I am trying to solve problem PLD on SPOJ, but I'm getting a WA on the 9th testcase.
My Approach:
I am implementing Manacher's Algorithm and I believe that if something wrong is there, then it can be wrong in this code.
if((k%2==0)&&(p[i]>=k)&&(temp[i]=='#'))
count++;
if((k%2==1)&&(p[i]>=k)&&(temp[i]!='#'))
count++;
But according to my approach if character is #, then the maximum length of palindromic string centered at it can be even only, so if p[i] >= k, then I am increasing count if we are finding a palindromic string of even length.
Similarly for characters [considering input character i.e other than #] centered at i-th location but for odd length strings.
#include<stdio.h>
#include<string.h>
char a[30002],temp[60010];
int p[60010];
int min(int a,int b)
{
if(a<b)
return a;
return b;
}
int main()
{
//freopen("input.txt","r+",stdin);
//freopen("a.txt","w+",stdout);
int k,len,z;
scanf("%d",&k);
getchar();
gets(a);
len=strlen(a);
//Coverting String
temp[0]='$';
temp[1]='#';
z=2;
for(int i=1;i<=len;i++)
{
temp[z++]=a[i-1];
temp[z++]='#';
}
len=z;
int r=0,c=0,check=0,idash,t,count=0;
for(int i=1;i<len;i++)
{
check=0;
idash=c-(i-c);
p[i]=r>i?min(r-i,p[idash]):0;
t=p[i];
while(temp[i+p[i]+1]==temp[i-1-p[i]])
p[i]++;
if(r<i+p[i])
{
c=i;
r=i+p[i];
}
if((k%2==0)&&(p[i]>=k)&&(temp[i]=='#'))
count++;
if((k%2==1)&&(p[i]>=k)&&(temp[i]!='#'))
count++;
}
printf("%d",count);
//getchar();
//getchar();
return 0;
}
You may want to take advantage of C++ short-circuit evaluation of logical expressions.
For example, rearrange the order so you check for '#' first:
if ((temp[i] == '#') && (k % 2 == 0) && (p[i] >= k))
In the above rearrangement, if the character is not '#', none of the other expressions are evaluated.
You may want to extract (p[i] >= k) to an outside if statement since it is common to both:
if (p[i] >= k)
{
if ((temp[i] == '#') && (k % 2 == 0)) ++count;
if ((temp[i] != '#') && (k % 2 == 1)) ++count;
}
The above modification will result in only one evaluation of the expression (p[i] >= k).
Also examine your for loop to see if there are statements or expressions that don't change or are repeated. If a statement or expression doesn't change inside the loop, it is called an invariant, and can be moved before or after the loop.
Statements or expressions that are duplicated (such as array index calculations) can be evaluated and stored in a temporary variable. Although good compilers may do this (depending on the optimization level), in your performance requirements, you may want to help out the compiler.
Another suggestion is to replace p[i] with a pointer to the location or a reference to the location. Again, this is to help out the compiler when the optimization is not set optimally:
int& p_slot_i = p[i]; // This syntax needs checking
// or
int * p_slot_i = &p[i];
//...
t = *p_slot_i;
while(temp[i + *p_slot_i + 1] == temp[i - 1 - *p_slot_i)
{
*p_slot_i++;
}
Lastly, Elimination of spaces, blank lines and curly braces DOES NOT AFFECT PROGRAM PERFORMANCE. A program that is one line or spaced across multiply lines will have the exact assembly translation and the exact performance. So please, add spaces, blank lines and curly braces to improve readability.
Edit 1: performance of min()
You may want to declare you min() function as inline to suggest to the compiler you want the function pasted where it is called, rather than calling the function. Function calls slow down a programs execution.
Related
#include <iostream>
int GCD()
{
int a,b,k;
cout<<"Enter a and b"<<endl;
cin>>a>>b;
cout<<endl;
if (a>b)
{
k=a;
}
else
{
k=b;
}
cout<<k<<endl;
do
{
k=k-1;
} while(a%k !=0 && b%k !=0);
cout<<k<<endl;
return 0;
}
Why programm like this doesnt work correctly? For example when i enter 125 and 5 answer is 25, but supposed to be 5? Am wrong with logic in while loop? As i understood problem is in modulus operator. When k hits 25 it says that 125%25=0 and 5%=25=0. How can i fix this?
