This is the code for reverse polish notation. When I run it on unix it works fine but codechef says wrong answer. Please help.
#include<stdio.h>
#include<string.h>
void push(char);
void pop();
char stack[400];
unsigned long top=0;
int main()
{
unsigned long len, test_case,i=0,j=0;
char expr[400];
scanf("%u\n",&test_case);
for(;j<test_case;j++)
{
scanf("%s\n",expr);
len=strlen(expr);
for(;i<len;i++)
{
if(expr[i]=='+'||expr[i]=='-'||expr[i]=='*'||expr[i]=='/'||expr[i]=='^')
push(expr[i]);
else if(expr[i]==')')
{
pop();
}
else if(expr[i]=='(')
continue;
else
printf("%c",expr[i]);
}
}
return 0;
}
void pop()
{
if(top==-1)
return;
else
{
printf("%c",stack[top]);
top=top-1;
}
}
void push(char x)
{
if(top==400)
return;
else
{
stack[++top]=x;
}
}
Change this line
for(;i<len;i++)
to
for(i=0;i<len;i++)
and then it will work fine. In codechef there are many testcases, so in your program it will work fine for 1st testcase but it will fail after that because the value of 'i' from the 2nd testcase will start from where it ends in the 1st case. So,you have to initialize the value of i in every testcase.
I went through the same situation and this is my solution to that problem:
#include<stdio.h>
#include<string.h>
char stack[400];
unsigned long top = 0;
void pop() {
if (top == -1) {
return;
} else {
printf("%c",stack[top]);
top = top - 1;
}
}
void push(char x) {
if (top == 400) {
return;
} else {
stack[++top] = x;
}
}
int main() {
unsigned long t, n, i = 0, j = 0;
char a[400];
scanf("%lu", &t);
while(t--) {
scanf("%s", &a);
n = strlen(a);
for (i = 0; i < n; i++) {
if (a[i] == '+' || a[i] == '-' || a[i] == '*' || a[i] == '/' || a[i] == '^') {
push(a[i]);
} else if (a[i] == ')') {
pop();
} else if (a[i] == '(') {
continue;
} else {
printf("%c", a[i]);
}
}
printf("\n");
}
return 0;
}
You missed writing some sections and there is another notation along with +, -, * and / i.e. ^ which was also required for the evaluation of the answer.
Only place that I noticed for the time being
`scanf("%u\n",&test_case);` //"%u" for unsigned not unsigned long
`scanf("%s\n",expr);` //%s : newline to not input are separated by white space
you have to initialize unsigned long i = 0 ; inside for loop as if you take two or more test cases i will initiate from where it left in last case for every next case .
Add
top=0;
stack[0]=0;
before ending for loop in function main and change
for(;i<len;i++) to for(i=0;i<len;i++).
Related
I have spent the last two hours trying to debug my code that is supposed to check if the input consists of well-formed brackets. What I mean by well formed is that ()()[] or ([()]) are acceptable but ((((() is not.
I'm not allowed to use any header file apart from <stdio.h>
#include <stdio.h>
void cross(char str[], int i, int j) {
str[i] = 'X';
str[j] = 'X';
}
int iscrossed(char str[]) {
int i = 0;
while (str[i] != '\0') {
if (str[i] != 'X')
return 0;
i++;
}
return 1;
}
int check(char str[]) {
int i = 1, j;
while (str[i] != '\0') {
if (str[i] == ')') {
for (j = i - 1; j >= 0; j--) {
if (str[j] == '(') {
cross(str, str[i], str[j]);
}
break;
}
} else
if (str[i] == ']') {
for (j = i - 1; j >= 0; j--) {
if (str[j] == '[') {
cross(str, str[i], str[j]);
}
break;
}
}
i++;
}
if (iscrossed(str) == 1)
return 1;
else
return 0;
}
int main() {
char str[20];
scanf("%s", str);
printf("%d\n", check(str));
}
For certain inputs the program prints a zero followed by a segmentation fault and for the others it just prints a zero.
Please keep in mind that I'm a beginner programmer so please don't include too much heavy stuff in your hints. I'd be grateful for any help on this.
Edit: It would be wonderful if your answer tells me the errors in my code, because that was my question in the first place.
