Simple recursive exercises

1 # include <stdio. h> 2 // # include <stdlib. h> 3 int jiecheng (int I) {4 // ① calculate n! 5 if (I = 1) return 1; 6 return I * jiecheng (I-1); 7} 8 int Fibonacci (int I) {9 // ② calculate the I-th Fibonacci Series 10 if (I = 1) return 1; 11 if (I = 0) return 0; 12 Return Fibonacci (I-1) + Fibonacci (I-2); 13} 14 int power (int n, int K) {15 // ③ calculate n ^ k16 if (k = 1) return N; 17 return N * power (n, k-1); 18} 19 int gcd (int m, int N) {20 // ④ use the Euclidean algorithm to calculate the maximum common approx. of M and N 21 if (M % N = 0) return N; 22 return gcd (n, m % N ); 23 24} 25 int digitsum (int n) {26 // ⑤ enter a non-negative integer and return the number Sum of 27 words if (n = 0) return 0; 28 return (N % 10) + digitsum (N/10); 29} 30 int digitalroot (int n) {31 // ⑥ integer n numeric root 32 If (n <10) return N; 33 return digitalroot (digitsum (n); 34} 35 int comb (n, R) {36 // 7 calculate the composite number C (n, R); composite number C (n, R) = (P (n, R)/R !) = (N! /(N-R )! * R !)); 37 // The following formula exists: C (n, m) = C (N, N-m) = C (N-1 m-1) + C (n-1, m) 38 If (n = r | r = 0) return 1; 39 return comb (n-1, R-1) + comb (n-1, R ); 40} 41 42 void main () {43 44 printf ("① % d \ n", jiecheng (5); 45 printf ("② % d \ n ", fibonacci (6); 46 printf ("③ % d \ n", power (2, 3); 47 printf ("④ % d \ n", gcd (18, 24 )); 48 printf ("⑤ % d \ n", digitsum (12345); 49 printf ("⑥ % d \ n", digitalroot (12345 )); 50 printf ("7%d \ n", comb (6, 4); 51 52}

Result: