factorial of big numbers with strings in c++ - c++

Question

I am doing a factorial program with strings because i need the factorial of Numbers greater than 250
I intent with:
string factorial(int n){
string fact="1";
for(int i=2; i<=n; i++){
b=atoi(fact)*n;
}
}
But the problem is that atoi not works. How can i convert my string in a integer.
And The most important Do I want to know if the program of this way will work with the factorial of 400 for example?

Answer 1

Not sure why you are trying to use string. Probably to save some space by not using integer vector? This is my solution by using integer vector to store factorial and print.Works well with 400 or any large number for that matter!
//Factorial of a big number
#include<iostream>
#include<vector>
using namespace std;
int main(){
int num;
cout<<"Enter the number :";
cin>>num;
vector<int> res;
res.push_back(1);
int carry=0;
for(int i=2;i<=num;i++){
for(int j=0;j<res.size();j++){
int tmp=res[j]*i;
res[j]=(tmp+carry)%10 ;
carry=(tmp+carry)/10;
}
while(carry!=0){
res.push_back(carry%10);
carry=carry/10;
}
}
for(int i=res.size()-1;i>=0;i--) cout<<res[i];
cout<<endl;
return 0;
}
Enter the number :400
Factorial of 400 :64034522846623895262347970319503005850702583026002959458684445942802397169186831436278478647463264676294350575035856810848298162883517435228961988646802997937341654150838162426461942352307046244325015114448670890662773914918117331955996440709549671345290477020322434911210797593280795101545372667251627877890009349763765710326350331533965349868386831339352024373788157786791506311858702618270169819740062983025308591298346162272304558339520759611505302236086810433297255194852674432232438669948422404232599805551610635942376961399231917134063858996537970147827206606320217379472010321356624613809077942304597360699567595836096158715129913822286578579549361617654480453222007825818400848436415591229454275384803558374518022675900061399560145595206127211192918105032491008000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000

Answer 2

There's a web site that will calculate factorials for you: http://www.nitrxgen.net/factorialcalc.php. It reports:
The resulting factorial of 250! is 493 digits long.
The result also contains 62 trailing zeroes (which constitutes to 12.58% of the whole number)
3232856260909107732320814552024368470994843717673780666747942427112823747555111209488817915371028199450928507353189432926730931712808990822791030279071281921676527240189264733218041186261006832925365133678939089569935713530175040513178760077247933065402339006164825552248819436572586057399222641254832982204849137721776650641276858807153128978777672951913990844377478702589172973255150283241787320658188482062478582659808848825548800000000000000000000000000000000000000000000000000000000000000
Many systems using C++ double only work up to 1E+308 or thereabouts; the value of 250! is too large to store in such numbers.
Consequently, you'll need to use some sort of multi-precision arithmetic library, either of your own devising using C++ string values, or using some other widely-used multi-precision library (GNU GMP for example).

Answer 3

The code below uses unsigned double long to calculate very large digits.
#include<iostream.h>
int main()
{
long k=1;
while(k!=0)
{
cout<<"\nLarge Factorial Calculator\n\n";
cout<<"Enter a number be calculated:";
cin>>k;
if (k<=33)
{
unsigned double long fact=1;
fact=1;
for(int b=k;b>=1;b--)
{
fact=fact*b;
}
cout<<"\nThe factorial of "<<k<<" is "<<fact<<"\n";
}
else
{
int numArr[10000];
int total,rem=0,count;
register int i;
//int i;
for(i=0;i<10000;i++)
numArr[i]=0;
numArr[10000]=1;
for(count=2;count<=k;count++)
{
while(i>0)
{
total=numArr[i]*count+rem;
rem=0;
if(total>9)
{
numArr[i]=total%10;
rem=total/10;
}
else
{
numArr[i]=total;
}
i--;
}
rem=0;
total=0;
i=10000;
}
cout<<"The factorial of "<<k<<" is \n\n";
for(i=0;i<10000;i++)
{
if(numArr[i]!=0 || count==1)
{
cout<<numArr[i];
count=1;
}
}
cout<<endl;
}
cout<<"\n\n";
}//while
return 0;
}
Output:
![Large Factorial Calculator
Enter a number be calculated:250
The factorial of 250 is
32328562609091077323208145520243684709948437176737806667479424271128237475551112
09488817915371028199450928507353189432926730931712808990822791030279071281921676
52724018926473321804118626100683292536513367893908956993571353017504051317876007
72479330654023390061648255522488194365725860573992226412548329822048491377217766
50641276858807153128978777672951913990844377478702589172973255150283241787320658
18848206247858265980884882554880000000000000000000000000000000000000000000000000
000000000000][1]

Answer 4

You can make atoi compile by adding c_str(), but it will be a long way to go till getting factorial. Currently you have no b around. And if you had, you still multiply int by int. So even if you eventually convert that to string before return, your range is still limited. Until you start to actually do multiplication with ASCII or use a bignum library there's no point to have string around.

