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  1. #include <stdio.h>
  2.  
  3. // 方法１
  4. int fib1(int n) {
  5.     if (n == 0) return 0;
  6.     if (n == 1) return 1;
  7.  
  8.     int a = 0, b = 1, c, i;
  9.     for (i = 2; i <= n; i++) {
  10.         c = a + b;
  11.         a = b;
  12.         b = c;
  13.     }
  14.     return b;
  15. }
  16.  
  17. // 方法２
  18. int fib2(int n) {
  19.     if (n == 0) return 0;
  20.  
  21.     int fib[n+1];
  22.     fib[0] = 0;
  23.     fib[1] = 1;
  24.  
  25.     int i;
  26.     for (i = 2; i <= n; i++) {
  27.         fib[i] = fib[i-1] + fib[i-2];
  28.     }
  29.     return fib[n];
  30. }
  31.  
  32. // 方法３
  33. int fib3(int n) {
  34.     if (n == 0) return 0;
  35.     if (n == 1) return 1;
  36.     return fib3(n-1) + fib3(n-2);
  37. }
  38.  
  39. int main(void) {
  40.     int n;
  41.  
  42.     printf("非負の整数 n を入力してください: ");
  43.     scanf("%d", &n);
  44.  
  45.     printf("fib1(%d) = %d（変数のみ）\n", n, fib1(n));
  46.     printf("fib2(%d) = %d（配列使用）\n", n, fib2(n));
  47.     printf("fib3(%d) = %d（再帰関数）\n", n, fib3(n));
  48.  
  49.     return 0;
  50. }
Success #stdin #stdout 0s 5324KB
comments (?)
stdin 
6
stdout 
非負の整数 n を入力してください: fib1(6) = 8（変数のみ）
fib2(6) = 8（配列使用）
fib3(6) = 8（再帰関数）
https://ideone.com/lweC4m
language:
C (gcc 8.3)
created:
15 days ago
visibility:
public

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