LAP algorithm not working with large numbers - c++

include
#include <fstream>
#include <algorithm>
using namespace std;
long int lenghtOfLongestAP(long int set[],long int n)
{
if (n <= 2) return n;
long int L[n][n];
long int llap = 2;
for (long int i = 0; i < n; i++)
L[i][n-1] = 2;
for (long int j=n-2; j>=1; j--)
{
int i = j-1, k = j+1;
while (i >= 0 && k <= n-1)
{
if (set[i] + set[k] < 2*set[j])
k++;
else if (set[i] + set[k] > 2*set[j])
{ L[i][j] = 2, i--; }
else
{
L[i][j] = L[j][k] + 1;
llap = max(llap, L[i][j]);
i--; k++;
}
}
while (i >= 0)
{
L[i][j] = 2;
i--;
}
}
return llap;
}
int main()
{
ofstream cout("Output.txt");
ifstream cin("cablecar-sub4-attempt3.txt");
int ab;
cin >> ab;
for (long int z = 0; z < ab; z++)
{
long int bs;
cin >> bs;
long int array[bs];
for(long int h = 0; h<bs; h++)
cin >> array[h];
sort(array, array + bs);
cout << "Case #" << z+1 << ": " << lenghtOfLongestAP(array, bs) << endl;
}
return 0;
}
This is my code. It is a LAP (Largest arithmetic progression) algorithm, so it finds the largest progression in an sorted array. I have the following set of data:
pastebin.com/77meKfKW
Strangely, the program crashes after case 30, which it shouldn't. What kind of problem might it be and how can I fix it?


This is probably a stack overflow. You are allocating your array on the stack with 267*267 entries which uses a lot of memory.
Try allocating the memory on the heap instead, or simply changing the array from being local to being global (with a fixed maximum value of n).
e.g.
change
long int lenghtOfLongestAP(long int set[],long int n)
{
if (n <= 2) return n;
long int L[n][n];
to
long int L[1000][1000]; // or whatever your maximum n might be
long int lenghtOfLongestAP(long int set[],long int n)
{
if (n <= 2) return n;

Related

print condition while integer overflow not working

#include <iostream>
using namespace std;
int getFectorial(int n)
{
int ans = 1;
for (int i = n; i >= 1; i--)
{
ans = ans * i;
}
return ans;
}
int printNcr(int n, int r)
{
if (getFectorial(n) > INT_MAX)
{
return 0;
}
return (getFectorial(n)) / ((getFectorial(r)) * (getFectorial(n - r)));
}
int main()
{
int n = 14;
for (int row = 0; row < n; row++)
{
for (int col = 0; col < row + 1; col++)
{
cout << printNcr(row, col) << " ";
}
cout << endl;
}
return 0;
}
When I give value of n more than 13th I want integer overflow condition should be working that given in printNcr() function, but it's not working and all line after 13th are printing wrong values instead of returning false.
How to make given INT_MAX condition work?

int oveflow cannot be reliably detected after it happens.
One way to detect upcoming int overflow in factorial:
int getFactorial(int n) {
if (n <= 0) {
return 1; // and maybe other code when n < 0
}
int limit = INT_MAX/n;
int ans = 1;
for (int i = 2; i <= n; i++) {
if (ans >= limit) {
return INT_MAX; // Or some other code
}
ans = ans * i;
}
return ans;
}
Another way is at startup, perform a one-time calculation for maximum n. With common 32-bit int, that limit is 12.
int getFactorial(int n) {
if (n > getFactorial_pre_calculated_limit) {
return INT_MAX;
}
...

