I have question that asks how values such as c are computed in terms of binary numbers. Im researching it but now but figured id ask here if anyone has somewhere they can send me or explain how this works.
int main()
{
int a 10, int b = 12, int c, int d;
int c = a << 2; //output 40
}
Well, I'm not answering with C++ code, as the question is not really related to the language.
The integer ten is written 10 in base 10 as it's equal to 1 * 10^1 + 0 * 10^0.
Binary is base 2, so let's try to write ten as a sum of powers of 2.
10 = 8 + 2
That is 2^3 + 2^1.
Let's switch to binary (using only two digits : 0 and 1).
2^3 is written 1000
2^1 is written 10
Their sum is 1010 in binary.
"<<" is the operation that shift left binary digits by a certain amount (beware of overflow).
So 1010 << 2 is 101000
That is in decimal 2^5 + 2^3 = 32 + 8 = 40
You can also think of "<< N" as being a multiplication by 2^N of an integer.
Related
I encountered a problem with bit arithmetic. It is bitwise NOT.
if A = 5; then ~A = ?
The binary of 5 is 101, the inverse is 010, and then converted to decimal is 0 * 2^2 + 1 * 2^1 + 0 * 2^0 = 2
But when I test in the IDE, the output is as follows:
System.out.println( ~5 );
Output:
-6
I don't know why. Thanks!!!
If you using a standard int, then after assignment your A to 5:
int A = 5;
Then your "A" would be not 101b, but 00000000000000000000000000000101b - all 32 bits.
After NEG operation, which inverse all bits, you will get:
A = 11111111111111111111111111111010
And this int-value is -6, in the 2-complement representation, used int the most of computers.
Is there efficient way to downscale number of elements in array by decimal factor?
I want to downsize elements from one array by certain factor.
Example:
If I have 10 elements and need to scale down by factor 2.
1 2 3 4 5 6 7 8 9 10
scaled to
1.5 3.5 5.5 7.5 9.5
Grouping 2 by 2 and use arithmetic mean.
My problem is what if I need to downsize array with 10 elements to 6 elements? In theory I should group 1.6 elements and find their arithmetic mean, but how to do that?
Before suggesting a solution, let's define "downsize" in a more formal way. I would suggest this definition:
Downsizing starts with an array a[N] and produces an array b[M] such that the following is true:
M <= N - otherwise it would be upsizing, not downsizing
SUM(b) = (M/N) * SUM(a) - The sum is reduced proportionally to the number of elements
Elements of a participate in computation of b in the order of their occurrence in a
Let's consider your example of downsizing 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 to six elements. The total for your array is 55, so the total for the new array would be (6/10)*55 = 33. We can achieve this total in two steps:
Walk the array a totaling its elements until we've reached the integer part of N/M fraction (it must be an improper fraction by rule 1 above)
Let's say that a[i] was the last element of a that we could take as a whole in the current iteration. Take the fraction of a[i+1] equal to the fractional part of N/M
Continue to the next number starting with the remaining fraction of a[i+1]
Once you are done, your array b would contain M numbers totaling to SUM(a). Walk the array once more, and scale the result by N/M.
Here is how it works with your example:
b[0] = a[0] + (2/3)*a[1] = 2.33333
b[1] = (1/3)*a[1] + a[2] + (1/3)*a[3] = 5
b[2] = (2/3)*a[3] + a[4] = 7.66666
b[3] = a[5] + (2/3)*a[6] = 10.6666
b[4] = (1/3)*a[6] + a[7] + (1/3)*a[8] = 13.3333
b[5] = (2/3)*a[8] + a[9] = 16
--------
Total = 55
Scaling down by 6/10 produces the final result:
1.4 3 4.6 6.4 8 9.6 (Total = 33)
Here is a simple implementation in C++:
double need = ((double)a.size()) / b.size();
double have = 0;
size_t pos = 0;
for (size_t i = 0 ; i != a.size() ; i++) {
if (need >= have+1) {
b[pos] += a[i];
have++;
} else {
double frac = (need-have); // frac is less than 1 because of the "if" condition
b[pos++] += frac * a[i]; // frac of a[i] goes to current element of b
have = 1 - frac;
b[pos] += have * a[i]; // (1-frac) of a[i] goes to the next position of b
}
}
for (size_t i = 0 ; i != b.size() ; i++) {
b[i] /= need;
}
Demo.
You will need to resort to some form of interpolation, as the number of elements to average isn't integer.
You can consider computing the prefix sum of the array, i.e.
0 1 2 3 4 5 6 7 8 9
1 2 3 4 5 6 7 8 9 10
yields by summation
0 1 2 3 4 5 6 7 8 9
1 3 6 10 15 21 28 36 45 55
Then perform linear interpolation to get the intermediate values that you are lacking, like at 0*, 10/6, 20/6, 30/5*, 40/6, 50/6, 60/6*. (Those with an asterisk are readily available).
0 1 10/6 2 3 20/6 4 5 6 40/6 7 8 50/6 9
1 3 15/3 6 10 35/3 15 21 28 100/3 36 45 145/3 55
Now you get fractional sums by subtracting values in pairs. The first average is
(15/3-1)/(10/6) = 12/5
I can't think of anything in the C++ library that will crank out something like this, all fully cooked and ready to go.
So you'll have to, pretty much, roll up your sleeves and go to work. At this point, the question of what's the "efficient" way of doing it boils down to its very basics. Which means:
1) Calculate how big the output array should be. Based on the description of the issue, you should be able to make that calculation even before looking at the values in the input array. You know the input array's size(), you can calculate the size() of the destination array.
