列印 Fibonacci 數列的函式

Fibonacci 數列

f1 = 1, f2 = 1, f3 = 2, f4 = 3, .... fn = fn-1 + fn-2

疊代 (iterative) 方式

1. 正的印
   unsigned long new, fn=1, fn1=1, fn2=0; printf("%lu %lu ",fn1, fn); while (fn < 2147483648) { new = fn + fn1; fn2 = fn1; fn1 = fn; fn = new; printf("%lu", fn); }

2. 倒過來印
   int i, iCount=1; unsigned long fn[100] = {1, 1}; while (fn[iCount++] < 2147483648) fn[iCount] = fn[iCount-1] + fn[iCount-2]; for (i=0; i<iCount; i++) printf("%lu ", fn[iCount-i-1]);

遞迴 (recursive) 方式

1. 正的印
   void printRecursiveForwardFibonacci(unsigned long fn1, unsigned long fn2) { if (fn1 > 2147483647) return; printf("%lu ", fn1); printfRecursiveForwardFibonacci(fn1 + fn2, fn1); } printRecursiveForwardFibonacci(1, 0);

2. 倒過來印
   void printRecursiveBackwardFibonacci(unsigned long fn1, unsigned long fn2) { if (fn1 > 2147483647) return; printfRecursiveBackwardFibonacci(fn1 + fn2, fn1); printf("%lu ", fn1); } printRecursiveBackwardFibonacci(1, 0);

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製作日期: 98/11/30 by 丁培毅 (Pei-yih Ting)
E-mail: pyting@cs.ntou.edu.tw