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I'm trying to get all prime numbers in the range of 2 and the entered value using this c++ code :
#include<iostream>
using namespace std;
int main() {
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
for (int b = 2; b <= num; b++) {
result = i % b;
if (result == 0) {
result = b;
break;
}
}
cout << result<< endl <<;
}
}
the problem is that I think am getting close to the logic, but those threes and twos keep showing up between the prime numbers. What am I doing wrong?
I've fixed your code and added comments where I did the changes
The key here is to understand that you need to check all the numbers smaller then "i" if one of them dividing "i", if so mark the number as not prime and break (the break is only optimization)
Then print only those who passed the "test" (originally you printed everything)
#include <iostream>
using namespace std;
#include<iostream>
using namespace std;
int main()
{
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
bool isPrime = true; // Assume the number is prime
for (int b = 2; b < i; b++) { // Run only till "i-1" not "num"
result = i % b;
if (result == 0) {
isPrime = false; // if found some dividor, number nut prime
break;
}
}
if (isPrime) // print only primes
cout << i << endl;
}
}
Many answers have been given which explains how to do it. None have answered the question:
What am I doing wrong?
So I'll give that a try.
#include<iostream>
using namespace std;
int main() {
int num = 0;
int result = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
for (int b = 2; b <= num; b++) { // wrong: use b < i instead of b <= num
result = i % b;
if (result == 0) {
result = b; // wrong: why assign result the value of b?
// just remove this line
break;
}
}
cout << result<< endl <<; // wrong: you need a if-condtion before you print
// if (result != 0) cout << i << endl;
}
}
You have multiple errors in your code.
Simplest algorithm (not the most optimal though) is for checking whether N is prim is just to check whether it doesn't have any dividers in range [2; N-1].
Here is working version:
int main() {
int num = 0;
cin >> num;
for (int i = 2; i <= num; i++) {
bool bIsPrime = true;
for (int b = 2; bIsPrime && b < i; b++) {
if (i % b == 0) {
bIsPrime = false;
}
}
if (bIsPrime) {
cout << i << endl;
}
}
}
I would suggest pulling out the logic of determining whether a number is a prime to a separate function, call the function from main and then create output accordingly.
// Declare the function
bool is_prime(int num);
Then, simplify the for loop to:
for (int i = 2; i <= num; i++) {
if ( is_prime(i) )
{
cout << i << " is a prime.\n";
}
}
And then implement is_prime:
bool is_prime(int num)
{
// If the number is even, return true if the number is 2 else false.
if ( num % 2 == 0 )
{
return (num == 2);
}
int stopAt = (int)sqrt(num);
// Start the number to divide by with 3 and increment it by 2.
for (int b = 3; b <= stopAt; b += 2)
{
// If the given number is divisible by b, it is not a prime
if ( num % b == 0 )
{
return false;
}
}
// The given number is not divisible by any of the numbers up to
// sqrt(num). It is a prime
return true;
}
I can pretty much guess its academic task :)
So here the think for prime numbers there are many methods to "get primes bf number" some are better some worse.
Erosthenes Sieve - is one of them, its pretty simple concept, but quite a bit more efficient in case of big numbers (like few milions), since OopsUser version is correct you can try and see for yourself what version is better
void main() {
int upperBound;
cin >> upperBound;
int upperBoundSquareRoot = (int)sqrt((double)upperBound);
bool *isComposite = new bool[upperBound + 1]; // create table
memset(isComposite, 0, sizeof(bool) * (upperBound + 1)); // set all to 0
for (int m = 2; m <= upperBoundSquareRoot; m++) {
if (!isComposite[m]) { // if not prime
cout << m << " ";
for (int k = m * m; k <= upperBound; k += m) // set all multiplies
isComposite[k] = true;
}
}
for (int m = upperBoundSquareRoot; m <= upperBound; m++) // print results
if (!isComposite[m])
cout << m << " ";
delete [] isComposite; // clean table
}
Small note, tho i took simple implementation code for Sive from here (writing this note so its not illegal, truth be told wanted to show its easy to find)
I'm writing a program that adds two large integers (up to 20 digits) together. I've had no problems so far in storing the two numbers as strings then sorting them into two arrays.
So far, I have half of the addition part working. When the sum of the two digits does not exceed double digits, it works fine.
The issue arises when the sum of the arrays hits double digits. I'm trying to work in the carry over, but it messes with the digits (adding in where it shouldn't.) In addition to that, I'm not sure how to get the carry to appear ahead of the final digits. For example: 9+9 outputs to 8.
Here's my code (please excuse all the letter variables in the for loops.)
#include <iostream>
#include <string>
using namespace std;
int main()
{
string str1;
string str2;
int array1[20];
int array2[20];
int array3[20];
string num3[20];
int i;
int j = 0;
int k;
int l;
int m = 0;
int n;
int o;
int carry = 0;
cout<<"Please enter the first number: "<<endl;
cin>>str1;
for (int i = str1.length() - 1; i >= 0; i--)
{
array1[j] = str1[i];
j++;
}
for (int k = str1.length()-1; k >=0; k--)
{
array1[k] = static_cast<int>(str1[k]) - static_cast<int>('0');
}
cout<<"Please enter the second number: "<<endl;
cin>>str2;
for (int l = str2.length() - 1; l >= 0; l--)
{
array2[m] = str2[l];
m++;
}
for (int n = str2.length()-1; n >=0; n--)
{
array2[n] = static_cast<int>(str2[n]) - static_cast<int>('0');
}
//Where the addition begins
for (int o = 0; o < str1.length(); o++)
{
if (array1[o] + array2[o] > 9)
{
array3[o] = array1[o] + array2[o] + carry;
array3[o] = array3[o] % 10;
carry = 1;
}
else
{
array3[o] = array1[o] + array2[o] + carry;
carry = 0;
}
cout<<array3[o];
}
return 0;
}
I think one thing I have to fix is how this line of code works:
array3[o] = array3[o] % 10;
Which keeps a second digit from appearing in the output. I would imagine if I disabled it once we reach the final numbers in the arrays, it would allow the final carry to show up. Unfortunately, everything I've tried hasn't worked.
Again, thank you!
Try this:
for (int o = 0; o < str1.length(); o++)
{
if (array1[o] + array2[o] + carry > 9)
{
array3[o] = array1[o] + array2[o] + carry;
array3[o] = array3[o] % 10;
carry = 1;
}
else
{
array3[o] = array1[o] + array2[o] + carry;
carry = 0;
}
cout<<array3[o];
}
Modify your for loop for Addition. In condition you need to add carry also
if (array1[o] + array2[o] + carry > 9)
The final for loop will be as below:
for (int o = 0; o < str1.length(); o++)
{
if (array1[o] + array2[o] + carry > 9)
{
array3[o] = array1[o] + array2[o] + carry;
array3[o] = array3[o] % 10;
carry = 1;
}
else
{
array3[o] = array1[o] + array2[o] + carry;
carry = 0;
}
cout<<array3[o];
}
My suggestions:
You can fill up the numbers from the input string in one loop. No need to use two loops.
for (int i = str1.length() - 1; i >= 0; i--)
{
array1[j] = str1[i] - '0';
j++;
}
Similarly for the other loop.
When computing the total you have iterate until the length of the longest string. If the first input is 12 and the second input is 4567, you have to make sure that your iteration stops at 4, not at 2.
The algorithm for computing the sum can be simplified to:
for (int o = 0; o < len+1; o++)
{
array3[o] = array1[o] + array2[o] + carry;
carry = array3[o]/10;
array3[o] %= 10;
}
where len is the maximum of the lengths.
Here's the final code I came up with:
#include <iostream>
#include <string>
using namespace std;
void printNumber(int array[])
{
// Skip the leading zeros.
int i = 19;
for ( ; i >= 0; i-- )
{
if ( array[i] > 0 )
{
break;
}
}
for ( ; i >= 0; i--)
{
cout << array[i];
}
}
int main()
{
string str1;
string str2;
int array1[20] = {0};
int array2[20] = {0};
int array3[20] = {0};
int i;
int j = 0;
int k;
int l;
int m = 0;
int n;
int o;
int carry = 0;
int len = 0;
cout<<"Please enter the first number: "<<endl;
cin>>str1;
len = str1.length();
for (int i = str1.length() - 1; i >= 0; i--)
{
array1[j] = str1[i] - '0';
j++;
}
cout<<"Please enter the second number: "<<endl;
cin>>str2;
if ( len < str2.length() )
{
len = str2.length();
}
for (int l = str2.length() - 1; l >= 0; l--)
{
array2[m] = str2[l] - '0';
m++;
}
//Where the addition begins
for (int o = 0; o < len+1; o++)
{
array3[o] = array1[o] + array2[o] + carry;
carry = array3[o]/10;
array3[o] %= 10;
}
// Print the result.
