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:

 1 /**
 2  * @(#)Main.java
 3  *
 4  *
 5  * @author 
 6  * @version 1.00 2014/12/21
 7  */
 8 import java.util.*;
 9 import java.math.*;
10 
11 public class Main {
12     //public static BigInteger a [] = new BigInteger[10001];
13     public static void main(String [] args){
14         Scanner input = new Scanner(System.in);
15         int n;
16         while(input.hasNext()){
17             n = input.nextInt();
18             BigInteger a [] = new BigInteger[2];
19             a[0] = new BigInteger("1");
20             for(int i = 1; i<=n ; i++){
21                 String s = Integer.toString(i);
22                 a[i&1] = new BigInteger(s);
23             //        a[i].valueOf(i);
24                 a[i&1] = a[i&1].multiply(a[1-(i&1)]);
25             }
26         
27             System.out.println(a[n&1]);
28         }
29     }
30     
31     
32 }

c++代码:

  1 #include <cstdio>
  2 #include <cstring>
  3 #include <algorithm>
  4 using namespace std ;
  5 
  6 typedef long long LL ;
  7 
  8 #define rep( i , a , b ) for ( int i = a ; i < b ; ++ i )
  9 #define For( i , a , b ) for ( int i = a ; i <= b ; ++ i )
 10 #define rev( i , a , b ) for ( int i = a ; i >= b ; -- i )
 11 
 12 const int M = 10000 ;
 13 
 14 struct BigInt {
 15     int digit[10000] ;
 16     int length ;
 17 
 18     BigInt () {
 19         length = 0 ;
 20         memset ( digit , 0 , sizeof digit ) ;
 21     }
 22 
 23     BigInt ( LL number ) {
 24         length = 0 ;
 25         memset ( digit , 0 , sizeof digit ) ;
 26         while ( number ) {
 27             digit[length ++] = number % M ;
 28             number /= M ;
 29         }
 30     }
 31 
 32     int operator [] ( const int index ) const {
 33         return digit[index] ;
 34     }
 35 
 36     int& operator [] ( const int index ) {
 37         return digit[index] ;
 38     }
 39 
 40     BigInt fix () {
 41         while ( length && digit[length - 1] == 0 ) -- length ;
 42         return *this ;
 43     }
 44 
 45     BigInt operator + ( const BigInt& a ) const {
 46         BigInt c ;
 47         c.length = max ( length , a.length ) + 1 ;
 48         int add = 0 ;
 49         rep ( i , 0 , c.length ) {
 50             add += a[i] + digit[i] ;
 51             c[i] = add % M ;
 52             add /= M ;
 53         }
 54         return c.fix () ;
 55     }
 56 
 57     BigInt operator - ( const BigInt& a ) const {
 58         BigInt c ;
 59         c.length = max ( a.length , length ) ;
 60         int del = 0 ;
 61         rep ( i , 0 , c.length ) {
 62             del += digit[i] - a[i] ;
 63             c[i] = del ;
 64             del = 0 ;
 65             if ( c[i] < 0 ) {
 66                 int tmp = ( c[i] - 1 ) / M + 1 ;
 67                 c[i] += tmp * M ;
 68                 del -= tmp ;
 69             }
 70         }
 71         return c.fix () ;
 72     }
 73 
 74     BigInt operator * ( const BigInt& a ) const {
 75         BigInt c ;
 76         c.length = a.length + length ;
 77         rep ( i , 0 , length ) {
 78             int mul = 0 ;
 79             For ( j , 0 , a.length ) {
 80                 mul += digit[i] * a[j] + c[i + j] ;
 81                 c[i + j] = mul % M ;
 82                 mul /= M ;
 83             }
 84         }
 85         return c.fix () ;
 86     }
 87 
 88     void show () {
 89         printf ( "%d" , digit[length - 1] ) ;
 90         rev ( i , length - 2 , 0 ) printf ( "%04d" , digit[i] ) ;
 91         printf ( "
" ) ;
 92     }
 93 
 94 } ;
 95 
 96 
 97 int main ()
 98 {
 99     int n;
100     while (~scanf("%d" , &n)) {
101         BigInt big[2];
102         big[0] = BigInt(1);
103         for(int i = 1 ; i<=n ; i++)
104             big[i&1] = big[1-(i&1)] * BigInt(i);
105         big[n&1].show();
106     }
107     return 0 ;
108 }