卡特兰数
卡特兰数又称卡塔兰数,英文名Catalan number,是组合数学中一个常出现在各种计数问题中出现的数列。由以比利时的数学家欧仁·查理·卡塔兰 (1814–1894)命名,其前几项为 : 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 9694845, 35357670, 129644790, 477638700, 1767263190, 6564120420, 24466267020, 91482563640, 343059613650, 1289904147324, 4861946401452, ...
原理:
h(n)= h(0)*h(n-1)+h(1)*h(n-2) + ... + h(n-1)h(0) (n>=2)
例如:h(2)=h(0)*h(1)+h(1)*h(0)=1*1+1*1=2
h(3)=h(0)*h(2)+h(1)*h(1)+h(2)*h(0)=1*2+1*1+2*1=5
h(n)=h(n-1)*(4*n-2)/(n+1);
递推关系的解为:
h(n)=C(2n,n)/(n+1) (n=0,1,2,...)
递推关系的另类解为:
h(n)=c(2n,n)-c(2n,n+1)(n=0,1,2,...)
应用:
代码:
1 #include <algorithm> 2 #include <iostream> 3 #include <sstream> 4 #include <cstdlib> 5 #include <cstring> 6 #include <cstdio> 7 #include <string> 8 #include <bitset> 9 #include <vector> 10 #include <queue> 11 #include <stack> 12 #include <cmath> 13 #include <list> 14 #include <set> 15 using namespace std; 16 /***************************************/ 17 #define ll long long 18 #define int64 __int64 19 /***************************************/ 20 const int INF = 0x7f7f7f7f; 21 const ll LINF = (1LL<<60); 22 const double eps = 1e-8; 23 const double PIE=acos(-1.0); 24 const int d1x[]= {0,-1,0,1}; 25 const int d1y[]= {-1,0,1,0}; 26 const int d2x[]= {0,-1,0,1}; 27 const int d2y[]= {1,0,-1,0}; 28 const int fx[]= {-1,-1,-1,0,0,1,1,1}; 29 const int fy[]= {-1,0,1,-1,1,-1,0,1}; 30 inline int min_32(int (a),int (b)) 31 { 32 return (a)<(b)?(a):(b); 33 } 34 inline int max_32(int (a),int (b)) 35 { 36 return (a)>(b)?(a):(b); 37 } 38 inline long long min_64(long long (a),long long (b)) 39 { 40 return (a)<(b)?(a):(b); 41 } 42 inline long long max_64(long long (a),long long (b)) 43 { 44 return (a)>(b)?(a):(b); 45 } 46 /***************************************/ 47 void openfile() 48 { 49 freopen("data.in","rb",stdin); 50 freopen("data.out","wb",stdout); 51 } 52 /**********************华丽丽的分割线,以上为模板部分*****************/ 53 54 55 //卡特兰数 其前几项为 : 1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786, 208012, 742900, 2674440, 56 // 9694845, 35357670, 129644790, 477638700, 1767263190 57 58 59 int a[22]; 60 61 int main() 62 { 63 64 int n; 65 a[0]=1; 66 int i,j; 67 scanf("%d",&n); 68 for(i=1;i<=n;i++) 69 { 70 a[i]=(a[i-1]*(4*i-2))/(i+1); //公式 71 } 72 printf("%d\n",a[n]); 73 return 0; 74 }
屌丝终有逆袭日,*******。