In this question we have to maximize the modulo function. A string is given And we have remove the element and check after removing a element which number give the maximum modulo answer.
Link:- https://www.hackerearth.com/practice/data-structures/arrays/1-d/practice-problems/algorithm/maximize-modulo-2-0cb15ded/
My code is not able to pass all test case it shows runtime error or wrong answer in most test case.
#include "bits/stdc++.h"
using namespace std;
int countDigit(long long n)
{
int count = 0;
while (n != 0)
{
n = n / 10;
++count;
}
return count;
}
int main() {
int t;
cin>>t;
for(int i=0;i<t;i++){
int m,k;
cin>>m>>k;
string s;
cin>>s;
int ans = INT_MIN;
int n = countDigit(k);
if(n == m && stoi(s) < k){
cout<<stoi(s)<<endl;
continue;
}
else{
for(int i=0;i<m;i++){
string a = "";
for(int j=0;j<m;j++){
if(j == i){
continue;
}
else{
a += s[j];
}
}
int mo;
int num = stoi(a);
mo = num%k;
if(mo > ans){
ans = mo;
}
}
cout<<ans<<endl;
}
}
return 0;
}
Please tell me where i'm making the mistake or please tell me any better way to solve this question?
#include <iostream>
using namespace std;
int findPrime(unsigned long long int number);
bool isPrime(unsigned long long int number);
int main(){
const unsigned long long int bound = 600851475143;
cout << findPrime(bound) << endl;
}
int findPrime(unsigned long long int number){
int largestFactor = 0;
for(int i = 1; i <= sqrt(number); i++){
if(number % i == 0){
int possible = number/i;
if(isPrime(possible))
largestFactor = possible;
}
}
return largestFactor;
}
bool isPrime(unsigned long long int number){
for(int i = 0; i <= sqrt(number); i++){
if(number % i == 0){
return true;
}
}
return false;
}
The purpose of this code is to find the largest prime factor. I am getting Floating point exception: 8 when running it, it is in Visual Studio Code and I am running from terminal.
In the first iteration of this code
bool isPrime(unsigned long long int number){
for(int i = 0; i <= sqrt(number); i++){
if(number % i == 0){
return true;
}
}
return false;
}
Division by zero is done, but it is not allowed. The loop should start from 2, not 0, for primality checking.
#include <iostream>
#include <math.h>
using namespace std;
int findPrime(long long int number);
bool isPrime(long long int number);
int main(){
long long int bound = 600851475143;
cout << findPrime(bound) << endl;
}
int findPrime(long long int number){
long long int largestFactor = 0;
for(int i = 1; i*i <=number; i++){
if(number % i == 0){
long long int possible = number/i;
if(isPrime(possible)){
largestFactor = possible;
largestFactor = 1;
}
}
}
return number;
if(largestFactor == 0){
return number;
}
return largestFactor;
}
bool isPrime(long long int number){
for(int i = 2; i*i <= number; i++){
if(number % i == 0){
return false;
}
}
return true;
}
I ran this instead and I'm getting still getting floating point exception: 8, which is really weird. I changed to unsigned long long int but then the number became stored as a negative number -443946297.
Please help me to solve the query that this code runs infinitely at a particular line.
It does not give any output as at the end of the code I write the code to print the vector. Even after I assign any value to vector "result" manually still it is not giving any output. why is it so?