You can detect overflow by watching for negative value
int getFectorial(int n)
{
int ans = 1;
for (int i = n; i >= 1; i--)
{
ans = ans * i;
if (ans < 0) <<<<======
return -1;
}
return ans;
}
then
int printNcr(int n, int r)
{
if (getFectorial(n) < 0)
{
return 0;
}
return (getFectorial(n)) / ((getFectorial(r)) * (getFectorial(n - r)));
}
Please note though that strictly speaking this is undefined behavior. It would be better to simply fail if you know the result is going to be too big (Ie n > 13)
Or better do it like this
int getFectorial(int n)
{
long long ans = 1; <<<====
for (int i = n; i >= 1; i--)
{
ans = ans * i;
if (ans >INT_MAX) <<<<======
return -1;
}
return (int)ans;
}
or you could throw std::overflow_error
BTW the word is factorial not fectorial

prime seive algorithm giving a runtime error

I am trying to implement the Sieve of Eratosthenes algorithm but it giving a runtime error.
didn't get any output though. after providing the input,
#include<iostream>
using namespace std;
//Sieve Approach - Generate an array containing prime Numbers
void prime_sieve(int *p) {
//first mark all odd number's prime
for (int i = 3; i <= 10000; i += 2) {
p[i] = 1;
}
// Sieve
for (long long int i = 3; i <= 10000; i += 2) {
//if the current number is not marked (it is prime)
if (p[i] == 1) {
//mark all the multiples of i as not prime
for (long long int j = i * i; j <= 10000; j = j + i ) {
p[j] = 0;
}
}
}
//special case
p[2] = 1;
p[1] = p[0] = 0;
}
int main() {
int n;
cin >> n;
int p[10000] = {0};
prime_sieve(p);
//lets print primes upto range n
for (int i = 0; i <= n; i++) {
if (p[i] == 1) {
cout << i << " ";
}
}
return 0;
}
compiler didn't throwing any error also it is not providing the output also
program freezes for some seconds and then terminates

As mentioned in the comments, you are going out of bound.
There is also some confusion about the meaning of p[].
In addition, you are not using the value of n in the function, which leads to unnecessary calculations.
Here is a tested programme (up to n = 10000):
#include <iostream>
#include <vector>
#include <cmath>
//Sieve Approach - Generate an array containing prime Numbers less than n
void prime_sieve(std::vector<int> &p, long long int n) {
//first mark all odd number's prime
for (long long int i = 4; i <= n; i += 2) {
p[i] = 0;
}
// Sieve
for (long long int i = 3; i <= sqrt(n); i += 2) {
//if the current number is not marked (it is prime)
if (p[i] == 1) {
//mark all the multiples of i as not prime
for (long long int j = i * i; j <= n; j = j + i ) {
p[j] = 0;
}
}
}
//special cases
p[1] = p[0] = 0;
}
int main() {
long long int n;
std::cout << "Enter n: ";
std::cin >> n;
std::vector<int> p (n+1, 1);
prime_sieve(p, n);
//lets print primes upto range n
for (long long int i = 0; i <= n; i++) {
if (p[i] == 1) {
std::cout << i << " ";
}
}
return 0;
}

How do I resolve sigabrt error on codechef ide

Sigabrt runtime error occurs of a fatal error, because of an assert statement not returning true? Or use of excessive memory, I'm not able to figure out what I'm doing wrong here, help me out?
( problem 1343 C on codeforces) link
so here's the code.
#include <iostream>
#include <stdlib.h>
#include<vector>
using namespace std;
int check(int i,vector<int> a) {
if (a[i] > 0) {
return 1;
}
else return 0;
}
int main() {
int t;
cin >> t;
while (t--)
{
long int n;
cin >> n;
vector<int> a(n), b;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int i = 0;
while (i < n)
{
int max = a[i];
int s = check(i,a);
i++;
while (i<n && check(i,a)== s) {
if (a[i] > max)max = a[i];
i++;
}
b.push_back(max);
}
int s = 0;
for (int k = 0; k< b.size(); k++) {
s += b[i];
}
cout << s << endl;
}
}

