1.递归和非递归分别实现求第n个斐波那契数
#include <stdio.h>
#include <stdlib.h>
int fib1(int n){
int c = 0;
int a = 1;
int b = 1;
if (n <= 2) {
return 1;
}
while (n > 2) {
c = a + b;
a = b;
b = c;
n--;
}
return c;
}
int fib2(int n){
if (n <= 2)
return 1;
else
return fib2(n - 1) + fib2(n - 2);
}
int main(){
int n = 20;
printf("%d\n", fib1(n));
printf("%d\n", fib2(n));
system("pause");
return 0;
}
2.编写一个函数实现n^k,使用递归实现
#include <stdio.h>
#include <stdlib.h>
int fun(int n, int k)
{
if (k == 0)
{
return 1;
}
return n * fun(n, k - 1);
}
int main()
{
int n = 7;
int k = 3;
printf("n^k=%d\n", fun(n, k));
system("pause");
return 0;
}
3. 写一个递归函数DigitSum(n),输入一个非负整数,返回组成它的数字之和
#include <stdio.h>
#include <stdlib.h>
int DigitSum(int n){
if (n == 0) {
return 0;
}
return n % 10 + fun(n / 10);
}
int main(){
int n = 1729;
printf("%d\n", DigitSum(n));
system("pause");
return 0;
}