x) {
if (x % i == 0)
return false;
i++;
}
return true;
}
int main() {
int n;
int i = 0;
cout << "entrer un entier n : ";
cin >> n;
while (i < n) {
if (n % i == 0 && premier(i))
cout << i;
i++;
}
;
}
As Igor pointed out, i is zero the first time when n%i is done. Since you want only prime numbers and the smallest prime number is 2, I suggest you initialise i to 2 instead of 0.
You want to print all prime numbers less than n and has a function to check primality already.
Just
while (i < n){
if ( premier(i) == true )
cout<<i;
i++;
}
And while printing, add a some character to separate the numbers inorder to be able to distinguish them like
cout<<i<<endl;
P.S: I think you call this a C++ program. Not a script.
Edit: This might interest you.
I wrote the following dp code for finding the prime factors of a number.
#include <bits/stdc++.h>
#define max 1000001
using namespace std;
vector <int> prime;
vector<bool> isprime(max,true);
vector<bool> visited(max,false);
vector<int> data(max,-1);
void dp(int n,int last)
{
if(n >= max || visited[n])
return;
visited[n] = true;
for(int i = last;i<prime.size();i++)
{
if(n*prime[i] >= max || data[n*prime[i]] != -1)
return;
data[n*prime[i]] = prime[i];
dp(n*prime[i],i);
}
}
int main()
{
isprime[1] = false;
data[1] = 1;
for(int i = 4;i<max;i += 2)
isprime[i] = false;
for(int i = 3; i*i< max;i += 2)
{
for(int j = i*i; j < max;j += i)
isprime[j] = false;
}
prime.push_back(2);
data[2] = 2;
for(int i =3;i<max;i += 2)
if(isprime[i])
{
prime.push_back(i);
data[i] = i;
}
for(int i = 0;i<prime.size();i++)
{
dp(prime[i],i);
}
cout<<"...1\n";
for(int i = 2;i<=8000;i++)
{
cout<<i<<" :- ";
int temp = i;
while(temp!= 1)
{
cout<<data[temp]<<" ";
temp = temp/data[temp];
}
cout<<endl;
}
return 0;
}
Here, last is the last index of prime number n.
But I am getting segmentation fault for this, when I change max to 10001, it runs perfectly. I'm not getting why is this happening since the data-structures used are 1-d vectors which can hold values up to 10^6 easily.
I checked your program out using GDB. The segfault is taking place at this line:
if(n*prime[i] >= max || data[n*prime[i]] != -1)
In your first ever call to DP in your for loop, where you call dp(2,0), the recursive calls eventually generate this call: dp(92692,2585).
92692 * 2585 = 239608820
This number is larger than a 32 bit integer can hold, so the r-value generated by the integer multiplication of those two numbers overflows and becomes negative. nprime[i] becomes negative, so your first condition of the above loop fails, and the second is checked. data[n * prime[i]] is accessed, and since n*prime[i] is negative, your program accesses invalid memory and segfaults. To fix this, simply change n to a long long in your parameter list and you should be fine.
void dp(long long n, int last)
I'm trying to write a code shows all the numbers with the following characteristics:
the number itself is a prime number.
for each digit removed from the right the remaining number should still be a prime number.
Considering the number 293 for example: 293 itself is a prime number if we delete the digit on the right we have 29 which is still a prime number, and if we delete the right digit again we have 2 which is still prime.
I'm trying to write a code that gets the integer n<=8 from the user and shows all the n-digit numbers that have the characteristics stated above. My algorithm is to write a recursive function (show) that returns the vector v.
If n=1 then it just shows the numbers 2-3-5-7... if n!=1 it should call show(n-1) and multiply all the generated numbers by 10 and add them up with odd numbers... then it should check if the new number is prime. If so it should be added to the vector.
My problem is the code only works for n=1. Here is my code:
#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>
#include <iterator>
using namespace std;
bool isPrime(int a)
{
int i, p = 0;
if (a == 1)
return false;
else
{
for (i = a - 1; i > sqrt(a); i--)
if (a % i == 0)
p++;
if (p != 0)
return false;
else
return true;
}
}
vector<int> show(int n)
{
vector<int> v;
int i, j;
if (n == 1)
{
v.push_back(2);
v.push_back(3);
v.push_back(5);
v.push_back(7);
}
else
{
show(n - 1);
if (n != 1)
for (i = 0; i < v.size(); i++)
{
for (j = 1; j <= 9; j += 2)
if (isPrime((v.at(i) * 10) + j))
v.at(i) = (v.at(i) * 10) + j;
}
}
return v;
}
int main()
{
int n, s = 0, i;
cin >> n;
show(n);
for (i = 0; i < show(n).size(); i++)
cout << show(n).at(i) << endl;
system("pause");
return 0;
}
In addition to comments on question, take a look at the show(n - 1) line. Shouldn't you be saving the return value: v = show(n - 1)?
Also, it will be better to pass the vector as a reference, in this way you avoid the copy of vector's content (in your case it doesn't have much impact since vector will not grow too much, but imagine using large values for n).
I have to write a program that runs a loop for a coin toss. I am supported to enter a number into the console and have it run a loop of the coin toss for that many times. I need to use nested loops. I have been working on this for hours and cannot make it work.
The console i/o is supposed to look like below:
Enter the number of tosses to perform [0=exit]: 3
Heads
Tails
Heads
Enter the number of tosses to perform [0=exit]: 2
Tails
Tails
Enter the number of tosses to perform [0=exit]: 0
This is the code i have so far:
#include <iostream>
#include <ctime>
#include <cstdlib>
using namespace std;
int main ()
{
srand(time(0));rand();
int result = rand() % 2;
while (true)
{
int n; // this many tosses
cout << "How many tosses";
cin >> n;
cin.ignore (1000, 10);
if (n == 0)
break;
for (int i = 0; i < n; i++)
//random number generator
{
if (result == 0)
cout<< "Heads"<<endl;
else if (result == 1)
cout << "Tails"<<endl;
else if (result != 0 || result !=1)
return 0;
} //for
}//while
}//main
Your for loop doesn't have the part that you are actually trying to execute inside of {}. Try adding the braces around the part you want to loop and see if that fixes it for you.
I edited the indentation in your code to show you the only line that will actually be looping (the srand(time(0)))
You need brackets around the loop block, i.e.
for( int i = 0; i < n; i++ )
{
// Code goes here
}
As shown above, you need to initialize i.
Put the seeding of rand() before the while(...) loop.
You need to move int result = rand() % 2; inside your for loop! Otherwise you will get the same result every single time until you restart the application.
for (int i = 0; i < n; i++)
//random number generator
{
int result = rand() % 2;
if (result == 0)
cout<< "Heads"<<endl; /* to make your output look like your example you should removed the <<endl from here */
else if (result == 1)
cout << "Tails"<<endl; /* and here */
else if (result != 0 || result !=1)
return 0;
} //for
/* and put it here */
cout << endl;