满足斐波那契数列
#include <iostream>
using namespace::std;
using std::cout;
using std::cin;
int main()
{
int n,a1=0,a2=1,a3;
cin >> n;
if(n == 0)
{
std::cout << a1;
}
else
{
for(int i=0; i<n; i++)
{
a3=a1+a2;
a1=a2;
a2=a3;
}
std::cout << a3;
}
return 0;
}
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