Jobdu 1389 - 剑指offer(16)-变态跳台阶[递推] | Acm之家


一只青蛙一次可以跳上1级台阶,也可以跳上2级……它也可以跳上n级。求该青蛙跳上一个n级的台阶总共有多少种跳法。
当台阶数为n时,可以分为以下步骤来完成:
设第一次跳的台阶数为s,跳台阶方式数为T,则:
(1)s=1时,T(n) = T(n-1)
(2)s=2时,T(n) = T(n-2)
(n)s=n时,T(n) = T(0) = 1
所以总的跳台阶方式数T可以表示为:
T(n) = T(0) + T(1) + T(2) + … + T(n-1)
由于T(0) = T(1) = 1,所以T(n) = 2^(n-1)
http://blog.csdn.net/pengyan0812/article/details/46343171
(1) 如果青蛙选择跳上1级台阶,则剩下的台阶数目是n - 1, 跳上一个n – 1级的台阶的跳法数目是f(n - 1);
(2) 如果青蛙选择跳上2级台阶,则剩下的台阶数目是n - 2, 跳上一个n – 2级的台阶的跳法数目是f(n - 2);
(3) 如果青蛙选择跳上3级台阶,则剩下的台阶数目是n - 3, 跳上一个n – 3级的台阶的跳法数目是f(n - 3);
……
(n - 1) 如果青蛙选择跳上n - 1级台阶,则剩下的台阶数目是1, 跳上一个1级的台阶的跳法数目是f(1);
(n) 如果青蛙选择跳上n级台阶,则剩下的台阶数目是0, 跳上一个0级的台阶的跳法数目是f(0)。
因此可以得知,
f(n) = f(n-1) + f(n-2) + ... + f(2) + f(1) + f(0)                                                           [1]
f(n-1) = f(n-2) + f(n-3) + ... + f(1) + f(0)                                                                 [2]
……                                                                                                                     ......
f(2) = f(1) + f(0)                                                                                                     [n-2]
f(1) = f(0)                                                                                                              [n-1]
f(0) = 1                                                                                                                  [n]
由公式[1]和公式[2]可以推出
f(n) = f(n-1) + f(n-1) = 2 * f(n-1)。
对于公式[n]:f(0) = 1,我是这样理解的:青蛙在跳上一个只有1级的台阶时,只有跳上1级台阶这一个选择,而剩余台阶数目为0,因此可以得出f(1) = f(0) = 1,所以可推出以下公式:

X. 
06while(cin>>n)
07{  long long result=(long long)pow(2,n-1);
08  cout<<result<<endl;
09}
X. Precompute
  1. ** 
  2. * 计算出青蛙跳上一个n级的台阶总共有多少种跳法 
  3. * f(n) = f(n-1) + f(n-2) + ... + f(2) + f(1) + f(0) ---->  f(n) = 2*(f(n-1))   // n >= 2 
  4. * @param jumpMethods  用于保存跳台阶的跳法数目 
  5. * @return void 
  6. */  
  7. void getNumberOfSuperJumpSteps(long long * jumpMethods)  
  8. {  
  9.     int i;  
  10.     jumpMethods[1] = 1;  
  11.     for(i = 2;i <= 50;i++)  
  12.     {  
  13.        jumpMethods[i] = 2 * jumpMethods[i - 1];  
  14.     }  
  15. }  
  1.     while(EOF != scanf("%d",&n))  
  2.     {  
  3.         printf("%lld\n",jumpMethods[n]);  
  4.     }
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