You have some mistakes here:
The GCD is lower or equal to the lower number. Currently, you start checking with the larger number. You need to flip the if block to if (a<b). (not exactly an error, but you check much more numbers than needed)
You need to check if the inital k is the GCD. When using a do {} while() the first number you check is k-1. Use a simple while instead. Also the loop condition has a logic flaw.
while (!((a % k == 0) && (b % k == 0)))
{
k--;
}
Note that the brackets around the modulo are not neccessary, but improve readability a bit.
Your code will not compile under all compilers and you should not omit the namespace std::.
Your while statement has logic error. It needs to be
while(!(a%k == 0 && b%k == 0));
When k is equal to 25, 125%25==0 so in your while statement a%k !=0 part is equals to false so it exit your do-while but it needs to test if b%k is equal to 0 or not!
Also your implementation tends to execute slow when a and b is big. You can take a look efficent solutions.
This question already has answers here:
If statements without brackets
(3 answers)
Closed 10 months ago.
Does the indentation in an if else statement have any bearing on the execution of the code or is it just something to do for cleaner code?
Example from the book Accelerated C++ written by Andrew Koening:
while(c != cols) {
if(r == pad + 1 && c == pad + 1) {
cout << greet;
c += greet.size();
} else {
if(r == 0 || r == rows - 1 || c == 0 || c == cols - 1)
cout << "*";
else
cout << " ";
++c;
}
}
The prefix increment of c is executed regardless of whether r=0 or not, but I don’t understand why.
If the if statement turns true, an asterisk is printed. If not, a blank space is printed and c is incremented.
That’s how I am reading it, but c gets incremented regardless of what the values of r or c are.
This is what it says in the book, but there isn’t any explanation I could find:
Notice how the different indentation of
++c;
draws attention to the fact that it is executed regardless of whether we are in the
border.
Whitespace does not affect C++ runtime behavior. (Unlike certain other languages, like Python).
I should mention that in your else block, you do not use braces. So, only the first statement (cout << " ";) will be part of the else clause. The subsequent ++c; will execute regardless of the value of r and c.
Note that this last point is subjective, so take it with a grain of salt... As you can see, when braces are omitted from if ... else ... blocks, there is potential for confusion. Some would argue that it leads to more concise code, but many (including myself) would argue that you should always use braces. This is especially important when you work on a large team because code tends to grow over time. I've seen many cases in production code where an if statement was missing the braces and someone added a second line to the if clause without remembering to add braces. This didn't work as expected and caused wasted time debugging fails, just because the braces were omitted.
Both C and C++ are not affected by white space in their interpretation of the code. That does not mean the programmer should not care about its misuse.
The best way to illustrate what the above code actually represents is to explicitly define all of the inferred braces as below. Note that the if statement that had no braces only has one line of code affected by the 'if then' or 'else' clause.
This is one of the reasons that people try to insist on
'good coding practices' to ensure that other people are able to clearly interpret the flow and intent of the programmer.
while(c != cols) {
if(r == pad + 1 && c == pad + 1) {
cout << greet;
c += greet.size();
} else {
if(r == 0 || r == rows - 1 || c == 0 || c == cols - 1) {
cout << "*";
} else {
cout << " ";
}
++c;
}
}
In C++ the length of indentation does not affect the interpretation of the statements. Sometimes whitespace is needed to separate characters, e.g., in int a. Other times it is not needed, e.g. in a=b+c;.
The if statement is defined that after the condition if(condition) can only be one statement.
If we want more statements we have to group them with braces {...}
Unlike Python, C++ does not care about indentation.
But your else applies only on the first line. To apply to a block, it should be within { }
else
{
cout << " ";
++c;
}
Indentation is not your problem here.
int trees = 3;
int tree_x, tree_y;
for(int r = 0; r < m_townsize; r++)
{
for(int c = 0; c < m_townsize; c++)
{
if(r == 0 || c == 0 || r == (m_townsize - 1) || c == (m_townsize - 1))
m_town[r][c] = 'W';
while(trees > 0)
{
tree_x = random() % m_townsize;
tree_y = random() % m_townsize;
cout << tree_y << "," << tree_x << endl;
if(m_town[tree_y][tree_x] == ' ')
{
m_town[tree_y][tree_x] = 'T';
trees -= 1;
}
}
}
}
According the code I have written, if there is a space character at the coordinate of the tree, it should place a tree and lower the tree count by 1.
If there is not a space there, it should skip placing a tree, thus not decrementing. This should cause it to pick another set of coordinates and run through again.