I have debugged your code and also the modified code has been accepted for the above question.
Mistakes you made:
1. In the below loop, value at i'th index of vector<int> b is being added to long int s. Instead, b[k] should be added to long int s because the variable being used in the loop is k not i.
for (int k = 0; k< b.size(); k++) {
s += b[i];
}
2. In the question, range of variable n is given as (1 ≤ n ≤ 2.10^5). So, it is safe to use int n instead of long int n. Also, when I submitted my code on codeforces it gave me signed integer overflow error when I used long int n.
3. You need to use long long s instead of long int s because the value of each element of array A lies between (−10^9 ≤ a[i] ≤ 10^9 , ai ≠ 0) and when we add the elements it can easily surpass int and long int ranges.
4. Although, the answer got accepted when I used vector<int> a in the function
int check(int i,vector<int> a) {
if (a[i] > 0) {
return 1;
}
else return 0;
}
But as the user Scheff has said and is correct that it comes with a penalty in space and time, you should use call by reference i.e. vector<int> &a.
Modified Code:
#include <iostream>
#include <stdlib.h>
#include<vector>
using namespace std;
int check(int i, vector<int> &a) {
if (a[i] > 0) {
return 1;
}
else return 0;
}
int main() {
int t;
cin >> t;
while (t--)
{
int n;
cin >> n;
vector<int> a(n), b;
for (int i = 0; i < n; i++) {
cin >> a[i];
}
int i = 0;
while (i < n)
{
int max = a[i];
int s = check(i,a);
i++;
while ((i<n) && (check(i,a)== s)) {
if (a[i] > max)
max = a[i];
i++;
}
b.push_back(max);
}
long long s = 0;
for (int k = 0; k< b.size(); k++) {
s += b[k];
}
cout << s << endl;
}
}
Screenshot of Accepted Answer:

Improve on binary converting algorithm (include negative numbers)

I'm doing some C++ array homework. The goals is to convert decimal to binary (include negative numbers). Here's my code, it gets the job done, but I would like to see if anything can be improved, or any better algorithm (using binary shift maybe?).
#include <iostream>
using namespace std;
// doi tu thap phan sang nhi phan
void decToBinary(int n, int nhiphan[])
{
for (int i=0; i < 16; i++)
{
// to binary
nhiphan[i] = n % 2;
n = n / 2;
}
// inverse array
for (int i = 0, j = 15; i < j; i++, j--)
{
int temp = nhiphan[i];
nhiphan[i] = nhiphan[j];
nhiphan[j] = temp;
}
}
void reverse(int& a)
{
if (a == 0)
a++;
else a--;
}
void outArr(const int a[], int size) {
for (int i = 0; i < size; ++i)
cout << a[i];
}
int main()
{
int nhiphan[16];
int n;
do {
cout << "Nhap so (-255 <= n <= 255) chuyen doi sang nhi phan (16 bit): ";
cin >> n;
} while (n > 255 || n < -255);
if (n < 0) {//check negative
n *= -1;
decToBinary(n, nhiphan);
for (int i = 0; i < 16; i++)// 1's complement
reverse(nhiphan[i]);
// +1
if (nhiphan[15] == 0)//2's complement
nhiphan[15] = 1;
else
{
nhiphan[15] = 0;
int i = 15;
do {
reverse(nhiphan[i-1]);
i--;
} while (nhiphan[i-1] == 0);
}
}
else decToBinary(n, nhiphan);
outArr(nhiphan, 16);
return 0;
}

implementing merge sort in C++

I have studied the theory of the merge sort but don't have any idea of how to implement it in C++. My question is, merge sort creates arrays in recursion. But when implementing, how do we create arrays in runtime? or what is the general approach for this?
Thanks.

To answer the question: Creating dynamically sized arrays at run-time is done using std::vector<T>. Ideally, you'd get your input using one of these. If not, it is easy to convert them. For example, you could create two arrays like this:
template <typename T>
void merge_sort(std::vector<T>& array) {
if (1 < array.size()) {
std::vector<T> array1(array.begin(), array.begin() + array.size() / 2);
merge_sort(array1);
std::vector<T> array2(array.begin() + array.size() / 2, array.end());
merge_sort(array2);
merge(array, array1, array2);
}
}
However, allocating dynamic arrays is relatively slow and generally should be avoided when possible. For merge sort you can just sort subsequences of the original array and in-place merge them. It seems, std::inplace_merge() asks for bidirectional iterators.

