/* 思路:斐波那契数列 */ #include <iostream> using namespace std; int f(int n){ if(n == 1 || n == 2){ return 1; } int m1 = 1; int m2 = 1; int sum = 0; for(int i = 3; i <= n; ++i){ sum = m1 + m2; m1 = m2; m2 = sum; } return sum; } int main() { int n; cin >> n; cout << f(n) << endl; } // 64 位输出请用 printf("%lld")