#include<bits/stdc++.h>
using namespace std;
bool authorize(int strValue, int value, int M)
{
long int newValue = (strValue - (value * 131) % M);
if (newValue >= 48 && newValue <= 57)
return true;
if (newValue > 65 && newValue <= 90)
return true;
if (newValue >= 97 && newValue <= 122)
return true;
return false;
}
int hashingfunct(string str, int M)
{
long int P, F, sum = 0;
int len = str.length();
for (int i = 0; i < len; i++)
{
P = pow(131, len - i - 1);
F = (int)str[i];
sum += (F * P) % M;
}
sum = sum % M;
return sum;
}
int main()
{
int n = 5;
string str1, str2;
vector<vector<string> > events;
for (int i = 0; i < n; i++) {
cin >> str1 >> str2;
vector<string > temp;
temp.push_back(str1);
temp.push_back(str2);
events.push_back(temp);
}
for (int i = 0; i < n; i++) {
cout << events[i][0] << events[i][1];
}
/*
INPUT FORMAT:
setpassword 1
setpassword 2
setpassword 3
authorize 49
authorize 50
*/
vector<int> result;
int j = 0;
long int m = pow(10, 9);
long int M = m + 7;
long int value, strValue;
for (int i = 0; i < events.size(); i++)
{
strValue = stoi(events[i][1]);
if (events[i][0] == "setPassword") {
value = hashingfunct(events[i][1], M);
}
else if (strValue == value)
result[j++] = 1;
else if (authorize(strValue, value, M))
result[j++] = 1;
else
result[j++] = 0;
}
for (int i = 0; i < result.size(); i++) {
cout << result[i];
}
}
Your program has complete Undefined Behaviour.
Let's get started with the first problem. In the following check code
long int value, strValue; // not initialised
for (int i = 0; i < events.size(); i++)
{
// ...
// here it should have been "setpassword" (i.e. all are small letters)
if (events[i][0] == "setPassword")
{
// if the check fails the `value` never get initialised!
value = hashingfunct(events[i][1], M);
}
// If the `value` not been initialised, check happens with any garbage value here!
else if (strValue == value)
// ...other code
}
You are checking whether the string is "setPassword" instead of "setpassword" (i.e. see in the events vector, all the strings are small letters).
If that goes wrong, the value will never get initialized, meaning it holds any garbage value and hence conducting this check else if (strValue == value) can cause any behaviour to your program (aka Undefined Behaviour).
Secondly, the vector<int> result; is empty at the beginning. Therefore accessing elements via std::vector::operator[] later
result[j++] = 1;
// ...
result[j++] = 1;
// ...
result[j++] = 0;
triggers the access out of bounds (UB). There you need just result.emplace_back(/*value*/); or result.push_back(/*value*/);, and no need of redutant variable j.
In short, you need
#include <iostream>
#include <vector>
#include <string>
// ..other functions
int main()
{
std::vector<std::vector<std::string> > events {
{"setpassword", "1"}, // can be also user input, like in your example
{"setpassword", "2"},
{"setpassword", "3"},
{"authorize", "49" },
{"authorize", "50" }
};
std::vector<int> result;
const long int M = pow(10, 9) + 7;
long int value{ 0 }, strValue{ 0 }; // default initialization
for (const std::vector<std::string> row: events) // better use range-based loop
{
strValue = std::stoi(row[1]);
if (row[0] == "setpassword") {
value = hashingfunct(row[1], M);
if (strValue == value)
result.emplace_back(1);
else if (authorize(strValue, value, M))
result.emplace_back(1);
}
else
result.emplace_back(0);
}
}
As a side note,
Please do not use using namespacestd;
Why should I not #include <bits/stdc++.h>?
Corrected code
#include<bits/stdc++.h>
using namespace std;
bool authorize(long int strValue,long int value,int M){
long int value1=value*131;
long int newValue=(strValue-(value1%M))%M;
if(newValue>=48 && newValue<=57)
return true;
if(newValue>=65 && newValue<=90)
return true;
if(newValue>=97 && newValue<=122)
return true;
return false;
}
int hashingfunct(string str,int M){
long int P,F,sum=0;
int len=str.length();
for(int i=0;i<len;i++){
P=pow(131,len-i-1);
F=(int)str[i];
sum+=(F*P)%M;
}
sum=sum%M;
return sum;
}
int main(){
int n=5;
string str1,str2;
vector<vector<string> > events;
for (int i=0;i<n;i++){
cin>>str1>>str2;
vector<string > temp;
temp.push_back(str1);
temp.push_back(str2);
events.push_back(temp);
}
/*
setPassword cAr1
authorize 223691457
authorize 303580761
setPassword d
authorize 100
*/
vector<int> result;
int j=0;
long int m=pow(10,9);
long int M=m+7;
long int value,strValue;
for(int i=0;i<events.size();i++){
if(events[i][0]=="setPassword"){
value=hashingfunct(events[i][1],M);
continue;
}
strValue=stoi(events[i][1]);
if(strValue==value)
result.push_back(1);
else if(authorize(strValue,value,M))
result.push_back(1);
else
result.push_back(0);
}
for(int i=0;i<result.size();i++){
cout<<result[i];
}
}
This is Project Euler question 26.