printNumber(array3);
cout << endl;
return 0;
}
int main()
{
char A[20],B[20],C[22]={0};
int carry,len_a,len_b,x=20,i,j,a,b;
printf("First Number");
gets(A);
printf("Second Number");
gets(B);
len_a=strlen(A);
len_b=strlen(B);
for(i=len_a-1;i>=0;i--)
{
carry=0;
b=(int)B[i]-48;
a=(int)A[len_b-1]-48;
C[x]=C[x]+a+b;
if(C[x]>9)
{
C[x]=C[x]%10;
C[x-1]+=1;
}
x--;
len_b--;
}
int flag=0;
printf("Result :");
for(j=0;j<=20;j++)
{
if(C[j]!=0)
{
printf("%d",C[j]);
flag=1;
}
else if(C[j]==0 && flag==1)
printf("%d",C[j]);
}
if(flag==0)
printf("0");
getch();
return 0;
}
If I were you, would do exactly what I have done here:
inline bigint &bigint::operator+( const bigint & _expr )
{
vector<uint8_t> left = this->_digits;
vector<uint8_t> right = _expr._digits;
vector<uint8_t> sum;
uint8_t carry = 0;
process_operands( left, right ); // makes the two operands have the same length and fills them with leading zeros
for( auto lit = left.cbegin(), rit = right.cbegin(); lit != left.cend(), rit != right.cend(); ++lit, ++rit )
{
uint8_t temp_sum = ( *lit + *rit + carry ) % 10;
carry = ( *lit + *rit + carry ) / 10;
sum.push_back( temp_sum );
}
if( carry ) sum.push_back( carry );
this->_digits = sum;
return *this;
}
To make things look a little bit more clear:
bigint is my class for big integers, and looks something like this:
class bigint
{
private:
vector<uint8_t> _digits;
typedef vector<uint8_t>::size_type size_type;
bigint( vector<uint8_t> & in );
public:
bigint() : _digits() {}
bigint( const string &number );
// ...
};
So you should actually stop using the built-in arrays, since they are error-prone, and because we have better things offered by STL, like std::vector. I am using std::vector<uint8_t> to store the digits of my number, and so, it becomes easier to cycle through the digits: we can use either the range for (for(uint8_t & c : _digits) { }) or the iterators.
Attaching the leading zeros will become easier, since you ony have to do:
_digits.push_back( 0 );
in a for loop.
The program builds and runs, however after entering the first integer and pressing enter then the error pop up box appears, then after pressing ignore and entering the second integer and pressing enter the pop up box appears and after pressing ignore it returns the correct answer. I am at my wits end with this can somebody help me fix the pop up box thing.
#include "stdafx.h"
#include <iostream>
#include <cctype>
#include <string>
using namespace std;
#define numbers 100
class largeintegers {
public:
largeintegers();
void
Input();
void
Output();
largeintegers
operator+(largeintegers);
largeintegers
operator-(largeintegers);
largeintegers
operator*(largeintegers);
int
operator==(largeintegers);
private:
int integer[numbers];
int len;
};
void largeintegers::Output() {
int i;
for (i = len - 1; i >= 0; i--)
cout << integer[i];
}
void largeintegers::Input() {
string in;
int i, j, k;
cout << "Enter any number:";
cin >> in;
for (i = 0; in[i] != '\0'; i++)
;
len = i;
k = 0;
for (j = i - 1; j >= 0; j--)
integer[j] = in[k++] - 48;
}
largeintegers::largeintegers() {
for (int i = 0; i < numbers; i++)
integer[i] = 0;
len = numbers - 1;
}
int largeintegers::operator==(largeintegers op2) {
int i;
if (len < op2.len) return -1;
if (op2.len < len) return 1;
for (i = len - 1; i >= 0; i--)
if (integer[i] < op2.integer[i])
return -1;
else if (op2.integer[i] < integer[i]) return 1;
return 0;
}
largeintegers largeintegers::operator+(largeintegers op2) {
largeintegers temp;
int carry = 0;
int c, i;
if (len > op2.len)
c = len;
else
c = op2.len;
for (i = 0; i < c; i++) {
temp.integer[i] = integer[i] + op2.integer[i] + carry;
