Haut OJ 1245: large factorial of CDs --- high precision factorial

Problem description ：

cds： I heard that you can already use C Language seeking n! 了

ykc： Of course ！ Light yard oj I have already solved all the factorial problems AC 了 , for example 1048 Factorial table ,1050 The cumulative sum of factorials ,1089 The highest position of factorial, etc ……

cds： Oh ？ Really? , Then I'll give you a number n, Can you find its factorial immediately ？

ykc： good , That's all right. ！

cds： can n Very big

ykc： Don't worry , I use long long Just fine

cds： Good. ,n=80

ykc：&#￥%#woc

Input ：

Single instance test , Enter a natural number n（n<=2000)

Output ：

Output n The factorial

The sample input ：

80

Sample output ：

71569457046263802294811533723186532165584657342365752577109445058227039255480148842668944867280814080000000000000000000

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Cause analysis ：

1. utilize i To achieve 1 To n The factorial   ,temp Record the number obtained by multiplying each time ,digit Record the number of digits （ So that each number can be multiplied ）

2.  temp=a[j]*i+num;    num It is a number on a bit that multiplies and advances backward （num=temp/10;） The number on the bit that should be carried * i after , Need to add low and progressive num.

3. hold num/10 such as 4 The factorial And finally   4*（2*3）, namely 24, here num=2 Save to digit That is to say 1 Location

Build another while Yes, it will appear later num There are several , So use a cycle to save each .

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Solution ：

#include <stdio.h>
int main()
{
	int a[20001];// Store the number obtained by each bit  
	int temp,digit,n,i,j=0;//temp The number of each time    digit The number of digits each time   
	scanf("%d",&n);
	a[0]=1;// from 1 Start to multiply  
	digit=1;// The number of digits starts from the first  
	for(i=2;i<=n;i++)
	{
		int num=0;
		for(j=0;j<digit;j++)                   //for Purpose ： Multiply each digit of a number by i,
		{
			temp=a[j]*i+num;                     
			a[j]=temp%10;                       // Each digit of a number is stored in an array 
			num=temp/10;
		}
		while(num)           
		{
			a[digit]=num%10;                          // Continue to store the number after the last carry  
                                                      // The previous carry is in the second for Stored in      
			num=num/10;
			digit++;
		}
	}
	for(i=digit-1;i>=0;i--)// Output each bit in reverse order  
		printf("%d",a[i]);
	printf("\n");
	return 0;
}

Multi instance input ：

#include<bits/stdc++.h>
using namespace std;
#define ll long long

int main()
{
    ll i,j,n;

    while(cin>>n)
    {
        ll digit=1,a[100000];        //a Array ,a[0]=1,digit Need to put it in .

         a[0]=1;
 
       for(j=2;j<=n;j++)
       {
             ll num=0,temp;       
           for(i=0;i<digit;i++)
          {
            temp=a[i]*j+num;
            a[i]=temp%10;
            num=temp/10;

          }
          while(num)
          {
              a[digit]=num%10;
              num/=10;
              digit++;

          }
       }
        for(i=digit-1;i>=0;i--)
            printf("%d",a[i]);
        printf("\n");

    }

}