Train Problem II

问题陈述:

  HDOJ Problem - 1023

 

问题解析:

  卡特兰数(Catalan)的应用

  基本性质:

  f(n) = f(1)f(n-1) + f(2)f(n-2) + ... + f(n-2)f(2) + f(n-1)f(1); 

  f(n) = C(2n, n) / (n+1) = C(2n-2, n-1) / n;

    C= (4n-2)/(n+1) Cn-1

 

代码详解:

I: C++  

 1 #include <iostream>

 2 #include <cstdio>

 3 #include <memory.h>

 4 

 5 using namespace std;

 6 

 7 #define MAX 101

 8 #define BASE 10000

 9 

10 void multiply(int a[], int len, int b) {

11     for(int i=len-1, carry=0; i>=0; i--) {

12         carry += b * a[i];

13         a[i] = carry % BASE;

14         carry /= BASE;

15     }

16 }

17 

18 void divide(int a[], int len, int b) {

19     for(int i=0, div=0; i<len; i++) {

20         div = div * BASE + a[i];

21         a[i] = div / b;

22         div %= b;

23     }

24 }

25 

26 int main()

27 {

28     int i, j, h[101][MAX];

29     memset(h[1], 0, MAX*sizeof(int));

30     for(i=2, h[1][MAX-1]=1; i<=100; i++) {

31         memcpy(h[i], h[i-1], MAX*sizeof(int));

32         multiply(h[i], MAX, 4*i-2);

33         divide(h[i], MAX, i+1);

34     }

35 

36     while(cin >> i && i>=1 && i <= 100) {

37         for(j=0; j<MAX && h[i][j]==0; j++);

38         printf("%d", h[i][j++]);

39         for(; j<MAX; j++)

40             printf("%04d", h[i][j]);

41         cout << endl;

42     }

43     return 0;

44 }

 II: Java

 1 import java.io.BufferedInputStream;

 2 import java.math.BigInteger;

 3 import java.util.Scanner;

 4 

 5 public class Main {

 6     

 7     public static void main(String[] args) {

 8         Scanner sc = new Scanner(new BufferedInputStream(System.in));

 9         int n;

10         while(sc.hasNext()) {

11             BigInteger catalan = BigInteger.valueOf(1);

12             n = sc.nextInt();

13             for(int i=1; i<=n; i++) {

14                 catalan = catalan.multiply(BigInteger.valueOf(4*i-2)).divide(BigInteger.valueOf(i+1));

15             }

16             System.out.println(catalan);

17         }

18         sc.close();

19     }

20 }

 

参考资料:

  维基百科

  CSDN

 

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