Answer 5

Your factorial depends on conversion to int, which will overflow pretty fast, so you want be able to compute large factorials that way. To properly implement computation on big numbers you need to implement logic as for computation on paper, rules that you were tought in primary school, but treat long long ints as "atoms", not individual digits. And don't do it on strings, it would be painfully slow and full of nasty conversions

Answer 6

If you are going to solve factorial for numbers larger than around 12, you need a different approach than using atoi, since that just gives you a 32-bit integer, and no matter what you do, you are not going to get more than 2 billion (give or take) out of that. Even if you double the size of the number, you'll only get to about 20 or 21.
It's not that hard (relatively speaking) to write a string multiplication routine that takes a small(ish) number and multiplies each digit and ripples the results through to the the number (start from the back of the number, and fill it up).
Here's my obfuscated code - it is intentionally written such that you can't just take it and hand in as school homework, but it appears to work (matches the number in Jonathan Leffler's answer), and works up to (at least) 20000! [subject to enough memory].
std::string operator*(const std::string &s, int x)
{
int l = (int)s.length();
std::string r;
r.resize(l);
std::fill(r.begin(), r.end(), '0');
int b = 0;
int e = ~b;
const int c = 10;
for(int i = l+e; i != e;)
{
int d = (s[i]-0x30) * x, p = i + b;
while (d && p > e)
{
int t = r[p] - 0x30 + (d % c);
r[p] = (t % c) + 0x30;
d = t / c + d / c;
p--;
}
while (d)
{
r = static_cast<char>((d % c) +0x30)+r;
d /= c;
b++;
}
i--;
}
return r;
}

Answer 7

In C++, the largest integer type is 'long long', and it hold 64 bits of memory, so obviously you can't store 250! in an integer type. It is a clever idea to use strings, but what you are basically doing with your code is (I have never used the atoi() function, so I don't know if it even works with strings larger than 1 character, but it doesn't matter):
covert the string to integer (a string that if this code worked well, in one moment contains the value of 249!)
multiply the value of the string
So, after you are done multiplying, you don't even convert the integer back to string. And even if you did that, at one moment when you convert the string back to an integer, your program will crash, because the integer won't be able to hold the value of the string.
My suggestion is, to use some class for big integers. Unfortunately, there isn't one available in C++, so you'll have to code it by yourself or find one on the internet. But, don't worry, even if you code it by yourself, if you think a little, you'll see it's not that hard. You can even use your idea with the strings, which, even tough is not the best approach, for this problem, will still yield the results in the desired time not using too much memory.

Answer 8

This is a typical high precision problem.
You can use an array of unsigned long long instead of string.
like this:
struct node
{
unsigned long long digit[100000];
}
It should be faster than string.
But You still can use string unless you are urgent.
It may take you a few days to calculate 10000!.
I like use string because it is easy to write.
#include <bits/stdc++.h>
#pragma GCC optimize (2)
using namespace std;
const int MAXN = 90;
int n, m;
int a[MAXN];
string base[MAXN], f[MAXN][MAXN];
string sum, ans;
template <typename _T>
void Swap(_T &a, _T &b)
{
_T temp;
temp = a;
a = b;
b = temp;
}
string operator + (string s1, string s2)
{
string ret;
int digit, up = 0;
int len1 = s1.length(), len2 = s2.length();
if (len1 < len2) Swap(s1, s2), Swap(len1, len2);
while(len2 < len1) s2 = '0' + s2, len2++;
for (int i = len1 - 1; i >= 0; i--)
{
digit = s1[i] + s2[i] - '0' - '0' + up; up = 0;
if (digit >= 10) up = digit / 10, digit %= 10;
ret = char(digit + '0') + ret;
}
if (up) ret = char(up + '0') + ret;
return ret;
}
string operator * (string str, int p)
{
string ret = "0", f; int digit, mul;
int len = str.length();
for (int i = len - 1; i >= 0; i--)
{
f = "";
digit = str[i] - '0';
mul = p * digit;
while(mul)
{
digit = mul % 10 , mul /= 10;
f = char(digit + '0') + f;
}
for (int j = 1; j < len - i; j++) f = f + '0';
ret = ret + f;
}
return ret;
}
int main()
{
freopen("factorial.out", "w", stdout);
string ans = "1";
for (int i = 1; i <= 5000; i++)
{
ans = ans * i;
cout << i << "! = " << ans << endl;
}
return 0;
}

Answer 9

Actually, I know where the problem raised At the point where we multiply , there is the actual problem ,when numbers get multiplied and get bigger and bigger.
this code is tested and is giving the correct result.
#include <bits/stdc++.h>
using namespace std;
#define mod 72057594037927936 // 2^56 (17 digits)
// #define mod 18446744073709551616 // 2^64 (20 digits) Not supported
long long int prod_uint64(long long int x, long long int y)
{
return x * y % mod;
}
int main()
{
long long int n=14, s = 1;
while (n != 1)
{
s = prod_uint64(s , n) ;
n--;
}
}
Expexted output for 14! = 87178291200

Answer 10

The logic should be:
unsigned int factorial(int n)
{
unsigned int b=1;
for(int i=2; i<=n; i++){
b=b*n;
}
return b;
}
However b may get overflowed. So you may use a bigger integral type.
Or you can use float type which is inaccurate but can hold much bigger numbers.
But it seems none of the built-in types are big enough.