I would like to save the abundant values in an array, assuming there are no more than 200 abundant numbers smaller than 28123, which is the upper limit provided in the question.
This is not the complete code, but my program stops defining values at abundantarr[200]. Why am I seeing a limit to the values in this array?
#include <iostream>
#include <cmath>
using namespace std;
bool IsitAbundant(int a);
int main()
{
int abundantarr[200] = {0};
int counter = 0;
int totalsum = 0;
for (int u = 1; u < 28123; u++)
{
if (IsitAbundant(u))
{
abundantarr[counter] = u;
cout << abundantarr[counter] << endl;
counter++;
}
}
return 0;
}
bool IsitAbundant (int a)
{
int sum = 0;
for (int i = 1; i < a; i++)
{
if (a % i == 0)
{
sum+= i;
}
}
if (sum > a)
{
return true;
}
else
{
return false;
}
}
I am currently reading "Programming: Principles and Practice Using C++", in Chapter 4 there is an exercise in which:
I need to make a program to calculate prime numbers between 1 and 100 using the Sieve of Eratosthenes algorithm.
This is the program I came up with:
#include <vector>
#include <iostream>
using namespace std;
//finds prime numbers using Sieve of Eratosthenes algorithm
vector<int> calc_primes(const int max);
int main()
{
const int max = 100;
vector<int> primes = calc_primes(max);
for(int i = 0; i < primes.size(); i++)
{
if(primes[i] != 0)
cout<<primes[i]<<endl;
}
return 0;
}
vector<int> calc_primes(const int max)
{
vector<int> primes;
for(int i = 2; i < max; i++)
{
primes.push_back(i);
}
for(int i = 0; i < primes.size(); i++)
{
if(!(primes[i] % 2) && primes[i] != 2)
primes[i] = 0;
else if(!(primes[i] % 3) && primes[i] != 3)
primes[i]= 0;
else if(!(primes[i] % 5) && primes[i] != 5)
primes[i]= 0;
else if(!(primes[i] % 7) && primes[i] != 7)
primes[i]= 0;
}
return primes;
}
Not the best or fastest, but I am still early in the book and don't know much about C++.
Now the problem, until max is not bigger than 500 all the values print on the console, if max > 500 not everything gets printed.
Am I doing something wrong?
P.S.: Also any constructive criticism would be greatly appreciated.
I have no idea why you're not getting all the output, as it looks like you should get everything. What output are you missing?
The sieve is implemented wrongly. Something like
vector<int> sieve;
vector<int> primes;
for (int i = 1; i < max + 1; ++i)
sieve.push_back(i); // you'll learn more efficient ways to handle this later
sieve[0]=0;
for (int i = 2; i < max + 1; ++i) { // there are lots of brace styles, this is mine
if (sieve[i-1] != 0) {
primes.push_back(sieve[i-1]);
for (int j = 2 * sieve[i-1]; j < max + 1; j += sieve[i-1]) {
sieve[j-1] = 0;
}
}
}
would implement the sieve. (Code above written off the top of my head; not guaranteed to work or even compile. I don't think it's got anything not covered by the end of chapter 4.)
Return primes as usual, and print out the entire contents.
Think of the sieve as a set.
Go through the set in order. For each value in thesive remove all numbers that are divisable by it.
#include <set>
#include <algorithm>
#include <iterator>
#include <iostream>
typedef std::set<int> Sieve;
int main()
{
static int const max = 100;
Sieve sieve;
for(int loop=2;loop < max;++loop)
{
sieve.insert(loop);
}
// A set is ordered.
// So going from beginning to end will give all the values in order.
for(Sieve::iterator loop = sieve.begin();loop != sieve.end();++loop)
{
// prime is the next item in the set
// It has not been deleted so it must be prime.
int prime = *loop;
// deleter will iterate over all the items from
// here to the end of the sieve and remove any
// that are divisable be this prime.
