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问题陈述:
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;
Cn = (4n-2)/(n+1) Cn-1
代码详解:
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 }
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原文地址:http://www.cnblogs.com/michaelwong/p/4346692.html