if (temp.integer[i] > 9) {
temp.integer[i] %= 10;
carry = 1;
} else
carry = 0;
}
if (carry == 1) {
temp.len = c + 1;
if (temp.len >= numbers)
cout << "***OVERFLOW*****\n";
else
temp.integer[i] = carry;
} else
temp.len = c;
return temp;
}
largeintegers largeintegers::operator-(largeintegers op2) {
largeintegers temp;
int c;
if (len > op2.len)
c = len;
else
c = op2.len;
int borrow = 0;
for (int i = c; i >= 0; i--)
if (borrow == 0) {
if (integer[i] >= op2.integer[i])
temp.integer[i] = integer[i] - op2.integer[i];
else {
borrow = 1;
temp.integer[i] = integer[i] + 10 - op2.integer[i];
}
} else {
borrow = 0;
if (integer[i] - 1 >= op2.integer[i])
temp.integer[i] = integer[i] - 1 - op2.integer[i];
else {
borrow = 1;
temp.integer[i] = integer[i] - 1 + 10 - op2.integer[i];
}
}
temp.len = c;
return temp;
}
largeintegers largeintegers::operator*(largeintegers op2) {
largeintegers temp;
int i, j, k, tmp, m = 0;
for (i = 0; i < op2.len; i++) {
k = i;
for (j = 0; j < len; j++) {
tmp = integer[j] * op2.integer[i];
temp.integer[k] = temp.integer[k] + tmp;
temp.integer[k + 1] = temp.integer[k + 1] + temp.integer[k] / 10;
temp.integer[k] %= 10;
k++;
if (k > m) m = k;
}
}
temp.len = m;
if (temp.len > numbers) cout << "***OVERFLOW*****\n";
return temp;
}
using namespace std;
int main() {
int c;
largeintegers num1, num2, result;
num1.Input();
num2.Input();
num1.Output();
cout << " + ";
num2.Output();
result = num1 + num2;
cout << " = ";
result.Output();
cout << "\n\n";
num1.Output();
cout << " - ";
num2.Output();
result = num1 - num2;
cout << " = ";
result.Output();
cout << "\n\n";
num1.Output();
cout << " * ";
num2.Output();
result = num1 * num2;
cout << " = ";
result.Output();
cout << "\n\n";
c = num1 == num2;
num1.Output();
switch (c) {
case -1:
cout << " is less than ";
break;
case 0:
cout << " is equal to ";
break;
case 1:
cout << " is greater than ";
break;
}
num2.Output();
cout << "\n\n";
system("pause");
}
It seems you are falling victim to the difference between C-style strings and C++ strings. C-style strings are a series of chars followed by a zero (or null) byte. C++ strings are objects that contain a series of characters (usually char, but eventually this will be an assumption you should break) and that know their own length. C++ strings can contain null bytes in the middle of themselves without problem.
To loop through all of the characters of a C++-style string, you can do one of a number of things:
You can use the .size() or .length() members of a string variable to find the number of characters in it, as in for (int i=0; i<str.size(); i++) { char c = str[i];
You can use .begin() and .end() to get iterators to the beginning and end of the string, respectively. A for loop in the form for (std::string::iterator it=str.begin(); it!=str.end(); ++it) will loop you through the members of the string by accessing *it.
If you're using C++11, you can use the for loop construct as follows: for (auto c: str) where c will be of the type of a character of the string str.
In the future, to solve problems like these, you can try using the debugger to see what happens when your program crashes or hits an exception. You likely would find that inside of largeintegers::Input() you running into either a memory access violation or some other problem.
Finally, as a future-looking criticism, you should not use C-style arrays (where you say int integer[ numbers ];) in favor of using C++-style containers, such as vector. A vector is a series of objects (such as ints) that can expand as needed.
How would I add two binary numbers in C++? What is the correct logic?
