用非递归的函数调用形式求斐波那契数列第n项

用非递归的函数调用形式求斐波那契数列第n项

main()
{
int n,n1;
n=scanf("%d",&n);
int a[1000];
a[0]=1;a[1]=1;
for(n1=2;n1<1000;n1++){
a[n1]=a[n1-1]+a[n1-2];
}
printf(("%4d",a[n]);

}

public class FibTest { public static void main(String[] arags){ long begin = System.currentTimeMillis(); System.out.println(fib(10)); long end = System.currentTimeMillis(); System.out.println(end - begin); } public static long fib(int n){ if(n < 3) return 1; else{ long a = 1; long b = 1; for(int i = 2 ; i < n-1 ;i++){ b = a + b; a = b - a; System.out.println("a = "+a +" b = "+b); } return a + b; } } }

^#include <iostream> using namespace std; int main() { int n,a=1,b=2; cout<<"Please input n:"<<endl; cin>>n; for(int i=0;i<n-1;i++) { a=b^a; b=b+a^b; a=a^a; } cout<<a<<endl; return 0; } 我所谓的斐波那契数列是1,2,3,5,8.。。。第一项是1的。 另外，^不是数学符号，不要include<cmath>.

#include <stdio.h> main() {     unsigned long fibonacci[100] = {0,1};     char i,n;     printf("\n第几项:");     scanf("%d",&n);     if(n > 100) n = 99;     for(i = 2;i <= n;i++) fibonacci[i] = fibonacci[i - 2] + fibonacci[i - 1];     printf("斐波那契数列的第%d项是:%u。\n\n\n",n,fibonacci[n - 1]); }