How to convert any base to base 10 for big numbers - c++

I want to convert a big number (100,000 digits) from any base to base 10. I'm using this code:
this is reading file method
string* ReadFile(string fileName) {
ifstream inFile(fileName);
string line;
string* myArray = new string[1000000];
if (inFile.is_open()) {
int count = 0;
while (getline(inFile, line)) {
myArray[count] = line;
count++;
}
inFile.close();
}
else
{
cout << "file is not open";
}
return myArray;}
convert every character to base 10 method
int val(char c){
if (c >= '0' && c <= '9')
return (int)c - '0';
else
return (int)c - 'A' + 10; }
any base to decimal
int Base2Dec(string* arr) {
int base = stoi(arr[0]);
string str = "";
int size = 0;
while (!arr[size].empty())
++size;
for (int i = size; i > 0; i--) {
str += arr[i];
}
int len = str.length();
int power = 1;
int num = 0;
for (int i = len - 1; i >= 0; i--)
{
if (val(str[i]) >= base)
{
printf("Invalid Number");
}
num += val(str[i]) * power;
power = power * base;
}
return num;}
and main method
int main(){
string* A = ReadFile("A2.txt"); cout << Base2Dec(A);
string* B = ReadFile("B.txt"); Base2Dec(B);
cin.get();}
it works but not for 100,000 digits ?! I know int type is limited and I am not allowed to use int64 so, I want to know what the solution is?!

Related

Return all codes - String

Assume that the value of a = 1, b = 2, c = 3, ... , z = 26. You are given a numeric string S. Write a program to return the list of all possible codes that can be generated from the given string.
For most of the cases this code works but it gives wrong output for inputs which have numbers greater than 26. For eg: 12345.
#include <iostream>
#include <string.h>
using namespace std;
using namespace std;
int atoi(char a)
{
int i=a-'0';
return i;
}
char itoa(int i)
{
char c='a'+i-1;
return c;
}
int getCodes(string input, string output[10000]) {
if(input.size()==0)
{
return 1;
}
if(input.size()==1)
{
output[0]=output[0]+itoa(atoi(input[0]));
return 1;
}
string result1[10000],result2[10000];
int size2;
int size1=getCodes(input.substr(1),result1);
if(input.size()>1)
{
if(atoi(input[0])*10+atoi(input[1])>10&&atoi(input[0])*10+atoi(input[1])<27)
{
size2=getCodes(input.substr(2),result2);
}
}
for(int i=0;i<size1;i++)
{
output[i]=itoa(atoi(input[0]))+result1[i];
}
for(int i=0;i<size2;i++)
{
output[i+size1]=itoa(atoi(input[0])*10+atoi(input[1]))+result2[i];
}
return size1+size2;
}
int main(){
string input;
cin >> input;
string output[10000];
int count = getCodes(input, output);
for(int i = 0; i < count && i < 10000; i++)
cout << output[i] << endl;
return 0;
}
if i give input 12345, the output is:
"
abcde
awde
lcde
l"
instead of :
"
abcde
awde
lcde"
i got it fellow members. i did not initialised the size2 variable to zero. also i didn't use >= operator.
int getCodes(string input, string output[10000]) {
if(input.size()==0)
{
output[0]="";
return 1;
}
if(input.size()==1)
{
output[0]=itoa(atoi(input[0]));
return 1;
}
string result1[10000],result2[10000];
int size2=0;
int size1=getCodes(input.substr(1),result1);
if(input.size()>1)
{
if(atoi(input[0])*10+atoi(input[1])>=10&&atoi(input[0])*10+atoi(input[1])<27)
{
size2=getCodes(input.substr(2),result2);
}
}
int k=0;
for(int i=0;i<size1;i++)
{
output[k++]=itoa(atoi(input[0]))+result1[i];
}
for(int i=0;i<size2;i++)
{
output[k++]=itoa(atoi(input[0])*10+atoi(input[1]))+result2[i];
}
return k;
}
this is the final code for getCodes function. Thanks everyone :)
You can do that more simply with something like this:
#include <utility>
#include <string>
#include <vector>
#include <iostream>
using namespace std;
void getCodesRec(unsigned int num, string& current, vector<string>& result)
{
// First and last chars for the codes
static constexpr char FIRST_CHAR = 'a';
static constexpr char LAST_CHAR = 'z';
if (num == 0)
{
// When there is no more number add the code to the results
result.push_back(current);
}
else
{
// Add chars to the existing code
unsigned int next = num;
unsigned int rem = next % 10;
unsigned int f = 1;
// While we have not gone over the max char number
// (in practice this loop will run twice at most for a-z letters)
while (next > 0 && rem <= (unsigned int)(LAST_CHAR - FIRST_CHAR) + 1)
{
next = next / 10;
if (rem != 0) // 0 does not have a replacement
{
// Add the corresponding char
current.insert(0, 1, FIRST_CHAR + char(rem - 1));
// Recursive call
getCodesRec(next, current, result);
// Remove the char
current.erase(0, 1);
}
// Add another number
f *= 10;
rem += f * (next % 10);
}
}
}
vector<string> getCodes(unsigned int num)
{
vector<string> result;
string current;
getCodesRec(num, current, result);
return result;
}
int main()
{
unsigned int num = 12345;
vector<string> codes = getCodes(12345);
cout << "Codes for " << num << endl;
for (string& code : codes)
{
cout << "* " << code << endl;
}
return 0;
}
Output:
Codes for 12345
* abcde
* lcde
* awde