Here a simple recursive solution:
#include <stdio.h>
int brace(const char **s, char cc)
{
while(1) {
if(**s == cc) { return 0; }
switch(**s) {
case '\0': return -1;
case '[': ++(*s); if(brace(s, ']')) { return -1; } ++(*s); break;
case '{': ++(*s); if(brace(s, '}')) { return -1; } ++(*s); break;
case '(': ++(*s); if(brace(s, ')')) { return -1; } ++(*s); break;
case ']':
case '}':
case ')': return -1;
default: ++(*s);
}
}
}
int check_brace(const char *s)
{
return brace(&s, '\0');
}
int main()
{
printf("%d\n", check_brace(" hekl(l o{ asdf } te)ts()({})"));
}
Returns -1 when somethings wrong. Otherwise 0.
There are multiple problems in your code:
you call cross(str, str[i], str[j]); instead of cross(str, i, j); when you find matches for parentheses and brackets.
the break statement should be moved inside the if block.
your method does not allow detection of nesting errors
your method would have undefined behavior if str is an empty string (which you cannot input via scanf())
Here is a modified version:
#include <stdio.h>
void cross(char str[], int i, int j) {
str[i] = str[j] = 'X';
}
int iscrossed(char str[]) {
int i = 0;
while (str[i] != '\0') {
if (str[i] != 'X')
return 0;
i++;
}
return 1;
}
int check(char str[]) {
int i = 0, j;
while (str[i] != '\0') {
if (str[i] == ')') {
for (j = i - 1; j >= 0; j--) {
if (str[j] == '(') {
cross(str, i, j);
break;
}
}
} else
if (str[i] == ']') {
for (j = i - 1; j >= 0; j--) {
if (str[j] == '[') {
cross(str, i, j);
break;
}
}
}
i++;
}
return iscrossed(str);
}
int main() {
char str[80];
if (scanf("%79s", str) == 1) {
printf("%d\n", check(str));
}
return 0;
}
Here is a simpler alternative:
#include <stdio.h>
const char *check(const char str[], int endc) {
while (str) {
int c = *str++;
switch (c) {
case '(': str = check(str, ')'); break;
case '[': str = check(str, ']'); break;
case '{': str = check(str, '}'); break;
case ')':
case ']':
case '}':
case '\0': return c == endc ? str : NULL;
}
}
return NULL;
}
int main() {
char str[80];
if (fgets(str, sizeof str, stdin)) {
printf("%d\n", check(str, '\0') != NULL);
}
return 0;
}
Pseudocode of a possible answer:
initialize char[] unclosed
int latest_unclosed_index = -1
for each char in string {
if char == opening_bracket {
latest_unclosed_index += 1
unclosed[latest_unclosed_index] = char
} else if char == closing_bracket {
if latest_unclosed_index < 0 {
return false
} else if char == closing_of(unclosed[latest_unclosed_index]) {
unclosed[latest_unclosed_index] = null
latest_unclosed_index -= 1
} else {
return false
}
}
}
if latest_unclosed_index == -1 {
return true
} else {
return false
}
This works by keeping an array of all unclosed opening brackets in the order that you encounter them in, and removing them whenever you encounter a closing bracket, as a sort of stack.
This solution has a complexity of O(n).
A problem with this implementation is that there is an unknown amount of brackets in string, which may cause the array to overflow if it isn't large enough.
Solution:
To be sure that this solution doesn't overflow, the size of the array should be at least half the size of the input string, and you'll have to check at each character if there are enough characters left in the input string to be able to completely close all brackets.
Use a list implementation (or write your own) instead of an array for unclosed.
If its ok for you to use stdlib.h then,
#include <stdio.h>
#include <stdlib.h>
#define bool int
// structure of a stack node
struct sNode {
char data;
struct sNode* next;
};
// Function to push an item to stack
void push(struct sNode** top_ref, int new_data);
// Function to pop an item from stack
int pop(struct sNode** top_ref);
// Returns 1 if character1 and character2 are matching left
// and right Brackets
bool isMatchingPair(char character1, char character2)
{
if (character1 == '(' && character2 == ')')
return 1;
else if (character1 == '{' && character2 == '}')
return 1;
else if (character1 == '[' && character2 == ']')
return 1;
else
return 0;
}
// Return 1 if expression has balanced Brackets
bool areBracketsBalanced(char exp[])
{
int i = 0;
// Declare an empty character stack
struct sNode* stack = NULL;
// Traverse the given expression to check matching
// brackets
while (exp[i])
{
// If the exp[i] is a starting bracket then push
// it
if (exp[i] == '{' || exp[i] == '(' || exp[i] == '[')
push(&stack, exp[i]);
// If exp[i] is an ending bracket then pop from
// stack and check if the popped bracket is a
// matching pair*/
if (exp[i] == '}' || exp[i] == ')'
|| exp[i] == ']') {
// If we see an ending bracket without a pair
// then return false
if (stack == NULL)
return 0;
// Pop the top element from stack, if it is not
// a pair bracket of character then there is a
// mismatch.
