/*
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")

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