Recieving "string subscript out of range" error

Doing this assignment where I parse information into a struct and I keep recieving an error saying "string subscript out of range" when trying to test. Cant figure out where I went wrong.
here is the struct
struct baby_type {
std::string name;
std::vector<int> yearsCount;
int total;
} ;
here is my function
baby_type Parse_Line(const std::string line )
{
baby_type baby;
std::string name;
std::string year;// = "0000";
std::string total;
int i = 0;
int k = 1;
int LengthOfOccurences = 0;
int DigitOfOccurences = 0;
while (line.at(i) != ',') {
name[i] = line.at(i);
i++;
}
baby.name = name;
i++;
while (k < 100) {
if (line.at(i) == ',') {
year.resize(LengthOfOccurences);
baby.yearsCount.at(k) = std::stoi(year);
//year = "0000";
i++;
k++;
DigitOfOccurences = 0;
LengthOfOccurences = 0;
}
else {
year.at(DigitOfOccurences) = line.at(i);
i++;
LengthOfOccurences++;
}
}
int m = 0;
int LengthOfLine = line.length();
while (i < LengthOfLine) {
total.at(m) = line.at(i);
m++;
i++;
}
baby.total = std::stoi(total);
return baby;
}
If you create empty std::string objects and then assign characters to specific positions. std::strings don't work that way.
In your first while loop, use
name.append(1,line.at(i));
The '1' is necessary because there is no simple std::append with just a character as parameter.

C++: Change of base function (i.e. hex to octal, decimal, etc.) - Output slightly off for hex values