Here is my effort, but it doesn't seem to be correct:
#include <iostream>
using namespace std;
int main()
{
int a[3];
int b[3];
int carry = 0;
int result[7];
a[0] = 1;
a[1] = 0;
a[2] = 0;
a[3] = 1;
b[0] = 1;
b[1] = 1;
b[2] = 1;
b[3] = 1;
for(int i = 0; i <= 3; i++)
{
if(a[i] + b[i] + carry == 0)
{
result[i] = 0;
carry = 0;
}
if(a[i] + b[i] + carry == 1)
{
result[i] = 0;
carry = 0;
}
if(a[i] + b[i] + carry == 2)
{
result[i] = 0;
carry = 1;
}
if(a[i] + b[i] + carry > 2)
{
result[i] = 1;
carry = 1;
}
}
for(int j = 0; j <= 7; j++)
{
cout<<result[j]<<" ";
}
system("pause");
}
Well, it is a pretty trivial problem.
How to add two binary numbers in c++. what is the logic of it.
For adding two binary numbers, a and b. You can use the following equations to do so.
sum = a xor b
carry = ab
This is the equation for a Half Adder.
Now to implement this, you may need to understand how a Full Adder works.
sum = a xor b xor c
carry = ab+bc+ca
Since you store your binary numbers in int array, you might want to understand bitwise operation.
You can use ^ for XOR,| operator for OR, & operator for AND.
Here is a sample code to calculate the sum.
for(i = 0; i < 8 ; i++){
sum[i] = ((a[i] ^ b[i]) ^ c); // c is carry
c = ((a[i] & b[i]) | (a[i] & c)) | (b[i] & c);
}
Since you were asking about C++, you deserve a C++ answer. Use bitsets:
#include <bitset>
#include <iostream>
int main() {
std::bitset<5> const a("1001");
std::bitset<5> const b("1111");
// m here is a mask to extract the lsb of a bitset.
std::bitset<5> const m("1");
std::bitset<5> result;
for (auto i = 0; i < result.size(); ++i) {
std::bitset<5> const diff(((a >> i)&m).to_ullong() + ((b >> i)&m).to_ullong() + (result >> i).to_ullong());
result ^= (diff ^ (result >> i)) << i;
}
std::cout << result << std::endl;
}
This works for arbitrarily long bit sets.
You could use "Bitwise OR" operation to reduce the code since
1 or 1 = 1
1 or 0 = 1
0 or 1 = 1
0 or 0 = 0
You could also convert both number to decimal sum and them go back to binary again.
Converting decimal to binary
int toBinary (unsigned int num, char b[32])
{
unsigned int x = INT_MIN; // (32bits)
int i = 0, count = 0;
while (x != 0)
{
if(x & num) // If the actual o bit is 1 & 1 = 1 otherwise = 0
{
b[i] = '1';
count++;
}
else b[i] = '0';
x >>=1; // pass to the left
i++;
}
return count;
}
There is a bug :
if(a[i]+b[i]+carry==1)
{
result[i]=1;
carry=0;
}
Also u might want to print in reverse
for(int j=6; j>=0; j--)
{
cout<<result[j]<<" ";
}
Your arrays are one item too small for your indexing.
int a[3] only has 3 elements, so a[3] = 1 is invalid (it has undefined behaviour) since it's accessing the 4th element, which doesn't exist.
Likewise for the other arrays.
That means that the whole program has undefined behaviour, i.e. it can do anything or nothing at all.
(What's probably happening in your case is that writing outside the arrays is overwriting the other variables.)
You're also not initialising the result array, so its content is just some random data.
Since you only update 4 of its elements but print all of them (and more), the output will be random data as well.
Following were the errors in your code and fixed code is also below"
int a[] was of size 3 so it cannot store at the 3rd index. use int a[4].
if(a[i]+b[i]+carry==1) wrong values were assigned in this check update result[i]=1; carry=0.
The sequence of checks is reversed.
The last carry was not stored in the result.
The addition result stored in the result array was in reverse order so printed it in reverse.
