一道水题，就是别忘了开long long和对32767取模，然后上代码：
#include <iostream> #include <cstdio> typedef long long ll; using namespace std; int main() { freopen("pell.in", "r", stdin); freopen("pell.out", "w", stdout); int n, k; ll a[1000000 + 5] = {1, 2}; for (int i = 2; i < 1e6 + 5; i++) { a[i] = (2 * a[i - 1] + a[i - 2]) % 32767; } cin >> n; while (n--) { cin >> k; cout << a[k - 1] << endl; } return 0; }