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 =  ;

 struct BigInt {

     int digit[] ;

     int length ;

     BigInt () {

         length =  ;

         memset ( digit ,  , sizeof digit ) ;

     }

     BigInt ( LL number ) {

         length =  ;

         memset ( digit ,  , 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 - ] ==  ) -- length ;

         return *this ;

     }

     BigInt operator + ( const BigInt& a ) const {

         BigInt c ;

         c.length = max ( length , a.length ) +  ;

         int add =  ;

         rep ( i ,  , 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 =  ;

         rep ( i ,  , c.length ) {

             del += digit[i] - a[i] ;

             c[i] = del ;

             del =  ;

             if ( c[i] <  ) {

                 int tmp = ( c[i] -  ) / M +  ;

                 c[i] += tmp * M ;

                 del -= tmp ;

             }

         }

         return c.fix () ;

     }

     BigInt operator * ( const BigInt& a ) const {

         BigInt c ;

         c.length = a.length + length ;

         rep ( i ,  , length ) {

             int mul =  ;

             For ( j ,  , 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 - ] ) ;

         rev ( i , length -  ,  ) printf ( "%04d" , digit[i] ) ;

         printf ( "\n" ) ;

     }

 } ;

 int main ()

 {

     int n;

     while (~scanf("%d" , &n)) {

         BigInt big[];

         big[] = BigInt();

         for(int i =  ; i<=n ; i++)

             big[i&] = big[-(i&)] * BigInt(i);

         big[n&].show();

     }

     return  ;

 }