a,b较小时 2000 #include using namespace std; const int N = 2010, mod = 1e9 + 7; int c[N][N]; int n; int main() { cin >> n; for(int i = 0; i x >> y; cout << c[x][y] << endl; } return 0; } a,b较大时 利用乘法逆元求 a