I need to create a generic function that changes from any starting base, to any final base. I have everything down, except my original function took (and takes) an int value for the number that it converts to another base. I decided to just overload the function. I am Ok with changing between every base, but am slightly off when using my new function to take in a string hex value.
The code below should output 1235 for both functions. It does for the first one, but for the second, I am currently getting 1347. Decimal to Hex works fine - It's just the overloaded function (Hex to anything else) that is slightly off.
Thanks.
#include <iostream>
#include <stack>
#include <string>
#include <cmath>
using namespace std;
void switchBasesFunction(stack<int> & myStack, int startBase, int finalBase, int num);
void switchBasesFunction(stack<int> & myStack, int startBase, int finalBase, string s);
int main()
{
stack<int> myStack;
string hexNum = "4D3";
switchBasesFunction(myStack, 8, 10, 2323);
cout << endl << endl;
switchBasesFunction(myStack, 16, 10, hexNum);
return 0;
}
void switchBasesFunction(stack<int> & myStack, int startBase, int finalBase, int num)
{
int totalVal = 0;
string s = to_string(num);
for (int i = 0; i < s.length(); i++)
{
myStack.push(s.at(i) - '0');
}
int k = 0;
while (myStack.size() > 0)
{
totalVal += (myStack.top() * pow(startBase, k++));
myStack.pop();
}
string s1;
while (totalVal > 0)
{
int temp = totalVal % finalBase;
totalVal = totalVal / finalBase;
char c;
if (temp < 10)
{
c = temp + '0';
s1 += c;
}
else
{
c = temp - 10 + 'A';
s1 += c;
}
}
for (int i = s1.length() - 1; i >= 0; i--)
{
cout << s1[i];
}
cout << endl << endl;
}
void switchBasesFunction(stack<int> & myStack, int startBase, int finalBase, string s)
{
int totalVal = 0;
for (int i = 0; i < s.length(); i++)
{
myStack.push(s.at(i) - '0');
}
int k = 0;
while (myStack.size() > 0)
{
totalVal += (myStack.top() * pow(startBase, k++));
myStack.pop();
}
string s1;
while (totalVal > 0)
{
int temp = totalVal % finalBase;
totalVal = totalVal / finalBase;
char c;
if (temp < 10)
{
c = temp + '0';
s1 += c;
}
else
{
c = temp - 10 + 'A';
s1 += c;
}
}
for (int i = s1.length() - 1; i >= 0; i--)
{
cout << s1[i];
}
cout << endl << endl;
}
Sorry, but I'm having issues understanding your code, so I thought I'd simplify it.
Here's the algorithm / code (untested):
void convert_to_base(const std::string& original_value,
unsigned int original_base,
std::string& final_value_str,
unsigned int final_base)
{
static const std::string digit_str =
"0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
if ((original_base > digit_str.length()) || (final_base > digit_str.length())
{
std::cerr << "Base value exceeds limit of " << digit_str.length() << ".\n";
return;
}
// Parse string from right to left, smallest value to largest.
// Convert to decimal.
unsigned int original_number = 0;
unsigned int digit_value = 0;
int index = 0;
for (index = original_value.length(); index > 0; --index)
{
std::string::size_type posn = digit_str.find(original_value[index];
if (posn == std::string::npos)
{
cerr << "unsupported digit encountered: " << original_value[index] << ".\n";
return;
}
digit_value = posn;
original_number = original_number * original_base + digit_value;
}
// Convert to a string of digits in the final base.
while (original_number != 0)
{
digit_value = original_number % final_base;
final_value_str.insert(0, 1, digit_str[digit_value]);
original_number = original_number / final_base;
}
}
*Warning: code not tested via compiler.**

Adding two strings mathematically?

