SpecialNumbers

1. Fibonacci Number

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233,377, 610 …

Formula:

F[0]=0F[1]=1F[i]=F[i−1]+F[i−2]F[n]=(1+5)n−(1−5)n2n5=[15(1+52)n]\begin{aligned} F[0] = 0 \\ F[1] = 1 \\ F[i] = F[i-1] + F[i-2] \\ F[n] = \frac{(1+\sqrt {5})^n-(1-\sqrt {5})^n}{2^n\sqrt 5} = \left[\frac{1}{\sqrt{5}}\left(\frac{1+\sqrt{5}}{2}\right)^n\right] \end{aligned}

2. Lucas Number

1, 3, 4, 7, 11, 18, 29, 47, 76, 123...

Formula:

L[n]=(1+52)n+(1−52)nL[n] = \left( \frac{1+\sqrt {5}}{2}\right)^n + \left( \frac{1-\sqrt {5}}{2}\right)^n

3. Catalan Number

1, 2, 5, 14, 42, 132, 429, 1430, 4862, 16796, 58786,208012…

Formula:

C[n]=C(2n,n)n+1C[n] = \frac{C(2n, n)}{n+1}

Application:

  1. 将 n + 2 边形沿弦切割成 n个三角形的不同切割数

  2. n + 1个数相乘, 给每两个元素加上括号的不同方法数

  3. n 个节点的不同形状的二叉树数(严《数据结构》P.155)

  4. 从n * n 方格的左上角移动到右下角不升路径数

4. StirlingNumber(Second Kind)

S(n, m)表示含n个元素的集合划分为m个集合的情况数

或者是n个有标号的球放到m 个无标号的盒子中, 要求无一为空, 其不同的方案数

Formula:

S(n,m)={0(m=0∣∣n<m)S(n−1,m−1)+m∗S(n−1,m)(n>m≥1)S(n,m)=1m!∑i=0m(−1)i∗C(m,i)∗(m−i)n\begin{aligned} S(n, m) = \begin{cases} 0 && (m = 0 || n < m) \\ S(n-1, m-1) + m*S(n-1, m) && (n > m \ge 1) \end{cases} \\ S(n, m) = \frac{1}{m!}\sum_{i=0}^m(-1)^i*C(m,i)*(m-i)^n \end{aligned}

Special Cases:

S(n,0)=0S(n,1)=1S(n,2)=2n−1−1S(n,3)=16(3n−3∗2n+3)S(n,n−1)=C(n,2)S(n,n)=1\begin{aligned} S(n, 0) = 0 \\ S(n, 1) = 1 \\ S(n, 2) = 2^{n-1} - 1 \\ S(n, 3) = \frac{1}{6}(3^n - 3*2^n+3) \\ S(n, n-1) = C(n, 2) \\ S(n, n) = 1 \end{aligned}

5. BellNumber

n 个元素集合所有的划分数

Formula:

B[n]=∑i=0nS(n,i)B[n] = \sum_{i=0}^n S(n, i)

6. Stirling's Approximation

n!=2πn(ne)nn! = \sqrt{2\pi n}(\frac{n}{e})^n

7. Sum of Reciprocal Approximation

EulerGamma = 0.57721566490153286060651209;

∑i=1n1i=ln⁡(n)+EulerGamma;(n→∞)\sum_{i=1}^n \frac{1}{i} = \ln(n) + EulerGamma; (n \rightarrow \infty)

8. Young Tableau

Young Tableau(杨式图表)是一个矩阵, 它满足条件:

如果格子[i, j]没有元素, 则[i+1, j]也一定没有元素

如果格子[i, j]有元素a[i, j],则[i+1, j]要么没有元素, 要么a[i+1, j] > a[i, j]

Y[n]代表n个数所组成的杨式图表的个数

Formula:

Y[1]=1Y[2]=2Y[n]=Y[n−1]+(n−1)∗Y[n−2];(n>2)\begin{aligned} Y[1] = 1 \\ Y[2] = 2 \\ Y[n] = Y[n-1] + (n-1) * Y[n-2]; (n>2) \end{aligned}

Sample:

n = 3;

  1.       整数划分

将整数n分成k份, 且每份不能为空, 任意两种分法不能相同

  1. 不考虑顺序

for(int p=1; p<=n ;p++)
     for(int i=p; i<=n ;i++)
          for(int j=k; j>=1 ;j--)
               dp[i][j] += dp[i-p][j-1];
cout<< dp[n][k] <<endl;

  1. 考虑顺序

dp[i][j] = dp[i-k][j-1]; (k=1..i)

  1. 若分解出来的每个数均有一个上限m

dp[i][j] = dp[i-k][ j-1];(k=1..m)

10. 错排公式

d[1]=0d[2]=1d[n]=(n−1)∗(d[n−1]+d[n−2])\begin{aligned} d[1] = 0 \\ d[2] = 1 \\ d[n] = (n-1) * (d[n-1] + d[n-2]) \end{aligned}
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