Delete the first group of consecutive negative
elements in the array.
void delNegative(int* arr, int& size) {
if (!check_prop(arr, size, neg)) return;
int ind_l, ind_f;//
if (arr[0] < 0) ind_f = 0;
if (arr[size - 1] < 0) ind_l = size - 1;
for (int i = 1; i < size; i++){
if (arr[i - 1] > 0 && arr[i] < 0) ind_f = i;
if (arr[i - 1] < 0 && arr[i] > 0) ind_l = i - 1;
}
if (ind_l < ind_f) ind_l = size - 1;
int ii = 0;
for (int i = 0; i < size; i++)
if (i > ind_l || i < ind_f) {
arr[ii] = arr[i];
ii++;
}
size -= ind_l - ind_f + 1;
}
There is this function, but it removes the last group of negative elements.
It needs to be redone for dynamic arrays
When I free memory, the error "Program has triggered a breakpoint" occur. Check the code below, I wonder where is wrong ?
int SSavep(char *visited, int t, int n, int m)
{
int* map = (int*)malloc(m*n * sizeof(int));
int* q = (int*)malloc(m*n * sizeof(int));
int count = 0, cur = 0;
int begin = 0, end = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
//set value for map
}
}
..........
if (t >= map[end]) {
free(map);
free(q);
return 0;
}
else{
free(map);
free(q);
return -1;
}
}
the entire code is as below:
static int dir[4][2] = { {0,1},{0,-1},{1,0},{-1,0} };
int SSavep(char *visited, int t, int n, int m)
{
int* map = (int*)malloc(m*n * sizeof(int));
int* q = (int*)malloc(m*n * sizeof(int));
int count = 0, cur = 0;
int begin = 0, end = 0;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (visited[i*n + j] == '.')
map[i*n + j] = 0;
else if (visited[i*n + j] == '*')
map[i*n + j] = -1;
else if (visited[i*n + j] == 'p') {
map[i*n + j] = -12;
end = i*n + j;
}
else {
map[i*n + j] = -9;
begin = i*n + j;
}
}
}
q[count++] = begin;
while (cur < count && q[cur] != end) {
int i = q[cur] / n;
int j = q[cur] % n;
for (int k = 0; k < 4; k++) {
int ni = i + dir[k][0];
int nj = j + dir[k][1];
if (ni < 0 || ni >= m || nj < 0 || nj >= n || map[ni*n + nj]>0 || map[ni*n + nj] == -1)
continue;
map[ni*n + nj] = map[i*n + j] + 1;
q[count++] = ni*n + nj;
}
cur++;
}
if (map[end] > 0 && t >= map[end]) {
free(map);
free(q);
return 0;
}
else{
free(map);
free(q);
return -1;
}
}
You are getting error on >> free(q);
for m=n=4
int* q = (int*)malloc(m*n * sizeof(int));
q == [m*n*sizeof(int)] == 4*4*4 == 64 bytes == int[16];
Because you wrote beyond the address space reserved for pointer variable 'q'.
check 'count' variable before free(q). I've got 1208, called with:
char* visited = new char[100 * 100];
memset(visited, 0, 10000);
int res = SSavep(visited, 0, 4, 4);
Btw this algo looks alot like some path finding examining neighboring cells on map and assigning weights, right? If so there are many opensource solutions, why not using them instead of reinventing the wheel? There are links to opensource solutions on wiki Path Finding page:
https://en.wikipedia.org/wiki/Pathfinding
check the bottom of the page for links.
I'm doing my os homework, and the requirement is to implement parallel merge sort using Pthread and using semaphore to lock and unlock them.
You can only look at the function names Multi____ and ignore Single_____, because I've already finished the single thread part.
I encountered a problem in the multithreaded part. I signal the master thread (sem[1]) in line 227, and it should go into the function 'void *MultiPartition'.
In this function, it gives value to arg[id * 2] and arg[id * 2 + 1].
For example, arg[1] will gives value to arg[2] and arg[3], and then it signals thread[2] and thread[3] to start by sem_post.
And doesn't seem to work.
So I use cout << "partition id = " << id << ", head = " << head << ", mid = " << mid << ", tail = " << tail << "\n"; in line 111 to check out what happens.
It looks really weird. It sometimes outputs
partition id = 1, head = 0, mid = 7, tail = 15
partition id = 2, head = 0, mid = 3, tail = 7
and it was stuck, but the program didn't exit. Means I need to press Ctrl^C to exit program.
Sometimes it outputs
partition id = 1, head = 0, mid = 7, tail = 15
partition id = 2, head = 0, mid = 3, tail = 7
partition id = 3, head = 8, mid = 11, tail = 15
partition id = 4, head = 0, mid = 1, tail = 3
and was stuck, too.
I'm curious where other threads going?
And if id = 4 is displayed, it will runs bubble id = 8 usually.
#include <iostream>
#include <pthread.h>
#include <semaphore.h>
#include <fstream>
#include <sys/time.h>
#include <unistd.h>
using namespace std;
//Pthread_create, pthread_exit *don't use pthread_join
//sem_init, sem_wait, sem_post, sem_getvalue, sem_destroy
//Enter input file name: test.txt
//MT sorting used x secs
//ST sorting used x secs
// g++ -o os_hw3.out os_hw3.cpp -pthread
typedef struct{
int head;
int mid;
int tail;
int id;
}arguments;
//declare global variables
sem_t sem[16]; // use id = 1 ~ 15
sem_t final; // the final semaphore signal that indicate all threads finished.
int* s1, * s2; // two array for single and multiple
arguments arg[16];
void swap(int* x, int* y) {
int temp;
temp = *x;
*x = *y;
*y = temp;
}
void *MultiMerge(void* argid) {
int id = *(int*)argid;
sem_wait(&sem[id]);
sem_wait(&sem[id]);
int head = arg[id].head, mid = arg[id].mid, tail = arg[id].tail;
//cout << "merge id = " << id << ", head = " << head << ", mid = " << mid << ", tail = " << tail << "\n";
int lenA = mid - head + 1;
int lenB = tail - (mid + 1) + 1;
int A[lenA];
int B[lenB];
for (int i = 0; i < lenA; i++) {
A[i] = *(s1 + head + i);
}
for (int j = 0; j < lenB; j++) {
B[j] = *(s1 + mid + 1 + j);
}
int i = 0, j = 0, k = 0;
while (i < lenA && j < lenB) {
if (A[i] < B[j]) {
*(s1 + head + k) = A[i];
i++;
}
else {
*(s1 + head + k) = B[j];
j++;
}
k++;
}
while (i < lenA) {
*(s1 + head + k) = A[i];
i++;
k++;
}
while (j < lenA) {
*(s1 + head + k) = B[j];
j++;
k++;
}
sem_post(&sem[id / 2]); // signal the upper level
if (id == 1) {
fstream fout;
fout.open("output1.txt", ios::out);
for (i = 0; i < arg[1].tail + 1; i++)
fout << *(s2 + i);
fout.close();
sem_post(&final);
}
}
void *MultiBubble(void *argid) {
int id = *(int*)argid;
sem_wait(&sem[id]);
//cout << "bubble id = " << id << ", head = " << arg[id].head << ", tail = " << arg[id].tail << "\n";
for (int i = arg[id].tail; i > 0; --i) {
for (int j = arg[id].head; j < i; ++j) {
if (*(s2 + j) > * (s2 + j + 1)) {
swap((s2 + j), (s2 + j + 1));
}
}
}
for (int i = arg[id].head; i <= arg[id].tail; i++) {
cout << *(s2 + i) << " ";
}
cout << "\n";
sem_post(&sem[id / 2]);
}
void *MultiPartition(void* argid) {
int id = *(int*)argid;
sem_wait(&sem[id]);
int head = arg[id].head, mid = arg[id].mid, tail = arg[id].tail;
cout << "partition id = " << id << ", head = " << head << ", mid = " << mid << ", tail = " << tail << "\n";
arg[id * 2].head = arg[id].head;
arg[id * 2].tail = arg[id].mid;
arg[id * 2].mid = (arg[id * 2].head + arg[id * 2].tail) / 2;
arg[id * 2 + 1].head = arg[id].mid + 1;