Sieve::iterator deleter = loop;
++deleter;
while(deleter != sieve.end())
{
if (((*deleter) % prime) == 0)
{
// If it is exactly divasable then it is not a prime
// So delete it from the sieve. Note the use of post
// increment here. This increments deleter but returns
// the old value to be used in the erase method.
sieve.erase(deleter++);
}
else
{
// Otherwise just increment the deleter.
++deleter;
}
}
}
// This copies all the values left in the sieve to the output.
// i.e. It prints all the primes.
std::copy(sieve.begin(),sieve.end(),std::ostream_iterator<int>(std::cout,"\n"));
}
From Algorithms and Data Structures:
void runEratosthenesSieve(int upperBound) {
int upperBoundSquareRoot = (int)sqrt((double)upperBound);
bool *isComposite = new bool[upperBound + 1];
memset(isComposite, 0, sizeof(bool) * (upperBound + 1));
for (int m = 2; m <= upperBoundSquareRoot; m++) {
if (!isComposite[m]) {
cout << m << " ";
for (int k = m * m; k <= upperBound; k += m)
isComposite[k] = true;
}
}
for (int m = upperBoundSquareRoot; m <= upperBound; m++)
if (!isComposite[m])
cout << m << " ";
delete [] isComposite;
}
Interestingly, nobody seems to have answered your question about the output problem. I don't see anything in the code that should effect the output depending on the value of max.
For what it's worth, on my Mac, I get all the output. It's wrong of course, since the algorithm isn't correct, but I do get all the output. You don't mention what platform you're running on, which might be useful if you continue to have output problems.
Here's a version of your code, minimally modified to follow the actual Sieve algorithm.
#include <vector>
#include <iostream>
using namespace std;
//finds prime numbers using Sieve of Eratosthenes algorithm
vector<int> calc_primes(const int max);
int main()
{
const int max = 100;
vector<int> primes = calc_primes(max);
for(int i = 0; i < primes.size(); i++)
{
if(primes[i] != 0)
cout<<primes[i]<<endl;
}
return 0;
}
vector<int> calc_primes(const int max)
{
vector<int> primes;
// fill vector with candidates
for(int i = 2; i < max; i++)
{
primes.push_back(i);
}
// for each value in the vector...
for(int i = 0; i < primes.size(); i++)
{
//get the value
int v = primes[i];
if (v!=0) {
//remove all multiples of the value
int x = i+v;
while(x < primes.size()) {
primes[x]=0;
x = x+v;
}
}
}
return primes;
}
In the code fragment below, the numbers are filtered before they are inserted into the vector. The divisors come from the vector.
I'm also passing the vector by reference. This means that the huge vector won't be copied from the function to the caller. (Large chunks of memory take long times to copy)
vector<unsigned int> primes;
void calc_primes(vector<unsigned int>& primes, const unsigned int MAX)
{
// If MAX is less than 2, return an empty vector
// because 2 is the first prime and can't be placed in the vector.
if (MAX < 2)
{
return;
}
// 2 is the initial and unusual prime, so enter it without calculations.
primes.push_back(2);
for (unsigned int number = 3; number < MAX; number += 2)
{
bool is_prime = true;
for (unsigned int index = 0; index < primes.size(); ++index)
{
if ((number % primes[k]) == 0)
{
is_prime = false;
break;
}
}
if (is_prime)
{
primes.push_back(number);
}
}
}
This not the most efficient algorithm, but it follows the Sieve algorithm.
below is my version which basically uses a bit vector of bool and then goes through the odd numbers and a fast add to find multiples to set to false. In the end a vector is constructed and returned to the client of the prime values.
std::vector<int> getSieveOfEratosthenes ( int max )
{
std::vector<bool> primes(max, true);
int sz = primes.size();
for ( int i = 3; i < sz ; i+=2 )
if ( primes[i] )
for ( int j = i * i; j < sz; j+=i)
primes[j] = false;
std::vector<int> ret;
ret.reserve(primes.size());
ret.push_back(2);
for ( int i = 3; i < sz; i+=2 )
if ( primes[i] )
ret.push_back(i);
return ret;
}
Here is a concise, well explained implementation using bool type:
#include <iostream>
#include <cmath>
void find_primes(bool[], unsigned int);
void print_primes(bool [], unsigned int);
//=========================================================================
int main()
{
const unsigned int max = 100;
bool sieve[max];
find_primes(sieve, max);
print_primes(sieve, max);
}
//=========================================================================
/*
Function: find_primes()
Use: find_primes(bool_array, size_of_array);
It marks all the prime numbers till the
number: size_of_array, in the form of the
indexes of the array with value: true.
