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  1. #include <stdio.h>
  2.  
  3. /* フィボナッチ数を求める関数（F₀ = F₁ = 1） */
  4. double fib(int n) {
  5.     if (n <= 1) return 1.0;
  6.  
  7.     double f0 = 1.0, f1 = 1.0, f;
  8.     for (int i = 2; i <= n; i++) {
  9.         f = f0 + f1;
  10.         f0 = f1;
  11.         f1 = f;
  12.     }
  13.     return f1;
  14. }
  15.  
  16. /* n^k を求める関数（再帰やpowなしで実装） */
  17. double power_nk(int n, int k) {
  18.     double result = 1.0;
  19.     for (int i = 0; i < k; i++) {
  20.         result *= n;
  21.     }
  22.     return result;
  23. }
  24.  
  25. /* 絶対値を求める関数（fabs を使わずに）*/
  26. double my_abs(double x) {
  27.     return (x < 0) ? -x : x;
  28. }
  29.  
  30. /* メイン関数 */
  31. int main(void) {
  32.     int k;
  33.  
  34.     printf("整数 k を入力してください: ");
  35.     scanf("%d", &k);
  36.  
  37.     int best_n = 1;
  38.     double best_diff = my_abs(fib(1) - power_nk(1, k));
  39.  
  40.     for (int n = 2; n <= 100; n++) {
  41.         double f = fib(n);
  42.         double p = power_nk(n, k);
  43.         double diff = my_abs(f - p);
  44.  
  45.         if (diff < best_diff) {
  46.             best_diff = diff;
  47.             best_n = n;
  48.         }
  49.     }
  50.  
  51.     printf("\n--- 結果 ---\n");
  52.     printf("差 |F_n - n^k| が最小になる n = %d\n", best_n);
  53.     printf("F_%d  = %.0f\n", best_n, fib(best_n));
  54.     printf("%d^%d = %.0f\n", best_n, k, power_nk(best_n, k));
  55.     printf("差     = %.0f\n", best_diff);
  56.  
  57.     return 0;
  58. }
Success #stdin #stdout 0.01s 5308KB
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Standard input is empty
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整数 k を入力してください: 
--- 結果 ---
差 |F_n - n^k| が最小になる n = 1
F_1  = 1
1^21947 = 1
差     = 0
https://ideone.com/fW66A6
language:
C (gcc 8.3)
created:
16 days ago
visibility:
public

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