arg[id * 2 + 1].tail = arg[id].tail;
arg[id * 2 + 1].mid = (arg[id * 2 + 1].head + arg[id * 2 + 1].tail) / 2;
sem_post(&sem[id * 2]);
sem_post(&sem[id * 2 + 1]);
}
void SingleMerge(int* s1, int head, int mid, int tail) {
int lenA = mid - head + 1;
int lenB = tail - (mid + 1) + 1;
int A[lenA];
int B[lenB];
for (int i = 0; i < lenA; i++) {
A[i] = *(s1 + head + i);
}
for (int j = 0; j < lenB; j++) {
B[j] = *(s1 + mid + 1 + j);
}
int i = 0, j = 0, k = 0;
while (i < lenA && j < lenB) {
if (A[i] < B[j]) {
*(s1 + head + k) = A[i];
i++;
}
else {
*(s1 + head + k) = B[j];
j++;
}
k++;
}
while (i < lenA) {
*(s1 + head + k) = A[i];
i++;
k++;
}
while (j < lenA) {
*(s1 + head + k) = B[j];
j++;
k++;
}
}
int SingleBubble(int* s1, int head, int tail) {
for (int i = tail; i > 0; --i) {
for (int j = head; j < i; ++j) {
if (*(s1 + j) > *(s1 + j + 1)) {
swap((s1 + j), (s1 + j + 1));
}
}
}
}
void SinglePartition(int* s1, int head, int tail, int times) {
if (head <= tail) {
int mid = (head + tail) / 2;
if (times < 3) {
SinglePartition(s1, head, mid, ++times);
SinglePartition(s1, mid + 1, tail, ++times);
}
else {
SingleBubble(s1, head, tail);
}
SingleMerge(s1, head, mid, tail);
}
}
int main() {
char filename[100];
int num;
struct timeval Tstart, Tend;
cout << "Enter the input file name: ";
cin >> filename;
fstream file, fout;
file.open(filename, ios::in);
if (!file) {
cout << "Read File Error.\n";
return -1;
}
else {
file >> num;
s1 = new int[num];
s2 = new int[num];
for (int i = 0; i < num; i++) {
file >> *(s1 + i);
*(s2 + i) = *(s1 + i);
}
file.close();
}
//SINGLE THREAD
gettimeofday(&Tstart, 0);
SinglePartition(s1, 0, num - 1, 0);
gettimeofday(&Tend, 0);
fout.open("output2.txt", ios::out);
for (int i = 0; i < num; i++)
fout << *(s1 + i) << " ";
fout.close();
double Tdifference = (Tend.tv_sec - Tstart.tv_sec) + (Tend.tv_usec - Tstart.tv_usec) / 1000000.0;
cout << "Single thread cost " << Tdifference << " s\n";
//MULTI THREAD
gettimeofday(&Tstart, 0);
arg[1].head = 0;
arg[1].tail = num - 1;
arg[1].mid = (arg[1].head + arg[1].tail) / 2;
pthread_t thread[16];
sem_init(&final, 0, 0);
sem_post(&sem[1]);
for (int i = 1; i < 16; i++){
arg[i].id = i;
sem_init(&sem[i], 0, 0);
if (i < 8) {
if(i == 1)
sem_post(&sem[1]); // call the master thread
pthread_create(&thread[i], NULL, MultiPartition, &arg[i].id);
}
else
pthread_create(&thread[i], NULL, MultiBubble, &arg[i].id);
}
for (int i = 7; i > 0; i--) {
pthread_create(&thread[i], NULL, MultiMerge, &arg[i].id);
}
sem_wait(&final);
gettimeofday(&Tend, 0);
Tdifference = (Tend.tv_sec - Tstart.tv_sec) + (Tend.tv_usec - Tstart.tv_usec) / 1000000.0;
cout << "Multi thread cost " << Tdifference << " s\n";
delete s1, s2;
for (int i = 0; i < 16; i++)
sem_destroy(&sem[i]);
sem_destroy(&final);
return 0;
}
I fixed it myself.
for (int i = 7; i > 0; i--) {
pthread_create(&thread[i], NULL, MultiMerge, &arg[i].id);
}
The above part will confront the following part.
for (int i = 1; i < 16; i++){
arg[i].id = i;
sem_init(&sem[i], 0, 0);
if (i < 8) {
if(i == 1)
sem_post(&sem[1]); // call the master thread
pthread_create(&thread[i], NULL, MultiPartition, &arg[i].id);
}
MultiMerge and MultiPartition both have
sem_wait(&sem[id]);
So if sem[id]'s value != 0, it doesn't know which function to do, in my opinion.