Based on the code here: http://cplusplus.happycodings.com/algorithms/code17.html
// Merge Sort
#include <iostream>
using namespace std;
int a[50];
void merge(int,int,int);
void merge_sort(int low,int high)
{
int mid;
if(low<high)
{
mid = low + (high-low)/2; //This avoids overflow when low, high are too large
merge_sort(low,mid);
merge_sort(mid+1,high);
merge(low,mid,high);
}
}
void merge(int low,int mid,int high)
{
int h,i,j,b[50],k;
h=low;
i=low;
j=mid+1;
while((h<=mid)&&(j<=high))
{
if(a[h]<=a[j])
{
b[i]=a[h];
h++;
}
else
{
b[i]=a[j];
j++;
}
i++;
}
if(h>mid)
{
for(k=j;k<=high;k++)
{
b[i]=a[k];
i++;
}
}
else
{
for(k=h;k<=mid;k++)
{
b[i]=a[k];
i++;
}
}
for(k=low;k<=high;k++) a[k]=b[k];
}
int main()
{
int num,i;
cout<<"*******************************************************************
*************"<<endl;
cout<<" MERGE SORT PROGRAM
"<<endl;
cout<<"*******************************************************************
*************"<<endl;
cout<<endl<<endl;
cout<<"Please Enter THE NUMBER OF ELEMENTS you want to sort [THEN
PRESS
ENTER]:"<<endl;
cin>>num;
cout<<endl;
cout<<"Now, Please Enter the ( "<< num <<" ) numbers (ELEMENTS) [THEN
PRESS ENTER]:"<<endl;
for(i=1;i<=num;i++)
{
cin>>a[i] ;
}
merge_sort(1,num);
cout<<endl;
cout<<"So, the sorted list (using MERGE SORT) will be :"<<endl;
cout<<endl<<endl;
for(i=1;i<=num;i++)
cout<<a[i]<<" ";
cout<<endl<<endl<<endl<<endl;
return 1;
}

I have completed #DietmarKühl s way of merge sort. Hope it helps all.
template <typename T>
void merge(vector<T>& array, vector<T>& array1, vector<T>& array2) {
array.clear();
int i, j, k;
for( i = 0, j = 0, k = 0; i < array1.size() && j < array2.size(); k++){
if(array1.at(i) <= array2.at(j)){
array.push_back(array1.at(i));
i++;
}else if(array1.at(i) > array2.at(j)){
array.push_back(array2.at(j));
j++;
}
k++;
}
while(i < array1.size()){
array.push_back(array1.at(i));
i++;
}
while(j < array2.size()){
array.push_back(array2.at(j));
j++;
}
}
template <typename T>
void merge_sort(std::vector<T>& array) {
if (1 < array.size()) {
std::vector<T> array1(array.begin(), array.begin() + array.size() / 2);
merge_sort(array1);
std::vector<T> array2(array.begin() + array.size() / 2, array.end());
merge_sort(array2);
merge(array, array1, array2);
}
}

I've rearranged the selected answer, used pointers for arrays and user input for number count is not pre-defined.
#include <iostream>
using namespace std;
void merge(int*, int*, int, int, int);
void mergesort(int *a, int*b, int start, int end) {
int halfpoint;
if (start < end) {
halfpoint = (start + end) / 2;
mergesort(a, b, start, halfpoint);
mergesort(a, b, halfpoint + 1, end);
merge(a, b, start, halfpoint, end);
}
}
void merge(int *a, int *b, int start, int halfpoint, int end) {
int h, i, j, k;
h = start;
i = start;
j = halfpoint + 1;
while ((h <= halfpoint) && (j <= end)) {
if (a[h] <= a[j]) {
b[i] = a[h];
h++;
} else {
b[i] = a[j];
j++;
}
i++;
}
if (h > halfpoint) {
for (k = j; k <= end; k++) {
b[i] = a[k];
i++;
}
} else {
for (k = h; k <= halfpoint; k++) {
b[i] = a[k];
i++;
}
}
// Write the final sorted array to our original one
for (k = start; k <= end; k++) {
a[k] = b[k];
}
}
int main(int argc, char** argv) {
int num;
cout << "How many numbers do you want to sort: ";
cin >> num;
int a[num];
int b[num];
for (int i = 0; i < num; i++) {
cout << (i + 1) << ": ";
cin >> a[i];
}
// Start merge sort
mergesort(a, b, 0, num - 1);
// Print the sorted array
cout << endl;
for (int i = 0; i < num; i++) {
cout << a[i] << " ";
}
cout << endl;
return 0;
}