2) So, you resize() the destination array up front. Now, you no longer need to worry about the time wasted in growing the size of the dynamic output array, incrementally, as you go through the input array, making your calculations.
3) So what's left is the actual work: iterating over the input array, and calculating the downsized values.
auto b=input_array.begin();
auto e=input_array.end();
auto p=output_array.begin();
Don't see many other options here, besides brute force iteration and calculations. Iterate from b to e, getting your samples, calculating each downsized value, and saving the resulting value into *p++.
How do I find max and min values that can be represented with
a 5-digit number that is in base 13 assuming only positive integers
are represented? then the answer needs to be in base 10.
does 5 digit number mean 5bits? Isn't the smallest number that
can be represented a zero and largest is 2^(N-1)?
This sounds like homework, but I'll bite anyway :)
5 digits probably means 5 digits, as in 12345.
Base 13 means there are 13 possible digits where we as humans are used to calculate with 10.
We could represent the extra 3 digits with A, B, C so that the full range of possible digits is 0123456789ABC. With this representation, it's clear that the smallest 5-digit value is 00000 and the largest CCCCC.
To convert CCCCC in base 13 to base 10 you do
((((C * 13) + C) * 13 + C) * 13 + C) * 13 + C
=
((((12 * 13) + 12 ) * 13 + 12 ) * 13 + 12 ) * 13 + 12
=
371,292
00000 is of course zero in any base.
I am trying to figure out how to solve this problem (Project Euler):
n! means n × (n − 1) × ... × 3 × 2 × 1
For example, 10! = 10 × 9 × ... × 3 × 2 × 1 = 3628800, and the sum of
the digits in the number 10! is 3 + 6 + 2 + 8 + 8 + 0 + 0 = 27.
Find the sum of the digits in the number 100!
Using BigInt is not an option, I am trying to figure out how to implement a solution using only c++.
I thought maybe splitting the big numbers to an array of about 7 digit's long or something like that and then deal with them but i still cant figure out how to do this either..
thanks in advance!
Try this
#include "iostream"
#include "vector"
int n,remainder,sum;
int main ()
{
std::vector <int> digits(5000);
std::cin>>n;
digits[0]=1;
digits[1]=1;
for (int k=2;k<n+1;k++) {
for (int i=1;i<=digits[0];i++) {
digits[i]=digits[i]*k+remainder;
remainder=0;
if (digits[i]>9) {
remainder=digits[i]/10;
digits[i]%=10;
if (i==digits[0])
digits[0]++;
}
}
}
for (int i=digits[0];i>=1;i--)
sum+=digits[i];
std::cout<<sum;
}
I have a code in C++ which convert 2 digits octal number to binary number. For testing validity of the code I used several online conversion site like
this and
this
When I enter 58 or 59 in as an octal value it says invalid octal values but when I enter 58 in my code it gives binary number as - 101000. Again for testing I enter 101000 as binary number in above website's calculator then they gives me result 50 as octal value.
I need some explanation why this is so.
Here is the C++ code -
#include <iostream.h>
#include <conio.h>
void octobin(int);
void main()
{
clrscr();
int a;
cout << "Enter a 2-digit octal number : ";
cin>>a;
octobin(a);
getch();
}
void octobin(int oct)
{
long bnum=0;
int A[6];
//Each octal digit is converted into 3 bits, 2 octal digits = 6 bits.
int a1,a2,quo,rem;
a2=oct/10;
a1=oct-a2*10;
for(int x=0;x<6;x++)
{
A[x]=0;
}
//Storing the remainders of the one's octal digit in the array.
for (x=0;x<3;x++)
{
quo=a1/2;
rem=a1%2;
A[x]=rem;
a1=quo;
}
//Storing the remainders of the ten's octal digit in the array.
for(x=3;x<6;x++)
{
quo=a2/2;
rem=a2%2;
A[x]=rem;
a2=quo;
}
//Obtaining the binary number from the remainders.
for(x=x-1;x>=0;x--)
{
bnum*=10;
bnum+=A[x];
}
cout << "The binary number for the octal number " << oct << " is " << bnum << "." << endl;
}
Octal numbers have digits that are all in the range [0,7]. Thus, 58 and 59 are not octal numbers, and your method should be expected to give erroneous results.
The reason that 58 evaluates to 101000 is because the first digit of the octal number expands to the first three digits of the binary number. 5 = 101_2. Same story for the second part, but 8 = 1000_2, so you only get the 000 part.
An alternate explanation is that 8 = 0 (mod 8) (I am using the = sign for congruency here), so both 8 and 0 will evaluate to 000 in binary using your code.
The best solution would be to do some input validation. For example, while converting you could check to make sure the digit is in the range [0,7]
You cannot use 58 or 59 as an input value. It's octal, for Christ's sake.
Valid digits are from 0 to 7 inclusive.
If you're encoding a number in base 8, none of the octets can be 8 or greater. If you're going to do the code octet by octet, there needs to be a test to see whether the octet is 8 or 9, and to throw an error. Right now your code isn't checking this so the 8 and 9 are overflowing to 10.
58 and 59 aren't valid octal values indeed ... the maximum digit you can use is yourbase-1 :
decimal => base = 10 => Digits from 0 t 9
hexadécimal => base = 16 => Digits from 0 to 15 (well, 0 to F)
Octal => base = 8 => Digits from 0 to 7