I delete
for (int i = 7; i > 0; i--) {
pthread_create(&thread[i], NULL, MultiMerge, &arg[i].id);
}
and add
MultiMerge(&arg[id].id);
in the bottom of MultiPartition to call MultiMerge, this can fixed my problem.
Given a set of numbers, check whether it can be partitioned into two subsets such that the sum of elements in both subsets is same or not
I am getting segmentation fault in C++(g++ 5.4) with a this problem.
This is where i submitted my solution in C++
https://practice.geeksforgeeks.org/problems/subset-sum-problem/0
I am checking if the array can be divided into two parts with equal sum. So I am just checking if there exists a subset with sum equal to half the sum of the array
I have implemented the below logic with dynamic programming
Let dp[i][j] denote yes or no whether a subset with sum j is possible to form with elements in the range [0, i](both inclusive) where i is 0-based index. I have done nothing new with this traditional problem. But I am getting segmentation fault. The program is giving correct output for small test cases. What mistake have I made
I haven't used any comments because I have done nothing new. Hope it is understandable.
#include <iostream>
#include <bits/stdc++.h>
#include<cstdio>
#define ll long long int
using namespace std;
bool isVowel(char c){
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
bool isLower(char c){
return 97 <= c && c <= 122;
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
cout << setprecision(10);
ll t, n;
cin >> t;
while (t--) {
cin >> n;
ll a[n];
ll sum = 0;
for (ll i = 0; i < n; i++) {
cin >> a[i];
sum += a[i];
}
if (sum % 2) {
cout << "NO" << '\n';
continue;
}
sum /= 2;
ll dp[n][sum + 1];
for (ll i = 0; i < n; i++) {
for(ll j = 0; j < sum + 1; j++) {
dp[i][j] = 0;
}
}
for (ll i = 0; i < n; i++) {
dp[i][a[i]] = 1;
dp[i][0] = 1;
}
for (ll i = 1; i < n; i++) {
for (ll j = 1; j < sum + 1; j++){
if (j - a[i] > 0) {
dp[i][j] = dp[i - 1][j - a[i]];
}
dp[i][j] |= dp[i - 1][j];
}
}
cout << (dp[n - 1][sum] ? "YES" : "NO") << '\n';
}
}
The segmentation fault is due to
ll dp[n][sum + 1];
Even though the constraints say 1 <= N<= 100, 0 <= arr[i]<= 1000, the test cases used are probably much larger, so ll dp[n][sum + 1] will end up taking some serious stack memory, use
bool dp[n][sum + 1];
It should work fine.
On a side note, avoid using ll randomly, use them according to the constraints.
I'm trying to do a pascal triangle but for some reason i have the error "Run-Time Check Failure #2 - Stack around the variable 'TP' was corrupted." Can someone help me, please?
#include <iostream>
using namespace std;
void main()
{
int TP[100][100] = { 0 }, n;
do
{
cout << "Digite a ordem do triangulo de pascal: ";
cin >> n;
} while (n < 0 || n > 100);
for (int j = 0; j < n; j++)
{
TP[j][0] = 1;
TP[j][j] = 1;
}
for (int i = 2; i < n + 1; i++)
for (int j = 1; TP[i][j] != 1; j++)
TP[i][j] = TP[i - 1][j - 1] + TP[i - 1][j];
for (int i = 0; i < n; i++)
{
cout << endl;
for (int j = 0; j != i + 1; j++)
cout << TP[i][j] << " ";
}
cout << endl << endl;
system("pause");
}
You go beyond the array boundaries in for (int i = 2; i < n + 1; i++) for i == n when n=100.