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that displays exactly 8 (eight) Fibonacci numbers starting and ending from user-specified numbers (program’s input). For example, if a user inputs index 3 and 10, then numbers (values) F3 - F10 are shown on the screen. Erroneous user’s input (e.g. negative number) or a smaller ending number than the first, should lead to warning and automatic repetition of the input.
To give you a small hint without doing all the work for you (as this seems to be some task for school, college, or university), here's how a Fibonacci number is defined:
f(0) = 0;
f(1) = 1;
f(n) = f(n - 1) + f(n - 2);
So in C++ this could be written like this:
int fibonacci(int n) {
if (n == 0)
return 0;
if (n == 1)
return 1;
return fibonacci(n - 1) + fibonacci(n - 2);
}
This of course can be further optimized and it's not necessarily the best approach. And it also includes possible errors, that might lead to stack overflows (hey, isn't that what this site is about? :)). So try to understand the code, then try to learn and improve it. Don't just copy & paste.
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I want to know about what is the use of that (index & 0x01) in the code?
if(((arr[index] >= 0) && (!(index & 0x01)))
|| ((arr[index] < 0) && (index & 0x01)))
{
outofplace = index;
}
A number is odd if and only if its last digit is odd, regardless of the base. So if we want to know the number's oddity, it's enough to check if the last bit is set.
index & 0x01
will be 1 if and only if index is odd.
If we have to deduce a general rule, we can say that for any non-negative number x,
x % y == (x & (y - 1))
provided that y is a positive power of 2.
This is a common hack in competitive coding. It is used because the competitive programmers think that bit-wise AND works faster than modulo.
In modern compilers, there is no performance difference at all. Read this thread.
There is no special reason in writing it as 0x01 instead of 1. Both compile to give the same assembly! Almost everyone (who uses this hack),= uses 1, because we have to type 3 characters extra in 0x01. :P
Here in this case, index & 0x1 is equivalent to index % 2 which is simply a condition to check if the number is odd. (Array indexes in C++ are always positive, unless you are going out of bound.)
As the other answers pointed out, while this is a well known pattern (see also this Q&A about that mask), it can be considered a premature optimization.
I'd like to suggest the following alternative to the posted code, which I find more readable. Your mileage may vary.
// Give a meaningful name.
constexpr auto is_odd = [] (auto x) -> bool {
return x % 2;
}
// Use it to simplify the condition.
if ( (arr[index] < 0) == is_odd(index) ) {
// Do something
}
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I am trying to figure out the following problem for an upcoming test. I have searched everywhere, and I understand the basics of recursion. What I don't understand for this particular question is the value of int n and int k respectively. I have the answer to this question as it is a practice question, but I have no idea how the answer was found.
// Precondition: n and k are non-negative integers
int f(int n, int k) {
if (k * n == 0)
return 1
else
return f(n - 1, k - 1) + f(n - 1, k)
}
What value is returned by the call f(4, 2)?
Just look at how it's called.
f(4,2) goes into 2nd block, calls f(3,1)+f(3,2)
f(3,1) calls f(2,0)+f(2,1) = 1+f(1,0)+f(1,1)=1+1+f(0,0)+f(0,1)=1+1+1+1=4
f(3,2) calls f(2,1)+f(2,2)= f(1,0)+f(1,1)+f(1,1)+f(1,2) and so on.
You should be able to work it out from here.
I am not sure what the problem is since
f(4,2)=f(3,1) + f(3,2)
=(f(2,0)+f(2,1) )+ (f(2,1) +f(2,2))
=(1 +(f(1,0)+f(1,1))+((f(1,0)+f(1,1))+(f(1,1)+f(1,2))
=(1 + 1 +(1+1)) +( 1 +(1+1) + (1+1) +1 + 1 ))
=11
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I want to make a function which takes an integer with the number like 113, and separates the one's digit "3" and and the hundreds and tens places "11" and returns the both of them in two separate integers.
x%10 for the first digit (from right) and x/10 for the rest.
#include <iostream>
#include <utility>
std::pair<int,int> split(int x)
{
return std::make_pair(x/10, x%10);
}
int main()
{
std::pair<int,int> z = split(113);
std::cout << z.first << " " << z.second;
}
I also used std::pair to return the result.
You want N % 10 to get the one's digit. For the other digits N / 10.
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I'm trying to convert the individual contents of a string to integers. I need to take each character from the string and convert it to an integer to add to another. This is not using C++11. Is there a simple way to do it?
if the characters are numbers then the numeral value of each is
num_value(c) = c - '0'
This is only possible because the characters representing numbers are in order in the ASCII table.. All you have to do is loop across the string.
"I need to take each character from the string and convert it to an integer to add to another"
In case you want to calculate the sum of digits stored in std::string object, you could do:
std::string myNum("567632");
int sum = 0;
for (size_t i = 0; i < myNum.size(); ++i)
sum += (myNum[i] - '0');
std::cout << sum;
which outputs 29 (i.e. 5 + 6 + 7 + 6 + 3 + 2)
How about std::accumulate ?
#include<string>
#include<algorithm>
//...
std::string myNum("123456789");
std::cout<<accumulate( myNum.begin(), myNum.end(), 0,
[](int sum,const char& x){return sum+=x-'0'; });
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I would like to know whether there is a way in which we can print the number alphabetically i.e
123 should be printed as one two three.
The only condition is that we should not reverse the number and we should not use array.
I only know these two ways:
"Reverse the number", that is, taking the last digit and cutting it off. For each cut-off digit, one can use an array to look up the correct string.
using switch and a lot of cases
Any ideas?
for hundreds place:
int hundreds = my_num / 100 //Needs "/", NOT "%"
if(hundreds == 0)
cout << "zero";
else if(hundreds == 1)
cout << "one";
//repeat for 2-9
This process could be tweaked to do the other digits as well. It is also worth mentioning that the if/else block a) could be done with a switch/case if preferred, and b) could pretty easily be made into a separate function to avoid having to repeat the block of code over and over, I just wrote out as much as I did for clarity's sake. Note that this assumes the number you're "translating" is an integer. With integers the "/" operator will return the full quotient WITHOUT the remainder, e.g. 123 / 100 = 1, not 1.23
Not necessarily the easiest route, but you can make a function, say DigitToWord which will take a digit 0, 1, 2, ...etc to its word with a switch statement. Then I recommend using a for loop over the number, continuously dividing by 10 and taking the mod for the loop:
int num; //my number i want to print
int div = pow(10, (int)log10(num)); //find the largest power of 10 smaller than num
while(num > 0) {
int remainder = num%div;
int digit = num/div;
DigitToWord();
num = remainder;
}