斐波納契數列是這樣的數列:
f1 = 1
f2 = 1
f3 = 2
f4 = 3
....
fn = fn-1 + fn-2
輸入一個整數n
求fn
一個整數n, n<= 40
一個整數fn
3
2
n<=40
#include<iostream>
using namespace std;
int main()
{
int f[41],n;
f[1]=1;
f[2]=1;
cin>>n;
for (int i=3;i<=n;i++)
f[i]=f[i-1]+f[i-2];
cout<<f[n];
return 0;
}