It implemenets the Sieve of Eratosthenes,
consisted of:
a loop through the first "sqrt(size_of_array)"
numbers starting from the first prime (2).
a loop through all the indexes < size_of_array,
marking the ones satisfying the relation i^2 + n * i
as false, i.e. composite numbers, where i - known prime
number starting from 2.
*/
void find_primes(bool sieve[], unsigned int size)
{
// by definition 0 and 1 are not prime numbers
sieve[0] = false;
sieve[1] = false;
// all numbers <= max are potential candidates for primes
for (unsigned int i = 2; i <= size; ++i)
{
sieve[i] = true;
}
// loop through the first prime numbers < sqrt(max) (suggested by the algorithm)
unsigned int first_prime = 2;
for (unsigned int i = first_prime; i <= std::sqrt(double(size)); ++i)
{
// find multiples of primes till < max
if (sieve[i] = true)
{
// mark as composite: i^2 + n * i
for (unsigned int j = i * i; j <= size; j += i)
{
sieve[j] = false;
}
}
}
}
/*
Function: print_primes()
Use: print_primes(bool_array, size_of_array);
It prints all the prime numbers,
i.e. the indexes with value: true.
*/
void print_primes(bool sieve[], unsigned int size)
{
// all the indexes of the array marked as true are primes
for (unsigned int i = 0; i <= size; ++i)
{
if (sieve[i] == true)
{
std::cout << i <<" ";
}
}
}
covering the array case. A std::vector implementation will include minor changes such as reducing the functions to one parameter, through which the vector is passed by reference and the loops will use the vector size() member function instead of the reduced parameter.
Here is a more efficient version for Sieve of Eratosthenes algorithm that I implemented.
#include <iostream>
#include <cmath>
#include <set>
using namespace std;
void sieve(int n){
set<int> primes;
primes.insert(2);
for(int i=3; i<=n ; i+=2){
primes.insert(i);
}
int p=*primes.begin();
cout<<p<<"\n";
primes.erase(p);
int maxRoot = sqrt(*(primes.rbegin()));
while(primes.size()>0){
if(p>maxRoot){
while(primes.size()>0){
p=*primes.begin();
cout<<p<<"\n";
primes.erase(p);
}
break;
}
int i=p*p;
int temp = (*(primes.rbegin()));
while(i<=temp){
primes.erase(i);
i+=p;
i+=p;
}
p=*primes.begin();
cout<<p<<"\n";
primes.erase(p);
}
}
int main(){
int n;
n = 1000000;
sieve(n);
return 0;
}
Here's my implementation not sure if 100% correct though :
http://pastebin.com/M2R2J72d
#include<iostream>
#include <stdlib.h>
using namespace std;
void listPrimes(int x);
int main() {
listPrimes(5000);
}
void listPrimes(int x) {
bool *not_prime = new bool[x];
unsigned j = 0, i = 0;
for (i = 0; i <= x; i++) {
if (i < 2) {
not_prime[i] = true;
} else if (i % 2 == 0 && i != 2) {
not_prime[i] = true;
}
}
while (j <= x) {
for (i = j; i <= x; i++) {
if (!not_prime[i]) {
j = i;
break;
}
}
for (i = (j * 2); i <= x; i += j) {
not_prime[i] = true;
}
j++;
}
for ( i = 0; i <= x; i++) {
if (!not_prime[i])
cout << i << ' ';
}
return;
}
I am following the same book now. I have come up with the following implementation of the algorithm.
