HDU 1042 大数计算

这道题一开始就采用将一万个解的表打好的话，虽然时间效率比较高，但是内存占用太大，就MLE

这里写好大数后，每次输入一个n，然后再老老实实一个个求阶层就好

java代码:

/**
 * @(#)Main.java
 *
 *
 * @author 
 * @version 1.00 2014/12/21
 */
import java.util.*;
import java.math.*;

public class Main {
    //public static BigInteger a [] = new BigInteger[10001];
    public static void main(String [] args){
        Scanner input = new Scanner(System.in);
        int n;
        while(input.hasNext()){
            n = input.nextInt();
            BigInteger a [] = new BigInteger[2];
            a[0] = new BigInteger("1");
            for(int i = 1; i<=n ; i++){
                String s = Integer.toString(i);
                a[i&1] = new BigInteger(s);
            //        a[i].valueOf(i);
                a[i&1] = a[i&1].multiply(a[1-(i&1)]);
            }
        
            System.out.println(a[n&1]);
        }
    }
    
    
}

c++代码:

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;

typedef long long LL ;

#define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )
#define For( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )

const int M = 10000 ;

struct BigInt {
    int digit[10000] ;
    int length ;

    BigInt () {
        length = 0 ;
        memset ( digit , 0 , sizeof digit ) ;
    }

    BigInt ( LL number ) {
        length = 0 ;
        memset ( digit , 0 , sizeof digit ) ;
        while ( number ) {
            digit[length ++] = number % M ;
            number /= M ;
        }
    }

    int operator [] ( const int index ) const {
        return digit[index] ;
    }

    int& operator [] ( const int index ) {
        return digit[index] ;
    }

    BigInt fix () {
        while ( length && digit[length - 1] == 0 ) -- length ;
        return *this ;
    }

    BigInt operator + ( const BigInt& a ) const {
        BigInt c ;
        c.length = max ( length , a.length ) + 1 ;
        int add = 0 ;
        rep ( i , 0 , c.length ) {
            add += a[i] + digit[i] ;
            c[i] = add % M ;
            add /= M ;
        }
        return c.fix () ;
    }

    BigInt operator - ( const BigInt& a ) const {
        BigInt c ;
        c.length = max ( a.length , length ) ;
        int del = 0 ;
        rep ( i , 0 , c.length ) {
            del += digit[i] - a[i] ;
            c[i] = del ;
            del = 0 ;
            if ( c[i] < 0 ) {
                int tmp = ( c[i] - 1 ) / M + 1 ;
                c[i] += tmp * M ;
                del -= tmp ;
            }
        }
        return c.fix () ;
    }

    BigInt operator * ( const BigInt& a ) const {
        BigInt c ;
        c.length = a.length + length ;
        rep ( i , 0 , length ) {
            int mul = 0 ;
            For ( j , 0 , a.length ) {
                mul += digit[i] * a[j] + c[i + j] ;
                c[i + j] = mul % M ;
                mul /= M ;
            }
        }
        return c.fix () ;
    }

    void show () {
        printf ( "%d" , digit[length - 1] ) ;
        rev ( i , length - 2 , 0 ) printf ( "%04d" , digit[i] ) ;
        printf ( "
" ) ;
    }

} ;


int main ()
{
    int n;
    while (~scanf("%d" , &n)) {
        BigInt big[2];
        big[0] = BigInt(1);
        for(int i = 1 ; i<=n ; i++)
            big[i&1] = big[1-(i&1)] * BigInt(i);
        big[n&1].show();
    }
    return 0 ;
}