However, if you look at this particular output it is running to the if-statement skipping the first option to replace it with a T--since it is a W--but still decrementing by 1. I don't get it. It should skip the statement all together, not skip just the first line. Netbeans tells me my brackets are right, so it shouldn't be an issue with the assignment belonging to the if and the decrement belonging to the while.
If I make a do-while loop it places a whole bunch. I don't know what's happening.
This output placed 2 trees.
You are walking over each coordinate.
If it is on the edge you put a 'W'. Then you randomly place a tree 'T'.
Then you proceed to the next coordinate.
This means you can place some trees in squares before you overwrite with a 'W'.
Finish all the walls before placing trees. Consider a more efficient way to place walls to, like doing each edge instead of loopimg over the middle abd doing nothing.
This is a piece of code I found in my textbook for using recursion to evaluate prefix expressions. I'm having trouble understanding this code and the process in which it goes through.
char *a; int i;
int eval()
{ int x = 0;
while (a[i] == ' ') i++;
if (a[i] == '+')
{ i++; return eval() + eval(); }
if (a[i] == '*')
{ i++; return eval() * eval(); }
while ((a[i] >= '0') && (a[i] <= '9'))
x = 10*x + (a[i++] - '0');
return x;
}
I guess I'm confused primarily with the return statements and how it eventually leads to solving a prefix expression. Thanks in advance!
The best way to understand recursive examples is to work through an example :
char* a = "+11 4"
first off, i is initialized to 0 because there is no default initializer. i is also global, so updates to it will affect all calls of eval().
i = 0, a[i] = '+'
there are no leading spaces, so the first while loop condition fails. The first if statement succeeds, i is incremented to 1 and eval() + eval() is executed. We'll evaluate these one at a time, and then come back after we have our results.
i = 1, a[1] = '1'
Again, no leading spaces, so the first while loop fails. The first and second if statements fail. In the last while loop, '1' is between 0 and 9(based on ascii value), so x becomes 0 + a[1] - '0', or 0 + 1 = 1. Important here is that i is incremented after a[i] is read, then i is incremented. The next iteration of the while loop adds to x. Here x = 10 * 1 + a[2] - '0', or 10 + 1 = 11. With the correct value of x, we can exit eval() and return the result of the first operand, again here 11.
i = 2, a[2] = '4'
As in the previous step, the only statement executed in this call of eval() is the last while loop. x = 0 + a[2] - '0', or 0 + 4 = 4. So we return 4.
At this point the control flow returns back to the original call to eval(), and now we have both values for the operands. We simply perform the addition to get 11 + 4 = 15, then return the result.
Every time eval() is called, it computes the value of the immediate next expression starting at position i, and returns that value.
Within eval:
The first while loop is just to ignore all the spaces.
Then there are 3 cases:
(a) Evaluate expressions starting with a + (i.e. An expression of the form A+B which is "+ A B" in prefix
(b) Evaluate expressions starting with a * (i.e. A*B = "* A B")
(c) Evaluate integer values (i.e. Any consecutive sequence of digits)
The while loop at the end takes care of case (c).
The code for case (a) is similar to that for case (b). Think about case (a):
If we encounter a + sign, it means we need to add the next two "things" we find in the sequence. The "things" might be numbers, or may themselves be expressions to be evaluated (such as X+Y or X*Y).
In order to get what these "things" are, the function eval() is called with an updated value of i. Each call to eval() will fetch the value of the immediate next expression, and update position i.
Thus, 2 successive calls to eval() obtain the values of the 2 following expressions.
We then apply the + operator to the 2 values, and return the result.
It will help to work through an example such as "+ * 2 3 * 4 5", which is prefix notation for (2*3)+(4*5).
So this piece of code can only eat +, *, spaces and numbers. It is supposed to eat one command which can be one of:
- + <op1> <op2>
- * <op1> <op2>
<number>
It gets a pointer to a string, and a reading position which is incremented as the program goes along that string.
char *a; int i;
int eval()
{ int x = 0;
while (a[i] == ' ') i++; // it eats all spaces
if (a[i] == '+')
/* if the program encounters '+', two operands are expected next.
The reading position i already points just before the place
from which you have to start reading the next operand
(which is what first eval() call will do).
After the first eval() is finished,
the reading position is moved to the begin of the second operand,
which will be read during the second eval() call. */
{ i++; return eval() + eval(); }
if (a[i] == '*') // exactly the same, but for '*' operation.
{ i++; return eval() * eval(); }
while ((a[i] >= '0') && (a[i] <= '9')) // here it eats all digit until something else is encountered.
x = 10*x + (a[i++] - '0'); // every time the new digit is read, it multiplies the previously obtained number by 10 and adds the new digit.
return x;
// base case: returning the number. Note that the reading position already moved past it.
