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.
Copy codeThe Code is as follows:
# Include <stdio. h>
// Print 2 * n-1 rows in total and print rows by row
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 * Numbers
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.
Copy codeThe Code is as follows:
# Include <stdio. h>
# Include <math. h>
# Include <time. h>
# Deprecision MAX 100
// Recursion
Int f1 (int n)
{
If (n = 1 | n = 0)
Return 1;
Return f1 (n-1) + f1 (n-2 );
}
// Recursive version to remove repeated computations
Int f2 (int n)
{
// Save the array of intermediate results
Static result [MAX] = {1, 1 };
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 (The 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) = % dn", f1 (40 ));
End = clock ();
Printf ("time used: % d msn", end-start );
Start = clock ();
Printf ("f (40) = % dn", f2 (40 ));
End = clock ();
Printf ("time used: % d msn", end-start );
Start = clock ();
Printf ("f (20) = % dn", f3 (20 ));
End = clock ();
Printf ("time used: % d msn", end-start );
Start = clock ();
Printf ("f (20) = % dn", f4 (20 ));
End = clock ();
Printf ("time used: % d msn", end-start );
}
Running result:
