Given the following code :
#include <iostream>
#include <vector>
using namespace std;
int main() {
int n;
cin >> n;
vector<int> v;
v.emplace_back(1);
v.emplace_back(2);
for(int i = 2; i < n; i++){ // Is this problem point?
int data = v.at(i-1) + v.at(i-2);
v.emplace_back(data);
}
cout << v.at(n);
return 0;
}
this code has Error that is out of range.
I think that for loop has problem.
but I'm not sure where is out of range.
Is vector container index starting 0?
In each iteration of for(int i = 2; i < n; i++) you add one element to v. After the loop v contains max(2, n) elements.
The problem is caused by
cout << v.at(n);
with n >= 2. The last valid index is v.size() - 1. But to print the last value of a vector you can simply use:
cout << v.back();
Related
#include <iostream>
using namespace std;
// finding the required sum of subarray
int main()
{
int n,s;
int i=0,j=0,st=-1,en=-1,sum=0;
cin>>s; //input required sum
cin>>n;
int a[n];
for(int i=0;i<n;i++){ // here if i only mention int again i //am getting the output or else the values of st and en are printing //out the same as i initialize
cin>>a[i];
}
while (j<n){
sum+=a[j];
while(sum>s){
sum-=a[i];
i++;
}
if(sum==s){
st=i+1;
en=j+1;
break;
}
j++;
}
cout<<st<<" "<<en<<" ";
return 0;
the output is -1 -1
and if i mention "int i" again in for loop of inputing array i a getting the answer.
i want to know the reason i already intialize i before why do i need to do it again
The problem statement is unclear, I'm assuming you simply want the indexes of the repeating numbers in array a. You are correct for the most part, using b[i] = i is the problem. If you understand what a vector is, then simply create a vector like this and push the indexes in the vector. For example,
vector<int> b;
and inside the a[i] == a[j] condition,
b.push_back(i);
then finally print out result like,
for(int i = 0 ; i < b.size() , i++)
cout << b[i] << " ";
If you're unfamiliar with vectors, simply use another variable cnt to update index of array b
int a[n], i, b[n], j, cnt = 0;
and inside the a[i] == a[j] condition,
b[cnt] = i;
cnt++;
and finally
for(int i = 0 ; i < cnt ; i++)
cout << b[i] << ' ';
As I'm new to c++ I get runtime error for first example(I mean I tested my program with 5 examples it actually happens automatically by a site for testing) of my program I know that's because of exceeding time for running it but I dunno how to fix this.
My program get n numbers from user and finds the largest one and prints it.
#include<iostream>
#include<curses.h>
using namespace std;
int main()
{
int n;
cin >> n;
int *p = new int(n);
for(int i = 1; i<=n; i++){
cin >> *(p+i);
}
int largest = *p;
for(int i = 1; i<=n; i++){
if(largest < *(p+i))
largest = *(p+i);
}
cout << largest;
return(0);
}
int *p=new int(n);
The line above allocates just a single int, and sets the value to n. It does not allocate an array of n integers.
That line should be:
int *p=new int[n];
And then delete [] p; to deallocate the memory.
But better yet:
#include <vector>
//...
std::vector<int> p(n);
is the preferred way to utilize dynamic arrays in C++.
Then the input loop would simply be:
for(int i=0;i<n; i++)
{
cin >> p[i];
}
That same input loop could have been used if you had used the pointer version.
Then you have this error:
for(int i=1;i<=n;i++)
Arrays (and vectors) are indexed starting from 0 with the upper index at n-1, where n is the total number of elements. That loop has an off-by-one error, where it exceeds the upper index on the last loop.
Basically any loop that uses <= as the limiting condition is suspect. That line should be:
for(int i=0; i<n; i++)
(Note that I changed the code above to fix this error).
However ultimately, that entire loop to figure out the largest can be accomplished with a single line of code using the std::max_element function:
#include <algorithm>
//...
int largest = *std::max_element(p, p + n);
and if using std::vector:
#include <algorithm>
//...
int largest = *std::max_element(p.begin(), p.begin() + n);
I've commented on suggested changes in this slightly modified version:
#include <iostream>
int main()
{
unsigned n; // don't allow a negative amount of numbers
if(std::cin >> n) { // check that "cin >> n" succeeds
int* p=new int[n]; // allocate an array of n ints instead of one int with value n
for(int i=0; i < n; ++i) { // corrected bounds [0,n)
if(not (std::cin >> p[i])) return 1; // check that "cin >> ..." succeeds
}
int largest = p[0];
for(int i=1; i < n; ++i) { // corrected bounds again, [1,n)
if(largest < p[i])
largest = p[i];
}
delete[] p; // free the memory when done
std::cout << largest << '\n';
}
}
Note that using *(p + i) does the same as using p[i]. The latter is often preferred.
This would work if all cin >> ... works, but shows some of the hazards when using raw pointers. If extracting the n ints failes, the program will return 1 and leak the memory allocated with new int[n].
A rewrite using a smart pointer (std::unique_ptr<int[]>) that automatically deallocates the memory when it goes out of scope:
#include <iostream>
#include <memory> // std::unique_ptr
int main()
{
unsigned n;
if(std::cin >> n) {
std::unique_ptr<int[]> p(new int[n]);
for(int i=0; i < n; ++i) { // corrected bounds [0,n)
if(not (std::cin >> p[i])) return 1; // will not leak "p"
}
int largest = p[0];
for(int i=1; i < n; ++i) {
if(largest < p[i])
largest = p[i];
}
std::cout << largest << '\n';
} // p is automatically delete[]ed here
}
However, it's often convenient to store an array and its size together and to do this, you could use a std::vector<int> instead. It comes with a lot of convenient member functions, like, size() - and also begin() and end() which lets you use it in range-based for loops.
#include <iostream>
#include <vector> // std::vector
int main()
{
unsigned n;
if(std::cin >> n) {
std::vector<int> p(n); // a vector of n ints
// a range-based for loop, "elem" becomes a refrence to each element in "p":
for(int& elem : p) {
if(not (std::cin >> elem)) return 1;
}
int largest = p[0];
for(int i = 1; i < p.size(); ++i) { // using the size() member function
if(largest < p[i])
largest = p[i];
}
std::cout << largest << '\n';
}
}
That said, you don't need to store any number in an array to figure out what the largest number is. Instead, just compare the input with the currently largest number.
#include <iostream>
#include <limits> // std::numeric_limits
int main()
{
unsigned n;
if(std::cin >> n) {
// initialize with the smallest possible int:
int largest = std::numeric_limits<int>::min();
while(n--) {
int tmp;
if(not (std::cin >> tmp)) return 1;
if(largest < tmp)
largest = tmp;
}
std::cout << largest << '\n';
}
}
I have no error message instead I only have unexpected behavior.
double get_optimal_value(int capacity, vector<int> weights, vector<int> values) {
int n = weights.size();
vector<pair<double, int>> valuePerWeight(n);
pair<double,int> x;
for(int i = 0; i < n; i++){
double v = values[i]/weights[i];
x = make_pair(values[i]/weights[i], weights[i]);
valuePerWeight.push_back(x);
}
for(int i = 0; i < n && capacity > 0; i++){
int amount = min(capacity, valuePerWeight[i].second);
value += valuePerWeight[i].first * amount;
capacity -= amount;
}
double value = 0.0;
return value;
}
I am creating a vector with values of type pair<double,int>. I create the pair using make_pair(some_double, some_int), then I call push_back with the pair.
Later in the function I index into the vector and do stuff using the pairs.
However an issue arises, when I index into my valuePerWeight vector and retrieve the attributes of the different pairs. They all end up being zero regardless of index and regardless of .first or .second.
Through printing a bunch of variables I have asserted the created pair is not {0,0} but as soon as I push_back into the vector and index the pair and look at it's .first and .second attributes both are equal to 0.
I can't seem to understand why this is, originally I was using push_back seen as below
valuePerWeight.push_back(make_pair(values[i]/weights[i], weights[i]));
instead of creating into a temporary variable x . However the same issue still stands.
Any help in the right direction would be greatly appreciated.
If there is any further clarification that I can give please ask me.
