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  ;
 }