60分求救

#include <bits/stdc++.h>
using namespace std;
long long f(int n) {
    if (n == 0 || n == 1) {
        return 1;
    }
    long long prev2 = 1, prev1 = 1, current;
    for (int i = 2; i <= n; ++i) {
        current = prev1 + prev2;
        prev2 = prev1;
        prev1 = current;
    }
    return current;
}

int main() {
    int n;
    cin >> n;
    cout << f(n) << endl;
    return 0;
}

从第七个样例过不去