If you'd like to see for some values below is a snippet which can be compiled
I use input
3 50
60 20
100 50
120 30
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
double get_optimal_value(int capacity, vector<int> weights, vector<int> values) {
double value = 0.0;
int n = weights.size();
vector<pair<double, int>> valuePerWeight(n);
pair<double,int> x;
for(int i = 0; i < n; i++){
double v = values[i]/weights[i];
cout << v << ' '<< weights[i] << '\n';
x = make_pair(values[i]/weights[i], weights[i]);
cout << x.first << ' ' << x.second << '\n';
valuePerWeight.push_back(x);
cout << valuePerWeight[i].first << ' ' << valuePerWeight[i].second << '\n';
}
for(int i = 0; i < n; i++){
cout << valuePerWeight[i].first;
cout << valuePerWeight[i].second;
cout << '\n';
}
sort(valuePerWeight.begin(), valuePerWeight.end());
for(int i = 0; i < n && capacity > 0; i++){
int amount = min(capacity, valuePerWeight[i].second);
value += valuePerWeight[i].first * amount;
capacity -= amount;
}
// for(auto vp: valuePerWeight){
// cout << vp.first << vp.second;
// cout << '\n';
// }
return value;
}
int main() {
int n;
int capacity;
std::cin >> n >> capacity;
vector<int> values(n);
vector<int> weights(n);
for (int i = 0; i < n; i++) {
std::cin >> values[i] >> weights[i];
}
double optimal_value = get_optimal_value(capacity, weights, values);
std::cout.precision(10);
std::cout << optimal_value << std::endl;
return 0;
}
The confusion here is due to the behavior of the constructor you use
vector<pair<double, int>> valuePerWeight(n);
This actually fills the vector with n default constructed pairs, which as you may surmise, are (0, 0). When you push_back, you push to the end of these, so you a totally get 2n pairs.
.reserve does something close to what you expected, not actually filling the vector, but is likely not needed for something not bottle-necking on vector resizing.
Short story, omit the (n) to just construct an empty vector.
Three more suggestions: accept the vectors as const& to save a copy, and look at emplace_back instead of making a pair yourself and pushing it. That's what it's meant for. Also, note the comment by churill - dividing two integers will result in integer division regardless of where you are assigning the result. Static cast one of them to a float or double (or multiply by 1.0 at the start) to ensure floating point division.
I'm trying to convert my array to a vector, yet I'm having trouble printing it.
It says in int main() in my for loop that v is undefined. When I define
vector v; inside int main() the program compiles and runs and yet prints nothing. What am I doing wrong?
#include <iostream>
#include <vector>
using namespace std;
vector<int> a2v(int x[], int n)
{
vector<int> v(n);
for(int i = 0; i < n; i++)
{
v.push_back(x[i]);
}
return(v);
}
int main()
{
vector<int> a2v(int x[], int n);
int array[] = {11,12,13,14,15,16,17,18};
a2v(array, 8);
for(int i = 0; i < v.size(); i++)
{
cout << v[i] <<" ";
}
cout << endl;
return(0);
}
This is your program corrected:
#include <iostream>
#include <vector>
using namespace std;
vector<int> a2v(int x[], int n)
{
vector<int> v(0);
v.reserve(n); //optional
for(int i = 0; i < n; i++)
{
v.push_back(x[i]);
}
return(v);
}
int main()
{
int array[] = {11,12,13,14,15,16,17,18};
auto v = a2v(array, 8);
for(size_t i = 0; i < v.size(); i++)
{
cout << v[i] <<" ";
}
cout << endl;
return(0);
}
There were 2 errors:
In the function a2v, you instantiated a vector of 0 with length n, and then you pushed back other elements;
You were not defining v inside the main, as the return value of a2v
The vector you want to read is the return of the a2v function.
But there is a lot more simpler than that to go from C-array to vector array , I put here the example in found the vector reference web page :
http://www.cplusplus.com/reference/vector/vector/vector/
int myints[] = {16,2,77,29};
std::vector<int> fifth (myints, myints + sizeof(myints) / sizeof(int) );
When you define a variable, such as your vector v, you always do it in a certain scope. The body of a function such as a2v, or main, is an example of a possible scope. So if you only do it in the function a2v, that's the only scope where it will be visible.
It says in int main() in my for loop that v is undefined.
Well it will be because there is no v in main()
a2v(array, 8);
The above function call returns a vector so you need to collect the returned vector like
vector<int> v=a2v(array,8)
Also,
vector<int> v(n);//creates a vector of size n
for(int i = 0; i < n; i++)
{
v.push_back(x[i]);//adds n more elements to the vector
}
The returned vector has 2n elements and not n
Finally you could directly create a vector from an array as
vector<int> v(arr,arr+n);//where n is the number of elements in arr
I'm creating this very simple C++ program.
the program asks the user to enter a few integers and stores them in an array.but when a specific integer(for example 50)is entered,the input is ended and then,all of the integers are displayed on the screen except for 50.
for example:
input:
1
2
88
50
output:
1
2
88
the error i'm getting is when i use cout to print the array,all of numbers are shown,including 50 and numbers i did'nt even entered.
this is my code so far:
#include<iostream>
int main() {
int num[100];
for(int i=0;i<=100;i++) {
cin >> num[i];
if (num[i]!=50) break;
}
for(int j=0;j<=100;j++) {
cout << num[j] << endl;
}
return 0;
}
Change the program the following way
#include<iostream>
int main()
{
const size_t N = 100;
int num[N];
size_t n = 0;
int value;
while ( n < N && std::cin >> value && value != 50 ) num[n++] = value;
for ( size_t i = 0; i < n; i++ ) std::cout << num[i] << std::endl;
return 0;
}
Here in the first loop variable n is used to count the actual number of entered values. And then this variable is used as the upper bound for the second loop.
As for your program then the valid range of indices for the first loop is 0-99 and you have to output only whose elements of the array that were inputed.
A do while loop is more suitable for your problem. The stop condition will check if the number fit inside the array (if k is not bigger than 100) and if number entered is 50.
#include<iostream>
using namespace std;
int main() {
int num[100];
int k = 0;
// A do while loop will be more suitable
do{
cin >> num[k++];
}while(k<100&&num[k-1]!=50);
for (int j = 0; j < k-1; j++) {
cout << num[j] << endl;
}
return 0;
}
Also, a better solution to get rid of 100 limitation is to use std::vector data structure that automatically adjust it's size, like this:
vector<int> num;
int temp;
do {
cin >> temp;
num.push_back(temp);
} while (temp != 50);
Note, you can use temp.size() to get the number of items stored.
You read up to 101 numbers, but if you enter 50 you break the loop and go for printing it. In the printing loop you go through all 101 numbers, but you actually may have not set all of them.
In the first loop count in a count variable the numbers you read until you meet 50 and in the printing loop just iterate count-1 times.
You have allocated an array of 100 integers on the stack. The values are not initialized to zero by default, so you end up having whatever was on the stack previously appear in your array.
You have also off-by-one in both of your loops, you allocated array of 100 integers so that means index range of 0-99.
As the question is tagged as C++, I would suggest that you leave the C-style array and instead use a std::vector to store the values. This makes it more flexible as you don't have to specify a fixed size (or manage memory) and you don't end up with uninitialized values.
Little example code (requires C++11 compiler):
#include <iostream>
#include <vector>
int main()
{
std::vector<int> numbers; // Store the numbers here
for(int i = 0; i < 100; ++i) // Ask a number 100 times
{
int n;
std::cin >> n;
if( n == 50 ) // Stop if user enters 50
break;
numbers.push_back(n); // Add the number to the numbers vector
}
for (auto n : numbers) // Print all the values in the numbers vector
std::cout << n << std::endl;
return 0;
}
There are just 2 changes in your code check it out :
int main()
{
int num[100],i; //initialize i outside scope to count number of inputs
for(i=0;i<100;i++) {
cin >> num[i];
if (num[i]==50) break; //break if the entered number is 50
}
for(int j=0;j<=i-1;j++)
{
cout << num[j] << endl;
}
return 0;
}
Okay, others already pointed out the two mistakes. You should use i < 100 in the loop conditions instead of i <= 100 and you have to keep track of how many elements you entered.
Now let me add an answer how I think it would be better.
#include <iostream>
#include <vector>
int main() {
std::vector<int> numbers; // Create an empty vector.
for (int temp; // a temp variable in the for loop.
numbers.size() < 100 && // check that we have less than 100 elements.
std::cin >> temp && // read in the temp variable,
// and check if the read was a success.
temp != 50) // lastly check that the value we read isn't 50.
{
numbers.push_back(temp); // Now we just add it to the vector.
}
for (int i = 0; i < numbers.size(); ++i)
std::cout << numbers[i]; // Now we just print all the elements of
// the vector. We only added correct items.
}
The above code doesn't even read anymore numbers after it found 50. And if you want to be able to enter any number of elements you just have to remove the check that we have less than 100 elements.
Now I commented the above code a bit much, if you compress it it'll reduce to just:
#include <iostream>
#include <vector>
int main() {
std::vector<int> numbers; // Create an empty vector.
for (int temp; numbers.size() < 100 && std::cin >> temp && temp != 50)
numbers.push_back(temp);
for (int i = 0; i < numbers.size(); ++i)
std::cout << numbers[i];
}
If you can use the C++11 standard it reduces to:
#include <iostream>
#include <vector>
int main() {
std::vector<int> numbers; // Create an empty vector.
for (int temp; numbers.size() < 100 && std::cin >> temp && temp != 50)
numbers.push_back(temp);
for (int element : numbers)
std::cout << element;
}
for (auto element : numbers) is new, it basically means for every int 'element' in 'numbers'.