问题
I want to calculate the sum of digits of N!.
I want to do this for really large values of N, say N(1500). I am not using .NET 4.0. I cannot use the BigInteger class to solve this.
Can this be solved by some other algorithm or procedure? Please help.
I want to do some thing like this Calculate the factorial of an arbitrarily large number, showing all the digits but in C#. However I am unable to solve.
回答1:
There is no special magic that allows you to calculate the sum of the digits, as far as I am concerned.
It shouldn't be that hard to create your own BigInteger class anyway - you only need to implement the long multiplication algorithm from 3rd grade.
回答2:
If your goal is to calculate the sum of the digits of N!, and if N is reasonably bounded, you can do the following without a BigInteger type:
- Find a list of factorial values online (table lookup will be much more efficient than calculating from scratch, and does not require
BigInteger) - Store as a string data type
- Parse each character in the string as an integer
- Add the resulting integers
回答3:
There are two performance shortcuts that you can use for whatever implementation you choose.
- Chop off any zeros from the numbers.
- If the number is evenly divisible by 5^n, divide it by 10^n.
in this way,
16*15*14*13*12*11*10*9*8*7*6*5*4*3*2 = 20,922,789,888,000
//-->
16*1.5*14*13*12*11*1*9*8*7*6*0.5*4*3*2 = 20,922,789,888 //Sum of 63
Also, it feels like there should be some algorithm without reverting to calculating it all out. Going to 18!, the sums of the digits are:
2,6,6,3,9,9,9,27,27,36,27,27,45,45,63,63,63
//the sums of the resulting digits are:
2,6,6,3,9,9,9,9,9,9,9,9,9,9,9,9,9
and notably, the sum of the digits of 1500! is 16749 (the sum of whose digits are 27)
回答4:
Here's some working code. Some components can be improved upon to increase efficiency. The idea is to use whatever multiplication algorithm I was told in school, and to store long integers as strings.
As an afterthought, I think it would be smarter to represent large numbers with List<int>() instead of string. But I'll leave that as an exercise to the reader.
Code Sample
static string Mult(string a, string b)
{
int shift = 0;
List<int> result = new List<int>();
foreach (int aDigit in a.Reverse().Select(c => int.Parse(c.ToString())))
{
List<int> subresult = new List<int>();
int store = 0;
foreach (int bDigit in b.Reverse().Select(c => int.Parse(c.ToString())))
{
int next = aDigit*bDigit + store;
subresult.Add(next%10);
store = next/10;
}
if (store != 0) subresult.Add(store);
subresult.Reverse();
for (int i = 0; i < shift; ++i) subresult.Add(0);
subresult.Reverse();
int newResult = new List<int>();
store = 0;
for (int i = 0; i < subresult.Count; ++i)
{
if (result.Count >= i + 1)
{
int next = subresult[i] + result[i] + store;
if (next >= 10)
newResult.Add(next % 10);
else newResult.Add(next);
store = next / 10;
}
else
{
int next = subresult[i] + store;
newResult.Add(next % 10);
store = next / 10;
}
}
if (store != 0) newResult.Add(store);
result = newResult;
++shift;
}
result.Reverse();
return string.Join("", result);
}
static int FactorialSum(int n)
{
string result = "1";
for (int i = 2; i <= n; i++)
result = Mult(i.ToString(), result);
return result.Sum(r => int.Parse(r.ToString()));
}
Code Testing
Assuming the code snippet above is in the same class as your Main method, call it thusly.
Input
static void Main(string[] args)
{
Console.WriteLine(FactorialSum(1500));
}
Output
16749
回答5:
Here's a port of the C++ code you reference in one of your comments. One thing to realize when porting from C++ to C# is that integers that are zero evaluate to false and integers that are non-zero evaluate to true when used in a Boolean comparison.
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace ArbitraryFactorial
{
class Program
{
const int max = 5000;
static void display(int[] arr)
{
int ctr = 0;
for (int i = 0; i < max; i++)
{
if (ctr == 0 && arr[i] != 0) ctr = 1;
if (ctr != 0)
Console.Write(arr[i]);
}
}
static void factorial(int[] arr, int n)
{
if (n == 0) return;
int carry = 0;
for (int i = max - 1; i >= 0; --i)
{
arr[i] = (arr[i] * n) + carry;
carry = arr[i] / 10;
arr[i] %= 10;
}
factorial(arr, n - 1);
}
static void Main(string[] args)
{
int[] arr = new int[max];
arr[max - 1] = 1;
int num;
Console.Write("Enter the number: ");
num = int.Parse(Console.ReadLine());
Console.Write("Factorial of " + num + " is: ");
factorial(arr, num);
display(arr);
}
}
}
回答6:
you can find the source code at : http://codingloverlavi.blogspot.in/2013/03/here-is-one-more-interesting-program.html
#include<stdio.h>
#include<conio.h>
#include<iostream.h>
#include<time.h>
#define max 5000
void multiply(long int *,long int);
void factorial(long int *,long int);
int main()
{
clrscr();
cout<<"PROGRAM TO CALCULATE FACTORIAL OF A NUMBER";
cout<<"\nENTER THE NUMBER\n";
long int num;
cin>>num;
long int a[max];
for(long int i=0;i<max;i++)
a[i]=0;
factorial(a,num);
clrscr();
//PRINTING THE FINAL ARRAY...:):):)
cout<<"THE FACTORIAL OF "<<num<<" is "<<endl<<endl;
long int flag=0;
int ans=0;
for(i=0;i<max;i++)
{
if(flag||a[i]!=0)
{
flag=1;
cout<<a[i];
ans=ans+a[i];
}
}
cout<<endl<<endl<<"the sum of all digits is: "<<ans;
getch();
return 1;
}
void factorial(long int *a,long int n)
{
long int lavish;
long int num=n;
lavish=n;
for(long int i=max-1;i>=0&&n;i--)
{
a[i]=n%10;
n=n/10;
}
for(i=2;i<(lavish);i++)
{
multiply(a,num-1);
num=num-1;
}
}
void multiply(long int *a,long int n)
{
for(long int i=0;i<max;i++)
a[i]=a[i]*n;
for(i=max-1;i>0;i--)
{
a[i-1]=a[i-1]+(a[i]/10);
a[i]=a[i]%10;
}
}
回答7:
You can't use these numbers at all without a BigInteger type.
No algorithm or procedure can squeeze numbers larger than 264 into a long.
You need to find a BigInteger implementation for .Net 3.5.
来源：https://stackoverflow.com/questions/6905286/sum-of-factorials-for-large-numbers