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Here if the condition occurs the value of i store in the array prime. But the index of the array is not increasing, all the values are storing at [0] index and destroy the previous value.
Tried all the other methods but didn't find anything.
I tried prime[x++] and prime[x+1], but they both didn't work for me. If someone gives me a solution then I'll be very thankful to you.
#include<iostream>
using namespace std;
int main()
{
int num = 20, prime[20], x;
for (int i = 1; i <= num; i++) {
if (i % 2 != 0) {
prime[x] = i;
}
}
for (int k = 1; k <= num; k++) {
cout << prime[x] << "\t";
}
}
You have the variable x un-initialized and you are using it, in the line
prime[x] = i;
assuming that it has been initialized. This invokes undefind behavior in your program and the result could not be predicted. Initialize it with the appropriate number to make the program to have a defined behavior.
Regarding prime numbers, see this SO post: Printing prime numbers from 1 through 100.
defining x as uninitialized is a undefined behaviour.
Initialize x as 0 (int x = 0;)
Try with bellow :
int x = 0;
for(int i=1; i<=num; i++){
if(i%2!=0){
prime[x] = i;
x++;
}
}
Now you have the number of prime array elements :
Now print prime array :
for(int k=0; k<x; k++){
cout << prime[k] << "\t";
}
The output (it seems your code detect odd numbers) :
1 3 5 7 9 11 13 15 17 19
Test the code online
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I was solving a problem https://codeforces.com/contest/489/problem/B
Its a simple brute force approach,In my terminal when i am giving input
#include<bits/stdc++.h>
using namespace std;
vector <int> b;
vector <int> g;
int main() {
int n;
cin >> n;
for (int i = 0; i < n; i++) {
int a;
cin >> a;
b.push_back(a);
}
int m;
cin >> m;
for (int i = 0; i < m; i++) {
int a;
cin >> a;
g.push_back(a);
}
sort(b.begin(), b.end());
sort(g.begin(), g.end());
int ans = 0;
bool visited[10000];
memset(visited, sizeof(visited), false);
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if(!visited[j])
if (abs(b[i] - g[j]) <= 1) {
visited[j] = true;
ans++;
break;
}
}
}
cout << ans;
}
4
1 4 6 2
5
5 1 5 7 9
I am getting correct output as 3 , This is the very first test case on codeforces also and codeforces showing output as 2 and as showing as wrong answer.
Please see here Proof ,I never faced such kind of problem in competitive
programming .
accepted solution solution
There was also announcement related to this ques read below here
This is a case of undefined behavior: if(!visited[j]) is undefined. visited is not initialized because the call memset(visited, sizeof(visited), false); is wrong. You are reading uninitialized variables.
The declaration of memset is
void *memset( void *dest, int ch, size_t count );
You are writting 0 times the value 10000 into visited. On your machine this memory was filled with zeros. But on other machines there can be different values.
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Question says:
You are given an integer N. You need to print the series of all prime numbers till N.
I want to know what's wrong with my code and suggestions will also be of great help.
#include<iostream>
using namespace std;
int main()
{
int N;
cin >> N;
int u;
for(int i = N; i > 0; i--)
{
u = 0;
for(int j = 2; j < N-1; j++)
{
if(i % j == 0)
{
u = 1;
}
}
if(u == 0)
{
cout << i << " ";
}
}
return 0;
}
First for future reference you should probably post that on code review, unless there is a specific problem then you should create a Minimal, Complete, and Verifiable post.
There is nothing inherently wrong other than you do not check that N>0 which could lead to an infinite loop, j should be going to i not N, and I think this would print 1 as a prime number which it is not. Here are some pointers:
Why are you going from N to 0? That seems a little counter intuitive compared from going from 2 to N (you can skip 1 since it is not prime)
If you are going to use a flag (u) then you should make it a bool which forces it to be either true or false
You do not need to do a flag, instead as soon as you find a divisor print the number and then break the inner loop like
for(j=2; j< N-1; j++){
if(i%j==0){
cout << i << " ";
break;
}
}
You do not need to have j go all the way to i, just the sqrt(i) since anything greater then the sqrt(i) that divides i must must be multiplied by some number smaller then the sqrt(i). So if i is not prime, then there must be a divisor below sqrt(i)
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I am currently going through a fibonacci practice problem on hackerrank and am having a malloc memory corruption error. This is the link to the problem I am doing:
https://www.hackerrank.com/contests/programming-interview-questions/challenges/fibonacci-returns/
Input is 0-10, each number separated by a new line.
For each input, the value at that point in the sequence is printed. It works for small inputs, but after 6 it gets the malloc error. It doesn't seem that the size of the sequence is an issue either, just how many are done in succession.
