卡特兰数 - 白夜叉降临

卡特兰数

Catalan数

　　中文:卡特兰数

　　原理：

　　令h(1)＝1,h(0)=1，catalan数满足递归式：

　　h(n)= h(1)*h(n-1) + h(2)*h(n-2) + ... + h(n-1)h(1) (其中n>=2)

　　另类递归式：

　　h(n)=((4*n-2)/(n+1))*h(n-1);

　　该递推关系的解为：

　　h(n+1)=C(2n,n)/(n+1) (n=1,2,3,...)

Catalan数的组合公式为 Cn=C(2n,n) / (n+1);

没什么好说的作为模板就行了

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <queue>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <string>
 8 #include <iterator>
 9 #include <cmath>
10 #include <deque>
11 #include <stack>
12 #include <cctype>
13 using namespace std;
14 
15 const int N = 150;
16 const int INF = 0xfffffff;
17 
18 typedef long long ll;
19 typedef long double ld;
20 
21 #define INFL 0x7fffffffffffffffLL
22 #define met(a, b) memset(a, b, sizeof(a))
23 #define rep(c, a, b) for (int c = a; c < b; c++)
24 #define nre(c, a, b) for (int c = a; c > b; c--)
25 
26 //h(n) = ( (4*n-2) / (n+1) ) * h(n-1);
27 
28 int ans[N][N];
29 
30 void init ();
31 
32 int main ()
33 {
34     int n;
35     init();
36     while (cin >> n)
37     {
38         int len = N - 1;
39         while (ans[n][len] == 0) len--;
40         while (len>=0) cout << ans[n][len--];
41         cout << endl;
42     }
43     return 0;
44 }
45 
46 void init ()
47 {
48     ans[1][0] = 1;
49     for (int i=2; i<101; i++)
50     {
51         int t = 0;
52         for (int j=0; j<N; j++)
53         {
54             int x = ans[i-1][j] * (4 * i - 2) + t;
55             ans[i][j] = x % 10, t = x / 10;
56         }
57         t = 0;
58         for (int j=N-1; j>=0; j--)
59         {
60             int x = ans[i][j] + t * 10;
61             ans[i][j] = x / (i + 1), t = x % (i + 1);
62         }
63     }
64 }

View Code

hdu 1131 卡特兰数简单变形 n！* calatan （n）

 1 #include <iostream>
 2 #include <algorithm>
 3 #include <queue>
 4 #include <cstring>
 5 #include <cstdio>
 6 #include <vector>
 7 #include <string>
 8 #include <iterator>
 9 #include <cmath>
10 #include <deque>
11 #include <stack>
12 #include <cctype>
13 using namespace std;
14 
15 const int N = 300;
16 const int INF = 0xfffffff;
17 
18 typedef long long ll;
19 typedef long double ld;
20 
21 #define INFL 0x7fffffffffffffffLL
22 #define met(a, b) memset(a, b, sizeof(a))
23 #define rep(c, a, b) for (int c = a; c < b; c++)
24 #define nre(c, a, b) for (int c = a; c > b; c--)
25 
26 //h(n) = ( (4*n-2) / (n+1) ) * h(n-1);
27 
28 int ans[110][N];
29 
30 void init ();
31 
32 int main ()
33 {
34     int n;
35     init();
36     while (cin >> n, n)
37     {
38         int len = N - 1;
39         while (ans[n][len] == 0) len--;
40         while (len>=0) cout << ans[n][len--];
41         cout << endl;
42     }
43     return 0;
44 }
45 
46 void init ()
47 {
48     ans[1][0] = 1;
49     for (int i=2; i<111; i++)
50     {
51         int t = 0;
52         for (int j=0; j<N; j++)
53         {
54             int x = ans[i-1][j] * (4 * i - 2)* i + t;
55             ans[i][j] = x % 10, t = x / 10;
56         }
57         t = 0;
58         for (int j=N-1; j>=0; j--)
59         {
60             int x = ans[i][j] + t * 10;
61             ans[i][j] = x / (i + 1), t = x % (i + 1);
62         }
63     }
64 }

View Code

posted on 2015-03-31 17:39 白夜叉降临阅读(179) 评论(0) 编辑收藏举报