// his happens for expressions like {(})
else if (!isMatchingPair(pop(&stack), exp[i]))
return 0;
}
i++;
}
// If there is something left in expression then there
// is a starting bracket without a closing
// bracket
if (stack == NULL)
return 1; // balanced
else
return 0; // not balanced
}
// Driver code
int main()
{
char exp[100] = "{()}[]";
// Function call
if (areBracketsBalanced(exp))
printf("Balanced \n");
else
printf("Not Balanced \n");
return 0;
}
// Function to push an item to stack
void push(struct sNode** top_ref, int new_data)
{
// allocate node
struct sNode* new_node
= (struct sNode*)malloc(sizeof(struct sNode));
if (new_node == NULL) {
printf("Stack overflow n");
getchar();
exit(0);
}
// put in the data
new_node->data = new_data;
// link the old list off the new node
new_node->next = (*top_ref);
// move the head to point to the new node
(*top_ref) = new_node;
}
// Function to pop an item from stack
int pop(struct sNode** top_ref)
{
char res;
struct sNode* top;
// If stack is empty then error
if (*top_ref == NULL) {
printf("Stack overflow n");
getchar();
exit(0);
}
else {
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
Features
Support nested parantheses like (())
Support bad nestings such as ([)]
Commented but not by me (see the below spoiler)
Note: i feel guilty that i have copied code from here
The algorithm
Pseudocode Like Block code!
If it's not ok for you to use stdlib.h,then someone may edit this code and remove the errors that occur when we remove that line(#include <stdlib.h>),I am not a c guy and i don't know to edit,i just copy pasted !
First attempt didn't worked with bad nesting, as MOehm wrote in the comments.
Storing the opened braces that not have been closed yet helps to recognize bad nesting. The last opened brace will determine which closing brace is need.
#include <stdio.h>
int check(char str[], int size)
{
char opened[size/2];
char close;
int i = 0;
int pos = 0;
int error = 0;
while((i < size) || (pos < size/2))
{
if((str[i] == '(') || (str[i] == '['))
{
opened[pos] = str[i];
pos++;
}
else if((str[i] == ')') || (str[i] == ']'))
{
if(str[i] == close)
{
opened[pos-1] = 0;
pos--;
}
else
{
error = 1;
break;
}
}
printf("%s\n", opened);
if(pos > 0)
{
switch(opened[pos-1])
{
case '(':
close = ')';
break;
case '[':
close = ']';
break;
}
}
else
close = 0;
i++;
}
return error;
}
int main() {
char str[20];
printf("%d\n", check(str, sizeof(str)));
return 0;
}
Write a program which accepts a valid infix expression and provide its corresponding postfix expression. Consider following assumptions:
• There are 5 operators (+,-,*,/ and ^).
• + and – follow right to left associativity
• * and / also follow right to left associativity
• ^ follow left to right associativity
#include<stdio.h>
#include<ctype.h>
char stack[20];
int top = -1;
void push(char x)
{
stack[++top] = x;
}
char pop()
{
if(top == -1)
return -1;
else
return stack[top--];
}
int priority(char x)
{
if(x=='^')
return 0;
if(x == '+' || x == '-')
return 1;
else if(x == '*' || x == '/')
return 2;
}
void main()
{
char exp[20];
char *e, x;
printf("Enter the expression :: ");
scanf("%s",exp);
e = exp;
while(*e != '\0')
{
if(isalnum(*e))
printf("%c",*e);
else if(*e=='^')
{
while(priority(stack[top]) >=priority(*e))
printf("%c",pop());
push(*e);
}
else
{
while(priority(stack[top]) > priority(*e))
printf("%c",pop());
push(*e);
}
e++;
}
while(top != -1)
{
printf("%c",pop());
}
}
This program gives an infinite loop upon execution.