#include <iostream>
using namespace std;
template <class T>
void merge_sort(T array[],int beg, int end){
if (beg==end){
return;
}
int mid = (beg+end)/2;
merge_sort(array,beg,mid);
merge_sort(array,mid+1,end);
int i=beg,j=mid+1;
int l=end-beg+1;
T *temp = new T [l];
for (int k=0;k<l;k++){
if (j>end || (i<=mid && array[i]<array[j])){
temp[k]=array[i];
i++;
}
else{
temp[k]=array[j];
j++;
}
}
for (int k=0,i=beg;k<l;k++,i++){
array[i]=temp[k];
}
delete temp;
}
int main() {
float array[] = {1000.5,1.2,3.4,2,9,4,3,2.3,0,-5};
int l = sizeof(array)/sizeof(array[0]);
merge_sort(array,0,l-1);
cout << "Result:\n";
for (int k=0;k<l;k++){
cout << array[k] << endl;
}
return 0;
}

The problem with merge sort is the merge, if you don't actually need to implement the merge, then it is pretty simple (for a vector of ints):
#include <algorithm>
#include <vector>
using namespace std;
typedef vector<int>::iterator iter;
void mergesort(iter b, iter e) {
if (e -b > 1) {
iter m = b + (e -b) / 2;
mergesort(b, m);
mergesort(m, e);
inplace_merge(b, m, e);
}
}

I know this question has already been answered, but I decided to add my two cents. Here is code for a merge sort that only uses additional space in the merge operation (and that additional space is temporary space which will be destroyed when the stack is popped). In fact, you will see in this code that there is not usage of heap operations (no declaring new anywhere).
Hope this helps.
void merge(int *arr, int size1, int size2) {
int temp[size1+size2];
int ptr1=0, ptr2=0;
int *arr1 = arr, *arr2 = arr+size1;
while (ptr1+ptr2 < size1+size2) {
if (ptr1 < size1 && arr1[ptr1] <= arr2[ptr2] || ptr1 < size1 && ptr2 >= size2)
temp[ptr1+ptr2] = arr1[ptr1++];
if (ptr2 < size2 && arr2[ptr2] < arr1[ptr1] || ptr2 < size2 && ptr1 >= size1)
temp[ptr1+ptr2] = arr2[ptr2++];
}
for (int i=0; i < size1+size2; i++)
arr[i] = temp[i];
}
void mergeSort(int *arr, int size) {
if (size == 1)
return;
int size1 = size/2, size2 = size-size1;
mergeSort(arr, size1);
mergeSort(arr+size1, size2);
merge(arr, size1, size2);
}
int main(int argc, char** argv) {
int num;
cout << "How many numbers do you want to sort: ";
cin >> num;
int a[num];
for (int i = 0; i < num; i++) {
cout << (i + 1) << ": ";
cin >> a[i];
}
// Start merge sort
mergeSort(a, num);
// Print the sorted array
cout << endl;
for (int i = 0; i < num; i++) {
cout << a[i] << " ";
}
cout << endl;
return 0;
}