#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;
vector<int> bigFib(1);
int main() {
/* Enter your code here. Read input from STDIN. Print output to STDOUT */
int x;
while(cin >> x){
if(bigFib.size()-1 >= x){
cout << bigFib[x] << endl;
}
else{
vector<int> fib(x);
fib[0] = 0;
fib[1] = 1;
for(int j = 2; j <= x; j++){
fib[j] = fib[j-1] + fib[j-2];
}
bigFib = fib;
cout << fib[x] << endl;
}
}
return 0;
}
I am pretty new to C++ and can't find the problem. Thanks for your time.
When you create std::vector of size N, you can access elements with index [0, N-1] - which is N elements. You create vector of size x and in your loop:
for(int j = 2; j <= x; j++){
fib[j] = fib[j-1] + fib[j-2];
}
and in this statement
cout << fib[x] << endl;
you try to access element with index equal to x, which is UB. If you do need to access index x create vector with at least x+1 size
In vector<int> fib(x); you declare a vector<int> that has x elements. Those elements are fib[0] through to fib[x - 1]. However, in for(int j = 2; j <= x; j++){ fib[j] = ... you assign to an element out of bounds.
Imagine if x is 1, then you'd expect your fib vector to contain only one element: fib[0]... yet your loop is assigning to fib[1]. Problem? Yup.
I reckon for(int j = 2; j <= x; j++){ should probably be for(int j = 2; j < x; j++){...
... and cout << fib[x] << endl; should be cout << fib[x - 1] << endl;
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Isn't the below code supposed to initialize all the array elements to 0 using the for loop?
#include<iostream>
using namespace std;
int main()
{
int a[10];
for(int i=0; i==9; i++){
a[i] = 0;
}
for(int i=0;i<10;i++){
cout << a[i] << endl;
}
}
Output:-
164752
6
0
0
-13120
0
-13211
0
0
0
Check here corrected-code
#include<iostream>
using namespace std;
int main()
{
int a[10];
for(int i=0; i<10; i++){ // <= i < 10 instead of i == 9
a[i] = 0;
}
for(int i=0;i<10;i++){
cout << a[i] << endl;
}
}
No, because the comparison in the for loop is not correct. It just checks if i is 9 at the beginning, which turns out to be false, and hence, the first for loop doesn't execute at all.
The correct condition should be i<=9 (or i<10) instead of i==9, which is what you have written in the 2nd loop.
In a for loop like for (int i = 0; i == 9; i++) the second expression (i == 9 is the condition for continuing to run the loop. Since i is 0 the first time through the loop, i == 9 is false, and the loop immediately ends; execution continues with the next statement. Change it to i < 10; that way, on the first time through the loop, when i is 0, the condition is true, and execution continues. In fact, it continues until i < 10 becomes false, which happens after the tenth execution of the body of the loop.
Corrected code:
#include <iostream>
int main()
{
int a[10];
for (int i = 0; i != 10; i++)
a[i] = 0;
for (int i = 0; i != 10; i++)
std::cout << a[i] << '\n';
}
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So, I need to get the sum of the first 5 even numbers from my array, this is the code that I've got so far, have no clue what to do next. It runs, but the result is incorrect
#include "stdafx.h"
#include <iostream>
#include <math.h>
using namespace std;
int main()
{
int niz[10];
cout << "Unesi 10 brojeva:";
for (int i = 0; i < 10; i++) {
cin >> niz[i];
}
int suma = 0;
int parni[5];
int j = 0;
for (int i = 0; i < 10; i++) {
if (niz[i] % 2 == 0) {
niz[i] == parni[j];
j++;
if (j == 5) {
break;
}
}
}
for (int i = 0; i < 5; i++) {
suma = parni[i] + suma;
}
cout << suma;
system("PAUSE");
return 0;
}
This line:
niz[i] == parni[j];
does nothing. (It tests if niz[i] happens to be equal to the current, uninitialized value of parni[j], and throws away the result of the comparison.)
You want to store niz[i] in parni[j], so do this:
parni[j] = niz[i];
Incidentally, there is a problem if there are fewer than 5 even numbers in the niz array. In that case, you still sum up all five entries of the parni array, using uninitialized values, which is Bad. One way to avoid this is to just sum up the even entries as you find them, without using a secondary array.
IE, do suma += niz[i] at the line in question, and get rid of parni altogether.
Unless you're really required to use arrays here, vectors will work much more nicely.
You can also use a couple of standard algorithms to make your life easier (especially std::copy_if and std::accumulate).
// for the moment I'll ignore the code to read the input from the user:
auto input = read_input();
auto pos = std::remove_if(input.begin(), input.end(),
[](int i) { return i % 2 != 0; });
// assume that `input` always contains at least 5 elements
pos = std::min(pos, input.begin() + 5);
sum = std::accumulate(input.begin(), pos, 0);