I get following output after entering any expression,
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When there is no element in the stack i.e. top = -1 in your code and you call priority on that it return garbage + IMO top should be returning the no. of elements in the stack for ease rather than position in array.
Apart from this the changed associativity will not serve the purpose as a-b is not equal to b-a usually.
Here is the corrected code:
#include<stdio.h>
#include<ctype.h>
char stack[20];
int top = 0; //CHANGE 1 -NOW IT MEANS NO. OF ELEMENTS IN THAT STACK
void push(char x)
{
stack[top++] = x;
}
void swap(char c[],int i, int j)
{
char temp;
temp =c[i];
c[i] = c[j];
c[j] = temp;
}
char pop()
{
return stack[--top]; //CHANGE 2
}
int priority(char x)
{
if(x=='^')
return 3; //CHANGE 3 - CONCEPTUAL
if(x == '+' || x == '-')
return 1;
else if(x == '*' || x == '/')
return 2;
return -1;
}
int main()
{
char exp[20];
char *e, x;
printf("Enter the expression :: ");
scanf("%s",exp);
//CHANGE INDUCED AS ASSOCIATIVITY HAS BEEN REVERSED FOR OPERATORS OTHER THAN '^'
int i=0,length=strlen(exp);
while(i<length/2) //REVERSING THE INPUT EXPRESSION
{
swap(exp,i,length-i-1);
++i;
}
e = exp;
while(*e != '\0')
{
if(*(e+1)=='^') //ASSOCIATIVITY OF THIS OPERATOR HAS BEEN KEPT AS USUAL HENCE REVERSING A SMALL PART JUST TO COMPUTE IT AS USUAL
{
swap(e,0,2);
}
x=*e; //CHANGE 4
if(isalnum(x))
printf("%c",x);
else
{
while((top>0)&&(priority(stack[top-1]) > priority(x))) //CHANGE 5
printf("%c",pop());
push(x);
}
e++;
}
while(top>0) //CHANGE 5
{
printf("%c",pop());
}
return 0;
}
INPUT:
a+b^c
cab^-
a+b*c^d
OUTPUT (with reversed associativity of operators except '^'):
1.
bc^a+
ab^c+
cd^b*a+
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Here is my code for the problem id-4 in spoj, it is running perfectly in ideone.com but in spoj its showing segmentation fault, I am unable to find out the bug.Please help. I have used stacks to implement it.Thanks in advance.
exp is the input string, and out is the output string, int l keeps track of the index of output string
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
//declarations
int top = 0;
void push(char ch);
char pop();
int prec(char c);
char stack[20];
char out[20];
int l = 0;
int main() {
char exp[20];
int x, t;
char temp;
int i = 0;
scanf("%d", &t); //no. of test cases
while (t--) {
l=0;
scanf("%s", exp);
stack[0] = '('; //initially pushing '('
x = strlen(exp);
exp[x] = ')'; //for completion of statement
for (i = 0; i < x + 1; i++) {
if (exp[i] == '+' || exp[i] == '-' ||
exp[i] == '/' || exp[i] == '*' || exp[i] == '^') { //operators
while (prec(exp[i]) < prec(stack[top])) { //checking precedence
out[l] = pop();
l++;
}
push(exp[i]);
} else
if (exp[i] == '(') {
push(exp[i]);
} else
if (exp[i] == ')') {
while (stack[top] != '(') {
out[l] = pop();
l++;
}
temp = pop(); //to throw out '('
} else {
out[l] = exp[i]; //variables
l++;
}
}
for (i = 0; i < l; i++) {
printf("%c", out[i]); //output
}
}
return 0;
}
void push(char c) { //push operation on stack
if (top >= 19) {
} else {
top++;
stack[top] = c;
}
return;
}
char pop() { //pop operation on stack
char t;
if (top <= -1) {
return 0;
} else {
t = stack[top];
top--;
return t;
}
}
int prec(char c) { //precedence check
if (c == 94) { return 5; }
else if (c == 47) { return 4; }
else if (c == 42) { return 3; }
else if (c == 43) { return 2; }
else if (c == 45) { return 1; }
else { return 0; }
}
The buffer to read the expression is very small: char exp[20]; and you do not protect scanf for buffer overflow.
