akoj-1053-大数阶乘

大数阶乘

Time Limit:3000MS? Memory Limit:65535K
Total Submit:154 Accepted:64

Description

我们都知道如何计算一个数的阶乘，可是，如果这个数很大呢，我们该如何去计算它并输出它？

Input

输入一个整数m( 0 < m < =5000)

Output

输出m的阶乘，并在输出结束之后输入一个换行符

Sample Input

50

Sample Output

30414093201713378043612608166064768844377641568960512000000000000

Hint

C 用数组
Java 用BigInteger

Source

NYIST

#include<stdio.h>
int a[1000001];
int main()
{

    int n;
    scanf("%d",&n);
    int len=1;
    a[1]=1;
    for(int i=2; i<=n; ++i)
    {
        int b=0;
        for(int j=1; j<=len; ++j)
        {
            int t=a[j]*i+b;
            a[j]=t%10;
            b=t/10;
            if(j==len&&b!=0)
                len++;
        }
    }
    for(int i=len; i>0; --i)
        printf("%d",a[i]);
    printf("n");
}

#include<stdio.h>
#include<string.h>
#define N 1200
int A[N];
int main()
{
    int n,i,j,temp,start,sc;
    while(1==scanf("%d",&n))
    {
        memset(A,sizeof(A));
        for(A[0]=1,i=1; i<=n; i++)
            for(sc=0,j=0; j<N; j++)
                temp=A[j]*i+sc,sc=temp/10,A[j]=temp%10;
        for(start=N-1; A[start]==0; --start);
        for(; start>=0; --start)
            printf("%d",A[start]);
        printf("n");
    }
}

 
#include<iostream>//最优代码
#include<iomanip>
using namespace std;
//存储20000以内的阶乘
int a[15470];
int main()
{
	//freopen("1.txt","r",stdin);
	//freopen("2.txt","w",stdout);
	int n;
	cin>>n;
	a[1]=1;
	a[0]=1;
	int up;
	for(int i=2;i<=n;++i)
	{
		up=0;
		for(int j=1;j<=a[0];++j)   //各个位相乘
		{
			a[j] *=i;
			a[j] +=up;
			up=a[j]/100000;
			a[j] %=100000;
		}
		if(up!=0)
		{
			a[0]++;
			a[a[0]]=up;
		}
	}
	if(a[0]==1) cout<<a[1];
		else
		{
			cout<<a[a[0]];
			for (int i=a[0]-1;i>0;i--)
			{
				cout<<setfill('0')<<setw(5)<<a[i];
			}
		}
}        

import java.math.BigInteger;
import java.util.Scanner;

public class ny28 {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int a = input.nextInt();
BigInteger s = BigInteger.valueOf(1);
for ( int i = 1;i<=a ; i++)
{
s = s.multiply(BigInteger.valueOf(i));
}
System.out.println(s);
}
}