斐波那契数列的形式为;
1,1,2,3,5,8,13,21,34,55,...
其通项为:
F(0)=1;
F(1)=1;
F(n)=F(n-1)+F(n-2)
C++程序:
/* 案例7 斐波那契数列2 求和 */ #include <iostream> using namespace std; int fib(int n) { if(n==0) { return 0; } else if(n==1) { return 1; } return fib(n-1)+fib(n-2); } int main() { int n,sum=0; cout<<"请输入待求的斐波那契数列项数n: "; cin>>n; cout<<"斐波那契数列的第"<<n<<"项是:"<<fib(n)<< endl; for (int i=1; i<=n; i++) { cout<<fib(i)<<"\t"; sum+=fib(i); } cout<<endl; cout<<"和:"<<sum<<endl; return 0; }