Sphere Online Judge specifies:
Input
t [the number of expressions <= 100]
expression [length <= 400]
[other expressions]
You should use a larger stack, at least 100, a larger buffer and use:
char exp[402];
...
scanf("%400s", exp);
Also test the return value from scanf("%d", &t), just in case they give you purposely erroneous input.
You have to check if top is greater or equal 0:
while( top >= 0 && prec(exp[i])<prec(stack[top])){
// ^^^^^^^^
....
while( top >= 0 && stack[top]!='('){
// ^^^^^^^^
The segmentation fault occurs, because you access to stack[top] and top is less than 0. Apart form this I recommend to increase the size of arrays exp and out coherently.
I have a program which convert the infix form to postfix form.
example if the input is :(-2+4-(3+3))
then the output is : 2-4+33+-
my expected ouput is : 2 - 4 + 3 3 + -
How should I do to change it , I am thinking somethings like extra 1 more space when I pop() the stack.
Here is the code:
#include <stdio.h>
#include <ctype.h>
#include<string.h>
#define SIZE 50
char s[SIZE];
int top = -1;
void push(char elem) {
s[++top] = elem;
}
char pop() {
return (s[top--]);
}
int pr(char elem) {
switch (elem) {
case '#':
return 0;
case '(':
return 1;
case '+':
case '-':
return 2;
case '*':
case '/':
return 3;
}
}
int main()
{
char infx[50], pofx[50], ch, elem;
int i = 0, k = 0;
printf("\n\nRead the Infix Expression ?");
scanf("%s", infx);
push('#');
while ((ch = infx[i++]) != '\0') {
if (ch=='(')
push(ch);
else if (isalnum(ch))
pofx[k++] = ch;
else if (ch == ')') {
while (s[top] != '(')
pofx[k++] = pop();
elem = pop();
} else {
while (pr(s[top]) >= pr(ch))
pofx[k++] = pop();
push(ch);
}
}
while (s[top] != '#')
pofx[k++] = pop();
pofx[k] = '\0';
printf("\n\nGiven Infix Expn: %s Postfix Expn: %s \n", infx,pofx);
}
Thank you
Well first off, I wasn't able to compile your code due to this line
int top = -1;
void push(char elem) {
s[++top] = elem;
}
It's very dangerous to initialize your arrays using a variable that's negative. When we go to pop() the character symbols in the array, it will return to its initial state at top = -1 which causes the compilation error. My suggestion is to change your code around to something like this, which will lead you to change somethings around with placing characters in your array:
int top = 0;
void push(char elem) {
s[top++] = elem;
}
As for printing your strings with spaces in between each character use something along the lines of:
for (int i = 0, i < stren(infix) + 1; i++)
{
printf ("%c ", infix[i]);
}
// Same goes for the postfix. More of a formatting of the print characters really...
static void collatz(int i)
{
int x=0,a=0,res=0,count=0;
int array[50];
array[0]=i;
while(array[count]!=0)
{
if(array[count]%2==0)
{
count++;
array[count]=i/2;
}
else
{
count++;
array[count]=3*array[count-1]-1;
}
}
}
int main()
{
int a;
scanf("%d",&a);
collatz(a);
system("pause");
return 0;
}
When I compile and run the code, I enter 8 as "a" and console crushes itself. I'm using dev c.
Sorry for my awful english but I hope I'm clear enough.
I count four errors:
The loop should terminate when array[count] is 1, not 0
You should check that count is less than 49
array[count]=i/2 should be array[count]=array[count-1]/2
array[count]=3*array[count-1]-1 should be array[count]=3*array[count-1]+1
So the code with these issues fixed, and some slight shortening, could be:
static void collatz(int i) {
int count=0;
int array[50];
array[0]=i;
while (count < 49) {
int j = array[count++];
if (j == 1)
break;
else if(j % 2 == 0)
array[count]=j/2;
else
array[count]=3*j+1;
}
}
first, the while loop is wrong, you should end the while when array[count] == 1, so you can use while(array[count] > 1) to test.
second, array[count]=i/2; is wrong, you shuld use array[count]=array[count-1]/2;.
Third, you should check the count < 50, because you declare int array[50];.
static void collatz(int i)
{
int x=0,a=0,res=0,count=0;
int array[50]= {0};
array[0]=i;
while(array[count] > 1 && count < 50)
{
if(array[count]%2==0)
{
count++;
array[count]=array[count-1]/2;
}
else
{
count++;
array[count]=3*array[count-1]-1;
}
}
}