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 ( "\n" ) ;

 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 }