here is the working piece of code:
#include <iostream>
#include <string>
using namespace std;
int main()
{
int a[4];
int b[4];
int carry=0;
int result[5];
a[0]=1;
a[1]=0;
a[2]=0;
a[3]=1;
b[0]=1;
b[1]=1;
b[2]=1;
b[3]=1;
for(int i=0; i<4; i++)
{
if(a[i]+b[i]+carry==3)
{
result[i]=1;
carry=1;
}
if(a[i]+b[i]+carry==2)
{
result[i]=0;
carry=1;
}
if(a[i]+b[i]+carry==1)
{
result[i]=1;
carry=0;
}
if(a[i]+b[i]+carry==0)
{
result[i]=0;
carry=0;
}
}
result[4]=carry;
for(int j=4; j>=0; j--)
{
cout<<result[j];
}
cout<<endl;
return 0;
}
#include <stdio.h>
int main()
{
long binary1, binary2;
int i = 0, remainder = 0, sum[20];
printf("Enter the first binary number: ");
scanf("%ld", &binary1);
printf("Enter the second binary number: ");
scanf("%ld", &binary2);
while (binary1 != 0 || binary2 != 0)
{
sum[i++] =(binary1 % 10 + binary2 % 10 + remainder) % 2;
remainder =(binary1 % 10 + binary2 % 10 + remainder) / 2;
binary1 = binary1 / 10;
binary2 = binary2 / 10;
}
if (remainder != 0)
sum[i++] = remainder;
--i;
printf("Sum of two binary numbers: ");
while (i >= 0)
printf("%d", sum[i--]);
getch();
return 0;
}
Repeatedly do
(x, y) <- ((x & y) << 1, x ^ y)
until x is 0. y is the answer.
you should do this
for(int i = 3; i >= 0; i--)
{
if(a[i] + b[i] + carry == 0)
{
result[i] = 0;
carry = 0;
}
else if(a[i]+b[i]+carry==1)
{
result[i]=1;
carry=0;
}
else if(a[i] + b[i] + carry == 2)
{
result[i] = 0;
carry = 1;
}
else if(a[i] + b[i] + carry > 2)
{
result[i] = 1;
carry = 1;
}
printf("%d",result[i]);
}
A non-conventional solution, but it works:
int main() {
int A[] = { 0, 0, 0, 1, 1, 0, 1, 0};
int B[] = { 0, 0, 0, 0, 1, 1, 0, 0};
int size = sizeof(A)/sizeof(*A);
int C[size+1];
int t = 0;
for(int i = size-1; i > -1; i--){
C[i+1] = A[i]+B[i]+t;
t = C[i+1]/2;
C[i+1] %= 2;
}
C[0] = t;
}
What if their sizes are not the same? Also, you would want to allow the user to input the binary numbers (in this case representing integers) as integers and not as elements of arrays. Here is a piece of code that accomplishes those :-)
#include <iostream>
using namespace std;
// Add two numbers in binary
void sumBinary(int num1, int num2, int* sum12){
int mod1 = 0;
int mod2 = 0;
int carry = 0;
int factor = 1;
int flag = 0;
*sum12 = 0;
while (!flag){
mod1 = num1 % 10;
mod2 = num2 % 10;
num1 /= 10;
num2 /= 10;
if ((carry + mod1 + mod2) == 2){
*sum12 += 0;
carry = 1;
}
else if ((carry + mod1 + mod2) == 3){
*sum12 += factor;
carry = 1;
}
else if ((carry + mod1 + mod2) == 0){
*sum12 += 0;
carry = 0;
}
else{
*sum12 += factor;
carry = 0;
}
factor *= 10;
if ((num1 == 0) && (num2 == 0)){
*sum12 += carry*factor;
flag = 1; }
}
}
void main(){
int num1, num2, sum12;
cout << "Enter the first binary integer number: ";
cin >> num1;
cout << "Enter the second binary integer number: ";
cin >> num2;
sumBinary(num1, num2, &sum12);
cout << "The sum in binary form is :" << sum12 << endl;
}
A simple way :
int getBit(string s, int index)
{
if(index >= 0) return (s[index] - '0');
else return 0;
}
string addBinary(string a, string b)
{
if(a.size() > b.size()) while(a.size() > b.size()) b = "0" + b;
else if(b.size() > a.size()) while(b.size() > a.size()) a = "0" + a;
int l = max(a.size()-1, b.size() - 1);
string result = "";
int s=0;
while(l>=0 || s==1)
{
s += getBit(a, l) + getBit(b, l) ;
result = char(s % 2 + '0') + result;
s /= 2;
l--;
}
return result;
}
int main(){
ios::sync_with_stdio(0); cin.tie(0);
int num1=12, num2=45, sum=0;
bool b1, b2, carry=0;
for(int i=0;i<32;i++){
b1=( 1<<i ) & num1;
b2=( 1<<i ) & num2;
sum = (b1 ^ b2 ^ carry) ? sum ^ (1<<i) : sum;
carry = ((b1 & b2) | (b1 & carry) | (b2 & carry));
}
cout<<sum;
return 0;
}
Easy to understand code
Add Two Binary Numbers (input datatype = int)
#include <iostream>
using namespace std;
int power(int a, int b)
{
int ans = 1;
while (b)
{
ans *= a;
b--;
}
return ans;
}
int main()
{
int n1, n2, carry = 0, ans = 0, rem1, rem2, remsum = 0, i;
cout << "Enter First Number : ";
cin >> n1;
cout << "Enter Second Number : ";
cin >> n2;
for (i = 0; n1 != 0 || n2 != 0; i++, n1 /= 10, n2 /= 10)
{
rem1 = n1 % 10;
rem2 = n2 % 10;
remsum = rem1 + rem2 + carry;
if (remsum == 2)
{
carry = 1;
remsum = 0;
}
else if (remsum == 3)
{
carry = 1;
remsum = 1;
}
else if (remsum == 0)
{
carry = 0;
remsum = 0;
}
else if (remsum == 1)
{
carry = 0;
remsum = 1;
}
ans = remsum * power(10, i) + ans;
}
ans = carry * power(10, i) + ans;
cout << ans;
return 0;
}
I am preparing the interview questions not for homework. There is one question about how to multiple very very long integer. Could anybody offer any source code in C++ to learn from? I am trying to reduce the gap between myself and others by learning other's solution to improve myself.