}
The example you are given uses a couple of global variables. They persist outside of the function's scope and must be initialized before calling the function.
i should be initialized to 0 so that you start at the beginning of the string, and the prefix expression is the string in a.
the operator is your prefix and so should be your first non-blank character, if you start with a number (string of numbers) you are done, that is the result.
example: a = " + 15 450"
eval() finds '+' at i = 1
calls eval()
which finds '1' at i = 3 and then '5'
calculates x = 1 x 10 + 5
returns 15
calls eval()
which finds '4' at i = 6 and then '5' and then '0'
calclulates x = ((4 x 10) + 5) x 10) + 0
returns 450
calculates the '+' operator of 15 and 450
returns 465
The returns are either a value found or the result of an operator and the succeeding results found. So recursively, the function successively looks through the input string and performs the operations until either the string ends or an invalid character is found.
Rather than breaking up code into chunks and so on, i'll try and just explain the concept it as simple as possible.
The eval function always skips spaces so that it points to either a number character ('0'->'9'), an addition ('+') or a multiply ('*') at the current place in the expression string.
If it encounters a number, it proceeds to continue to eat the number digits, until it reaches a non-number digit returning the total result in integer format.
If it encounters operator ('+' and '*') it requires two integers, so eval calls itself twice to get the next two numbers from the expression string and returns that result as an integer.
One hair in the soup may be evaluation order, cf. https://www.securecoding.cert.org/confluence/display/seccode/EXP10-C.+Do+not+depend+on+the+order+of+evaluation+of+subexpressions+or+the+order+in+which+side+effects+take+place.
It is not specified which eval in "eval() + eval()" is, well, evaluated first. That's ok for commutative operators but will fail for - or /, because eval() as a side effect advances the global position counter so that the (in time) second eval gets the (in space) second expression. But that may well be the (in space) first eval.
I think the fix is easy; assign to a temp and compute with that:
if (a[i] == '-')
{ i++; int tmp = eval(); return tmp - eval(); }
I am trying to make a program generating random numbers until it finds a predefined set of numbers (eg. if I had a set of my 5 favourite numbers, how many times would I need to play for the computer to randomly find the same numbers). I have written a simple program but don't understand the outcome which seems to be slightly unrelated to what I expected, for example the outcome does not necessarily contain all of the predefined numbers sometimes it does (and even that doesn't stop the loop from running). I think that the problem lies in the logical operator '&&' but am not sure. Here is the code:
const int one = 1;
const int two = 2;
const int three = 3;
using namespace std;
int main()
{
int first, second, third;
int i = 0;
time_t seconds;
time(&seconds);
srand ((unsigned int) seconds);
do
{
first = rand() % 10 + 1;
second = rand() % 10 + 1;
third = rand() % 10 + 1;
i++;
cout << first<<","<<second<<","<<third<< endl;
cout <<i<<endl;
} while (first != one && second != two && third != three);
return 0;
}
and here is out of the possible outcomes:
3,10,4
1 // itineration variable
7,10,4
2
4,4,6
3
3,5,6
4
7,1,8
5
5,4,2
6
2,5,7
7
2,4,7
8
8,4,9
9
7,4,4
10
8,6,5
11
3,2,7
12
I have also noticed that If I use the || operator instead of && the loop will execute until it finds the exact numbers respecting the order in which the variables were set (here: 1,2,3). This is better however what shall I do make the loop stop even if the order is not the same, only the numbers? Thanks for your answers and help.
The issue is here in your condition:
} while (first != one && second != two && third != three);
You continue while none of them is equal. But once at least one of them is equal, you stop/leave the loop.
To fix this, use logical or (||) rather than a logical and (&&) to link the tests:
} while (first != one || second != two || third != three);
Now it will continue as long as any of them doesn't match.
Edit - for a more advanced comparison:
I'll be using a simple macro to make it easier to read:
#define isoneof(x,a,b,c) ((x) == (a) || (x) == (b) || (x) == (c))
Note that there are different approaches you could use.
} while(!isoneof(first, one, two, three) || !isoneof(second, one, two, three) || !isoneof(third, one, two, three))
You have a mistake in your logical condition: it means "while all numbers are not equal". To break this condition, it is enough for one pair to become equal.
You needed to construct a different condition - either put "not" in front of it
!(first==one && second==two && third==three)
or convert using De Morgan's law:
first!=one || second!=two || third!=three