i was looking around the forums and i still couldnt find my answer to my problem.
I got two strings, that are just really an array of numbers. for example(i just choose random numbers
string input1="12345678909876543212";
string input2="12345";
I want to add these two string together but act them like there integers.
My goal is creating a class where i can add bigger numbers than (long long int) so it can exceed the largest long long int variable.
So i revese the string with no problem, so now there
input1="21234567890987654321"
input2="54321"
then i tried adding, let's say input1[0]+input2[0] (2+5) to a new string lets call it newString[0] where that would equal (7); but i cant find a good way to temporally convert the current number in the string so i can add it to the new string? can anyone help. I get sick and tired of atoi,stof,stod. they don't seem to work at all for me.
Any way i can make this function work.
I don't care about making the class yet, i just care about finding a way to add those two strings mathematically but still maintaining the newString's string format. Thank you for whoever can figure this out for me
Okay, so, assuming your only problem is with the logic, not the class design thing, I came up with this logic
fill up the inputs with 0s, checking the lengths, match the lengths
add like normal addition, keeping track of carry
finally remove leading zeros from result
So using std::transform with a lambda function on reverse iterators :-
char carry = 0;
std::transform(input1.rbegin(),input1.rend(),input2.rbegin(),
result.rbegin(),[&carry]( char x, char y){
char z = (x-'0')+(y-'0') + carry;
if (z > 9)
{
carry = 1;
z -= 10;
}
else
{
carry = 0;
}
return z + '0';
});
//And finally the last carry
result[0] = carry + '0';
//Remove the leading zero
n = result.find_first_not_of("0");
if (n != string::npos)
{
result = result.substr(n);
}
See Here
Edit "Can you comment on what your doing here"
+--------+--------------+------------+-------> Reverse Iterator
| | | |
std::transform( | input1.rbegin(), input1.rend(),input2.rbegin(),
result.rbegin(), [&carry]( char x, char y){
//This starts a lambda function
char z = (x-'0')+(y-'0') + carry; // x,y have ASCII value of each digit
// Substracr ASCII of 0 i.e. 48 to get the "original" number
// Add them up
if (z > 9) //If result greater than 9, you have a carry
{
carry = 1; // store carry for proceeding sums
z -= 10; // Obviously
}
else
{
carry = 0; //Else no carry was generated
}
return z + '0'; // Now you have "correct" number, make it a char, add 48
});
std::transform is present in header <algorithm>, see the ideone posted link.
Here's A Solution for adding two numbers represented as strings .
#include<iostream>
using namespace std;
string add(string a, string b)
{
int al=a.size()-1;
int bl=b.size()-1;
int carry=0;
string result="";
while(al>=0 && bl>=0)
{
int temp = (int)(a[al] - '0') + (int)(b[bl] - '0') + carry ;
carry = 0;
if(temp > 9 )
{
carry=1;
temp=temp-10;
}
result+=char(temp + '0');
al--;
bl--;
}
while(al>=0)
{
int temp = (int)(a[al] - '0') + carry ;
carry = 0;
if(temp>9)
{
carry=1;
temp=temp%10;
}
result+=char(temp + '0');
al--;
}
while(bl>=0)
{
int temp = (int)(b[bl] - '0') + carry ;
carry = 0;
if(temp>9)
{
carry=1;
temp=temp%10;
}
result+=char(temp + '0');
bl--;
}
if(carry)
result+="1";
string addition="";
for(int i=result.size()-1;i>=0;i--)
addition+=result[i]; // reversing the answer
return addition;
}
string trim(string a) // for removing leading 0s
{
string res="";
int i=0;
while(a[i]=='0')
i++;
for(;i<a.size();i++)
res+=a[i];
return res;
}
int main()
{
string a;
string b;
cin>>a>>b;
cout<<trim(add(a,b))<<endl;
}
I am not a very femilier with C++ but cant we do this?
int i = stoi( input1[0]);
int j = stoi( input2[0]);
int x = i+j;
Please note this can be done in C++11 Please refer [1] and 2 as well
You can convert a char to an int by subtracting '0' from it:
char sumdigit = (input1[0]-'0') + (input2[0]-'0') + '0';
atoi() would be a better to go, as far as converting input[0] to an int:
int temp = atoi(input.substr(0,1).c_str());
then use stringstream to convert back to string:
stringstream convert;
convert << temp;
string newString = convert.str();
Here is a solution, but this is so far from sensible that it is not even funny.
GCC 4.7.3: g++ -Wall -Wextra -std=c++0x dumb-big-num.cpp
#include <algorithm>
#include <cctype>
#include <iostream>
#include <sstream>
#include <stdexcept>
#include <string>
// dumb big num
// unsigned integer
class DBN {
public:
DBN() : num("0") {}
explicit DBN(const std::string& s) : num(s) {
for (const auto& c : num) {
if (!std::isdigit(c)) { throw std::invalid_argument("DBN::DBN"); } }
std::reverse(std::begin(num), std::end(num)); }
DBN operator+(const DBN& rhs) const {
DBN tmp(*this);
return tmp += rhs; }
DBN& operator+=(const DBN& rhs) {
std::string r;
const int m = std::min(num.size(), rhs.num.size());
int c = 0;
for (int i = 0; i < m; ++i) {
int s = (num[i] - '0') + (rhs.num[i] - '0') + c;
c = s / 10;
s %= 10;
r += static_cast<char>('0' + s); }
const std::string& ref = num.size() < rhs.num.size() ? rhs.num : num;
for (int i = m; i < ref.size(); ++i) {
int s = (ref[i] - '0') + c;
c = s / 10;
s %= 10;
r += static_cast<char>('0' + s); }
if (0 < c) { r += '1'; }
num = r;
return *this; }
friend std::ostream& operator<<(std::ostream& os, const DBN& rhs);
friend std::istream& operator>>(std::istream& os, DBN& rhs);
private:
std::string num;
};
std::ostream& operator<<(std::ostream& os, const DBN& rhs) {
std::string s(rhs.num);
std::reverse(std::begin(s), std::end(s));
return os << s;
}
std::istream& operator>>(std::istream& is, DBN& rhs) {
std::stringstream ss;
char c;
while (is && std::isspace(is.peek())) { is.ignore(); }
while (is) {
if (!std::isdigit(is.peek())) { break; }
is >> c;
ss << c; }
DBN n(ss.str());
rhs = n;
return is;
}
int main() {
DBN a, b, t;
while (std::cin >> a >> b) {
std::cout << a + b << "\n";
(t += a) += b;
}
std::cout << t << "\n";
}
Here it is a simple C++ code
string Sum(string a, string b)
{
if(a.size() < b.size())
swap(a, b);
int j = a.size()-1;
for(int i=b.size()-1; i>=0; i--, j--)
a[j]+=(b[i]-'0');
for(int i=a.size()-1; i>0; i--)
if(a[i] > '9')
{
int d = a[i]-'0';
a[i-1] = ((a[i-1]-'0') + d/10) + '0';
a[i] = (d%10)+'0';
}
if(a[0] > '9')
{
string k;
k+=a[0];
a[0] = ((a[0]-'0')%10)+'0';
k[0] = ((k[0]-'0')/10)+'0';
a = k+a;
}
return a;
}
cited from C - Adding the numbers in 2 strings together if a different length
answer, I write a more readable codeļ¼š
void str_reverse(char *beg, char *end){
if(!beg || !end)return;
char cTmp;
while(beg < end){
cTmp = *beg;
*beg++ = *end;
*end-- = cTmp;
}
}
#define c2d(c) (c - '0')
#define d2c(d) (d + '0')
void str_add(const char* s1, const char* s2, char* s_ret){
int s1_len = strlen(s1);
int s2_len = strlen(s2);
int max_len = s1_len;
int min_len = s2_len;
const char *ps_max = s1;
const char *ps_min = s2;
if(s2_len > s1_len){
ps_min = s1;min_len = s1_len;
ps_max = s2;max_len = s2_len;
}
int carry = 0;
int i, j = 0;
for (i = max_len - 1; i >= 0; --i) {
// this wrong-prone
int idx = (i - max_len + min_len) >=0 ? (i - max_len + min_len) : -1;
int sum = c2d(ps_max[i]) + (idx >=0 ? c2d(ps_min[idx]) : 0) + carry;
carry = sum / 10;
sum = sum % 10;
s_ret[j++] = d2c(sum);
}
if(carry)s_ret[j] = '1';
str_reverse(s_ret, s_ret + strlen(s_ret) - 1);
}
test code as below:
void test_str_str_add(){
char s1[] = "123";
char s2[] = "456";
char s3[10] = {'\0'};
str_add(s1, s2, s3);
std::cout<<s3<<std::endl;
char s4[] = "456789";
char s5[10] = {'\0'};
str_add(s1, s4, s5);
std::cout<<s5<<std::endl;
char s7[] = "99999";
char s8[] = "21";
char s9[10] = {'\0'};
str_add(s7, s8, s9);
std::cout<<s9<<std::endl;
}
output:
579
456912
100020

long integer multiplication

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;
}