Thanks so much!
Sorry if you think this is not the right place to ask such questions.
you can use GNU Multiple Precision Arithmetic Library for C++.
If you just want an easy way to multiply huge numbers( Integers ), here you are:
#include<iostream>
#include<string>
#include<sstream>
#define SIZE 700
using namespace std;
class Bignum{
int no[SIZE];
public:
Bignum operator *(Bignum& x){ // overload the * operator
/*
34 x 46
-------
204 // these values are stored in the
136 // two dimensional array mat[][];
-------
1564 // this the value stored in "Bignum ret"
*/
Bignum ret;
int carry=0;
int mat[2*SIZE+1][2*SIZE]={0};
for(int i=SIZE-1;i>=0;i--){
for(int j=SIZE-1;j>=0;j--){
carry += no[i]*x.no[j];
if(carry < 10){
mat[i][j-(SIZE-1-i)]=carry;
carry=0;
}
else{
mat[i][j-(SIZE-1-i)]=carry%10;
carry=carry/10;
}
}
}
for(int i=1;i<SIZE+1;i++){
for(int j=SIZE-1;j>=0;j--){
carry += mat[i][j]+mat[i-1][j];
if(carry < 10){
mat[i][j]=carry;
carry=0;
}
else{
mat[i][j]=carry%10;
carry=carry/10;
}
}
}
for(int i=0;i<SIZE;i++)
ret.no[i]=mat[SIZE][i];
return ret;
}
Bignum (){
for(int i=0;i<SIZE;i++)
no[i]=0;
}
Bignum (string _no){
for(int i=0;i<SIZE;i++)
no[i]=0;
int index=SIZE-1;
for(int i=_no.length()-1;i>=0;i--,index--){
no[index]=_no[i]-'0';
}
}
void print(){
int start=0;
for(int i=0;i<SIZE;i++)
if(no[i]!=0){
start=i;
break; // find the first non zero digit. store the index in start.
}
for(int i=start;i<SIZE;i++) // print the number starting from start till the end of array.