#include<iostream>
#include<string>
#include<vector>
#include<algorithm>
#include<cmath>
using namespace std;
inline void keep_window_open() { char ch; cin>>ch; }
int main ()
{
int max_no = 100;
vector <int> numbers (max_no - 1);
iota(numbers.begin(), numbers.end(), 2);
for (unsigned int ind = 0; ind < numbers.size(); ++ind)
{
for (unsigned int index = ind+1; index < numbers.size(); ++index)
{
if (numbers[index] % numbers[ind] == 0)
{
numbers.erase(numbers.begin() + index);
}
}
}
cout << "The primes are\n";
for (int primes: numbers)
{
cout << primes << '\n';
}
}
Here is my version:
#include "std_lib_facilities.h"
//helper function:check an int prime, x assumed positive.
bool check_prime(int x) {
bool check_result = true;
for (int i = 2; i < x; ++i){
if (x%i == 0){
check_result = false;
break;
}
}
return check_result;
}
//helper function:return the largest prime smaller than n(>=2).
int near_prime(int n) {
for (int i = n; i > 0; --i) {
if (check_prime(i)) { return i; break; }
}
}
vector<int> sieve_primes(int max_limit) {
vector<int> num;
vector<int> primes;
int stop = near_prime(max_limit);
for (int i = 2; i < max_limit+1; ++i) { num.push_back(i); }
int step = 2;
primes.push_back(2);
//stop when finding the last prime
while (step!=stop){
for (int i = step; i < max_limit+1; i+=step) {num[i-2] = 0; }
//the multiples set to 0, the first none zero element is a prime also step
for (int j = step; j < max_limit+1; ++j) {
if (num[j-2] != 0) { step = num[j-2]; break; }
}
primes.push_back(step);
}
return primes;
}
int main() {
int max_limit = 1000000;
vector<int> primes = sieve_primes(max_limit);
for (int i = 0; i < primes.size(); ++i) {
cout << primes[i] << ',';
}
}
Here is a classic method for doing this,
int main()
{
int max = 500;
vector<int> array(max); // vector of max numbers, initialized to default value 0
for (int i = 2; i < array.size(); ++ i) // loop for rang of numbers from 2 to max
{
// initialize j as a composite number; increment in consecutive composite numbers
for (int j = i * i; j < array.size(); j +=i)
array[j] = 1; // assign j to array[index] with value 1
}
for (int i = 2; i < array.size(); ++ i) // loop for rang of numbers from 2 to max
if (array[i] == 0) // array[index] with value 0 is a prime number
cout << i << '\n'; // get array[index] with value 0
return 0;
}
I think im late to this party but im reading the same book as you, this is the solution in came up with! Feel free to make suggestions (you or any!), for what im seeing here a couple of us extracted the operation to know if a number is multiple of another to a function.
#include "../../std_lib_facilities.h"
bool numIsMultipleOf(int n, int m) {
return n%m == 0;
}
int main() {
vector<int> rawCollection = {};
vector<int> numsToCheck = {2,3,5,7};
// Prepare raw collection
for (int i=2;i<=100;++i) {
rawCollection.push_back(i);
}
// Check multiples
for (int m: numsToCheck) {
vector<int> _temp = {};
for (int n: rawCollection) {
if (!numIsMultipleOf(n,m)||n==m) _temp.push_back(n);
}
rawCollection = _temp;
}
for (int p: rawCollection) {
cout<<"N("<<p<<")"<<" is prime.\n";
}
return 0;
}
Try this code it will be useful to you by using java question bank
import java.io.*;
class Sieve
{
public static void main(String[] args) throws IOException
{
int n = 0, primeCounter = 0;
double sqrt = 0;
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
System.out.println(“Enter the n value : ”);
n = Integer.parseInt(br.readLine());
sqrt = Math.sqrt(n);
boolean[] prime = new boolean[n];
System.out.println(“\n\nThe primes upto ” + n + ” are : ”);
for (int i = 2; i<n; i++)
{
prime[i] = true;
}
for (int i = 2; i <= sqrt; i++)
{
for (int j = i * 2; j<n; j += i)
{
prime[j] = false;
}
}
for (int i = 0; i<prime.length; i++)
{
if (prime[i])
{
primeCounter++;
System.out.print(i + ” “);
}
}
prime = new boolean[0];
}
}