Here's a way to implement it, using just arrays.
#include <iostream>
using namespace std;
//The merge function
void merge(int a[], int startIndex, int endIndex)
{
int size = (endIndex - startIndex) + 1;
int *b = new int [size]();
int i = startIndex;
int mid = (startIndex + endIndex)/2;
int k = 0;
int j = mid + 1;
while (k < size)
{
if((i<=mid) && (a[i] < a[j]))
{
b[k++] = a[i++];
}
else
{
b[k++] = a[j++];
}
}
for(k=0; k < size; k++)
{
a[startIndex+k] = b[k];
}
delete []b;
}
//The recursive merge sort function
void merge_sort(int iArray[], int startIndex, int endIndex)
{
int midIndex;
//Check for base case
if (startIndex >= endIndex)
{
return;
}
//First, divide in half
midIndex = (startIndex + endIndex)/2;
//First recursive call
merge_sort(iArray, startIndex, midIndex);
//Second recursive call
merge_sort(iArray, midIndex+1, endIndex);
merge(iArray, startIndex, endIndex);
}
//The main function
int main(int argc, char *argv[])
{
int iArray[10] = {2,5,6,4,7,2,8,3,9,10};
merge_sort(iArray, 0, 9);
//Print the sorted array
for(int i=0; i < 10; i++)
{
cout << iArray[i] << endl;
}
return 0;
}

This would be easy to understand:
#include <iostream>
using namespace std;
void Merge(int *a, int *L, int *R, int p, int q)
{
int i, j=0, k=0;
for(i=0; i<p+q; i++)
{
if(j==p) //When array L is empty
{
*(a+i) = *(R+k);
k++;
}
else if(k==q) //When array R is empty
{
*(a+i) = *(L+j);
j++;
}
else if(*(L+j) < *(R+k)) //When element in L is smaller than element in R
{
*(a+i) = *(L+j);
j++;
}
else //When element in R is smaller or equal to element in L
{
*(a+i) = *(R+k);
k++;
}
}
}
void MergeSort(int *a, int len)
{
int i, j;
if(len > 1)
{
int p = len/2 + len%2; //length of first array
int q = len/2; //length of second array
int L[p]; //first array
int R[q]; //second array
for(i=0; i<p; i++)
{
L[i] = *(a+i); //inserting elements in first array
}
for(i=0; i<q; i++)
{
R[i] = *(a+p+i); //inserting elements in second array
}
MergeSort(&L[0], p);
MergeSort(&R[0], q);
Merge(a, &L[0], &R[0], p, q); //Merge arrays L and R into A
}
else
{
return; //if array only have one element just return
}
}
int main()
{
int i, n;
int a[100000];
cout<<"Enter numbers to sort. When you are done, enter -1\n";
i=0;
while(true)
{
cin>>n;
if(n==-1)
{
break;
}
else
{
a[i] = n;
i++;
}
}
int len = i;
MergeSort(&a[0], len);
for(i=0; i<len; i++)
{
cout<<a[i]<<" ";
}
return 0;
}

This is my version (simple and easy):
uses memory only twice the size of original array.
[ a is the left array ] [ b is the right array ] [ c used to merge a and b ] [ p is counter for c ]
void MergeSort(int list[], int size)
{
int blockSize = 1, p;
int *a, *b;
int *c = new int[size];
do
{
for (int k = 0; k < size; k += (blockSize * 2))
{
a = &list[k];
b = &list[k + blockSize];
p = 0;
for (int i = 0, j = 0; i < blockSize || j < blockSize;)
{
if ((j < blockSize) && ((k + j + blockSize) >= size))
{
++j;
}
else if ((i < blockSize) && ((k + i) >= size))
{
++i;
}
else if (i >= blockSize)
{
c[p++] = b[j++];
}
else if (j >= blockSize)
{
c[p++] = a[i++];
}
else if (a[i] >= b[j])
{
c[p++] = b[j++];
}
else if (a[i] < b[j])
{
c[p++] = a[i++];
}
}
for (int i = 0; i < p; i++)
{
a[i] = c[i];
}
}
blockSize *= 2;
} while (blockSize < size);
}