cout<<no[i];
cout<<endl;
return;
}
};
int main(){
Bignum n1("100122354123451234516326245372363523632123458913760187501287519875019671647109857108740138475018937460298374610938765410938457109384571039846");
Bignum n2("92759375839475239085472390845783940752398636109570251809571085701287505712857018570198713984570329867103986475103984765109384675109386713984751098570932847510938247510398475130984571093846571394675137846510874510847513049875610384750183274501978365109387460374651873496710394867103984761098347609138746297561762234873519257610");
Bignum n3 = n1*n2;
n3.print();
return 0;
}
as you can see, it's multiply 2 huge integer :) ... (up to 700 digits)
Try this:
//------------DEVELOPED BY:Ighit F4YSAL-------------
#include<iostream>
#include<string>
#include<sstream>
#define BIG 250 //MAX length input
using namespace std;
int main(){
int DUC[BIG][BIG*2+1]={0},n0[BIG],n1[BIG],i,t,h,carry=0,res;
string _n0,_n1;
while(1){
//-----------------------------------get data------------------------------------------
cout<<"n0=";
cin>>_n0;
cout<<"n1=";
cin>>_n1;
//--------------------string to int[]----------------------------------------
for(i=_n0.length()-1,t=0;i>=0,t<=_n0.length()-1;i--,t++){
n0[i]=_n0[t]-'0';
}
i=0;
for(i=_n1.length()-1,t=0;i>=0,t<=_n1.length()-1;i--,t++){
n1[i]=_n1[t]-'0';
}
i=0;t=0;
//--------------------------produce lines of multiplication----------------
for(i=0;i<=_n1.length()-1;i++){
for(t=0;t<=_n0.length()-1;t++){
res=((n1[i]*n0[t])+carry);
carry=(res/10);
DUC[i][t+i]=res%10;
}
DUC[i][t+i]=carry;
carry=0;
}
i=0;t=0;res=0;carry=0;
//-----------------------------add the lines-------------------------------
for(i=0;i<=_n0.length()*2-1;i++){
for(t=0;t<=_n1.length()-1;t++){
DUC[BIG-1][i]+=DUC[t][i];
//cout<<DUC[t][i]<<"-";
}
res=((DUC[BIG-1][i])+carry);
carry=res/10;
DUC[BIG-1][i]=res%10;
//cout<<" ="<<DUC[BIG-1][i]<<endl;
}
i=0;t=0;
//------------------------print the result------------------------------------
cout<<"n1*n0=";
for(i=_n0.length()*2-1;i>=0;i--){
if((DUC[BIG-1][i]==0) and (t==0)){}else{cout<<DUC[BIG-1][i];t++;}
//cout<<DUC[BIG-1][i];
}
//-------------------------clear all-------------------------------------
for(i=0;i<=BIG-1;i++){
for(t=0;t<=BIG*2;t++){
DUC[i][t]=0;
}
n0[i]=0;n1[i]=0;
}
//--------------------------do it again-------------------------------------
cout<<"\n------------------------------------------------\n\n";
}
return 0;
}
This solution is good for very very big numbers but not so effective for factorial calculation of very big numbers. Hope it will help someone.
#include <iostream>
#include <string>
using namespace std;
string addition(string a, string b) {
string ans = "";
int i, j, temp = 0;
i = a.length() - 1;
j = b.length() - 1;
while (i >= 0 || j >= 0) {
if (i >= 0 && a[i])
temp += a[i] - '0';
if (j >= 0 && b[j])
temp += b[j] - '0';
int t = (temp % 10);
char c = t + '0';
ans = ans + c;
temp = temp / 10;
i--;
j--;
}
while (temp > 0) {
int t = (temp % 10);
char c = t + '0';
ans = ans + c;
temp = temp / 10;
}
string fnal = "";
for (int i = ans.length() - 1;i >= 0;i--) {
fnal = fnal + ans[i];
}
return fnal;
}
string multiplication(string a, string b) {
string a1, b1 = "0";
int i, j, temp = 0, zero = 0;
i = a.length() - 1;
int m1, m2;
while (i >= 0) {
a1 = "";
m1 = a[i] - '0';
j = b.length() - 1;
while (j >= 0) {
m2 = b[j] - '0';
temp = temp + m1*m2;
int t = temp % 10;
char c = t + '0';
a1 = a1 + c;
temp = temp / 10;
j--;
}
while (temp > 0) {
int t = (temp % 10);
char c = t + '0';
a1 = a1 + c;
temp = temp / 10;
}
string fnal = "";
// reverse string
for (int i = a1.length() - 1;i >= 0;i--) {
fnal = fnal + a1[i];
}
a1 = fnal;
//add zero
for (int p = 0;p < zero;p++)
a1 = a1 + '0';
b1 = addition(a1, b1);
i--;
zero++;
}
return b1;
}
// upto 50 is ok
int factorial(int n) {
string a = "1";
for (int i = 2;i <= n;i++) {
string b = to_string(i);
a = multiplication(a, b);
}
cout << a << endl;
return a.length();
}
int main() {
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
int n;
cin >> n;
//cout << factorial(n) << endl;
string a, b;
a = "1281264836465376528195645386412541764536452813416724654125432754276451246124362456354236454857858653";
b = "3767523857619651386274519192362375426426534237624548235729562582916259723586347852943763548355248625";
//addition(a, b);
cout << multiplication(a, b);
return 0;
}