Several solutions for printing diamond and Fibonacci Series

1. Write a program and print * Diamond

 

Releases the number of spaces to be printed on line I and the number of numbers *, and prints the rows in sequence using the for loop.

# Include <stdio. h> // print 2 * n-1 rows in total, and print void print1 (int n) {int I, j; for (I = 1; I <= N; I ++) {// print 1 to n rows for (j = 1; j <= n-I; j ++) // print n-I spaces printf (""); For (j = 1; j <= 2 * I-1; j ++) // print 2 * I-1 * digits printf ("*"); printf ("\ n");} For (; I <2 * n; I ++) {// print n + 1 to 2 * n-1 rows, same as (2 * n-I) rows for (j = 1; j <= N-(2 * n-I); j ++) printf (""); For (j = 1; j <= 2*(2 * n-I)-1; j ++) printf ("*"); printf ("\ n") ;}} void main () {int N; // n is the number of digits on the side of the Diamond. printf ("Enter N:"); scanf ("% d", & N); print1 (n );}

 

2,The Fibonacci sequence, also known as the Golden series, refers to a series like this:

1, 1, 2, 3, 5, 8, 13, 21 ,...... In mathematics, the Fibonacci sequence is defined as follows in a recursive method: F (0) = 0, F (1) = 1, F (n) = f (n-1) + f (n-2) (N> = 2, N *). write a program and output the value of F (20.

 

Here we provide four different solutions. Note that there is a big difference between Recursion and improved recursion efficiency.

# Include <stdio. h> # include <math. h> # include <time. h> # define max 100 // recursive int F1 (int n) {If (n = 1 | n = 0) return 1; return F1 (n-1) + F1 (n-2);} // recursive version for removing repeated calculation of int F2 (int n) {// Save the array static result of the intermediate result [Max] = }; if (n = 1 | n = 0) return 1; if (result [n-1] = 0) result [n-1] = F2 (n-1 ); if (result [N-2] = 0) result [N-2] = F2 (n-2); return result [n-1] + result [N-2];} // use an array to save the intermediate result (from Chen Xiaojie) int F3 (int n) {int A [Max], I; A [1] = 1; A [0] = 1; for (I = 2; I <= N; I ++) A [I] = A [I-1] + A [I-2]; return A [n];} // iteration int F4 (int n) {int I = 2, A = 1, B = 1, sum = 1; while (I <= N) {sum = a + B; A = B; B = sum; I ++;} return sum;} void main () {long start, end; Start = clock (); printf ("F (40) = % d \ n", F1 (40); End = clock (); printf ("Time: % d Ms \ n ", end-Start); Start = clock (); printf (" F (40) = % d \ n ", F2 (40 )); end = clock (); printf ("time used: % d Ms \ n", end-Start); Start = clock (); printf ("F (20) = % d \ n ", F3 (20); End = clock (); printf (" time used: % d Ms \ n ", end-Start ); start = clock (); printf ("F (20) = % d \ n", F4 (20); End = clock (); printf ("Time: % d Ms \ n ", end-Start );}

Running result: