This documentation is automatically generated by online-judge-tools/verification-helper
This project is maintained by tsutaj
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#include "../math_017_modint.cpp"
#include "../math_018_ragrange_polynomial.cpp"
// Verified on Apr 23, 2019
// Educational Codeforces Round 63 E: Guess the Root
// Judge: https://codeforces.com/contest/1155/problem/E
// TLE が厳しいので逆元は前計算しないと通りません
// あと f(0) から f(N) までわかっていれば O(N) でできるのでそちらの戦略のほうがよいかも
void ECR065_E() {
const int MOD = 1000003;
using mint = ModInt<MOD>;
const int DEG = 11;
auto eval_expr = [&](vector<mint> &expr, mint x) {
mint base(1), res(0);
for(int i=0; i<DEG; i++) {
res += expr[i] * base;
base *= x;
}
return res;
};
vector<mint> x, y, expr;
vector<int> raw_expr = {1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0};
for(int i=0; i<DEG; i++) {
expr.emplace_back(mint(raw_expr[i]));
}
for(int i=1; i<=DEG; i++) {
x.push_back(mint(i));
cout << "? " << i << endl;
int res;
cin >> res;
// res = eval_expr(expr, i).v;
y.push_back(mint(res));
}
LagrangePolynomial<mint> lp(x, y, MOD);
for(int i=0; i<MOD; i++) {
if(lp.interpolate(i) == mint(0)) {
cout << "! " << i << endl;
return;
}
}
cout << "! " << -1 << endl;
}
// Verified (部分点, N <= 3000) on Apr 23, 2019
// AtCoder Regular Contest 033 F (旧 D): 見たことのない多項式
// Judge: https://arc033.contest.atcoder.jp/tasks/arc033_4
void ARC033_F_PART() {
const int MOD = 1000000007;
using mint = ModInt<MOD>;
int N; cin >> N;
if(N > 3000) exit(1);
vector<mint> x(N+1);
vector<mint> y(N+1);
for(int i=0; i<=N; i++) {
x[i] = mint(i);
cin >> y[i];
}
int T; cin >> T;
LagrangePolynomial<mint> lp(x, y);
cout << lp.interpolate(T) << endl;
}
int main() {
// ECR065_E();
ARC033_F_PART();
return 0;
}
#line 1 "math/verify/verify_math_018_ragrange_polynomial.cpp"
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
#line 1 "math/math_017_modint.cpp"
// ModInt begin
using ll = long long;
template<ll mod>
struct ModInt {
ll v;
ll mod_pow(ll x, ll n) const {
return (!n) ? 1 : (mod_pow((x*x)%mod,n/2) * ((n&1)?x:1)) % mod;
}
ModInt(ll a = 0) : v((a %= mod) < 0 ? a + mod : a) {}
ModInt operator+ ( const ModInt& b ) const {
return (v + b.v >= mod ? ModInt(v + b.v - mod) : ModInt(v + b.v));
}
ModInt operator- () const {
return ModInt(-v);
}
ModInt operator- ( const ModInt& b ) const {
return (v - b.v < 0 ? ModInt(v - b.v + mod) : ModInt(v - b.v));
}
ModInt operator* ( const ModInt& b ) const {return (v * b.v) % mod;}
ModInt operator/ ( const ModInt& b ) const {return (v * mod_pow(b.v, mod-2)) % mod;}
bool operator== ( const ModInt &b ) const {return v == b.v;}
bool operator!= ( const ModInt &b ) const {return !(*this == b); }
ModInt& operator+= ( const ModInt &b ) {
v += b.v;
if(v >= mod) v -= mod;
return *this;
}
ModInt& operator-= ( const ModInt &b ) {
v -= b.v;
if(v < 0) v += mod;
return *this;
}
ModInt& operator*= ( const ModInt &b ) {
(v *= b.v) %= mod;
return *this;
}
ModInt& operator/= ( const ModInt &b ) {
(v *= mod_pow(b.v, mod-2)) %= mod;
return *this;
}
ModInt pow(ll x) { return ModInt(mod_pow(v, x)); }
// operator int() const { return int(v); }
// operator long long int() const { return v; }
};
template<ll mod>
ModInt<mod> pow(ModInt<mod> n, ll k) {
return ModInt<mod>(n.mod_pow(n.v, k));
}
template<ll mod>
ostream& operator<< (ostream& out, ModInt<mod> a) {return out << a.v;}
template<ll mod>
istream& operator>> (istream& in, ModInt<mod>& a) {
in >> a.v;
return in;
}
// ModInt end
#line 1 "math/math_018_ragrange_polynomial.cpp"
ll mod_pow(ll n, ll k, ll mod) {
ll res = 1;
for(; k>0; k>>=1) {
if(k & 1) (res *= n) %= mod;
(n *= n) %= mod;
}
return res;
}
// N 未満の範囲で、i の逆元 (mod P) を配列に覚える
template <typename NumType>
vector<NumType> get_inv_table(int N, int P) {
vector<NumType> res;
for(int i=0; i<N; i++) {
res.emplace_back(NumType(mod_pow(i, P-2, P)));
}
return res;
}
// ラグランジュ補間
// 観測された (x_i, y_i) を元に多項式を補間
// 元の式が N 次式で N+1 個の点が観測されている場合は元の式が復元可能
// 観測された点を N とすると O(N^2) で動く
template <typename NumType>
struct LagrangePolynomial {
vector<NumType> x, y, f_table, inv_table;
bool use_inv_table;
LagrangePolynomial() : x(), y() {}
LagrangePolynomial(vector<NumType> x_,
vector<NumType> y_,
int P=-1)
: x(x_), y(y_), use_inv_table(P > 0) {
int N = x.size();
for(int i=0; i<N; i++) {
f_table.emplace_back(f(i, x[i]));
}
// get_inv_table がないと CE になるので、
// コピペが面倒ならここを消す (最悪)
if(use_inv_table) {
inv_table = get_inv_table<NumType>(P, P);
}
}
NumType f(int i, NumType p) {
int N = x.size();
NumType res(1);
for(int k=0; k<N; k++) {
if(i == k) continue;
res *= NumType(p - x[k]);
}
return res;
}
NumType interpolate(NumType p) {
int N = x.size();
NumType res(0);
for(int i=0; i<N; i++) {
if(use_inv_table) {
res += y[i] * f(i, p) * inv_table[ int(f_table[i]) ];
}
else {
res += y[i] * f(i, p) / f_table[i];
}
}
return res;
}
};
#line 7 "math/verify/verify_math_018_ragrange_polynomial.cpp"
// Verified on Apr 23, 2019
// Educational Codeforces Round 63 E: Guess the Root
// Judge: https://codeforces.com/contest/1155/problem/E
// TLE が厳しいので逆元は前計算しないと通りません
// あと f(0) から f(N) までわかっていれば O(N) でできるのでそちらの戦略のほうがよいかも
void ECR065_E() {
const int MOD = 1000003;
using mint = ModInt<MOD>;
const int DEG = 11;
auto eval_expr = [&](vector<mint> &expr, mint x) {
mint base(1), res(0);
for(int i=0; i<DEG; i++) {
res += expr[i] * base;
base *= x;
}
return res;
};
vector<mint> x, y, expr;
vector<int> raw_expr = {1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0};
for(int i=0; i<DEG; i++) {
expr.emplace_back(mint(raw_expr[i]));
}
for(int i=1; i<=DEG; i++) {
x.push_back(mint(i));
cout << "? " << i << endl;
int res;
cin >> res;
// res = eval_expr(expr, i).v;
y.push_back(mint(res));
}
LagrangePolynomial<mint> lp(x, y, MOD);
for(int i=0; i<MOD; i++) {
if(lp.interpolate(i) == mint(0)) {
cout << "! " << i << endl;
return;
}
}
cout << "! " << -1 << endl;
}
// Verified (部分点, N <= 3000) on Apr 23, 2019
// AtCoder Regular Contest 033 F (旧 D): 見たことのない多項式
// Judge: https://arc033.contest.atcoder.jp/tasks/arc033_4
void ARC033_F_PART() {
const int MOD = 1000000007;
using mint = ModInt<MOD>;
int N; cin >> N;
if(N > 3000) exit(1);
vector<mint> x(N+1);
vector<mint> y(N+1);
for(int i=0; i<=N; i++) {
x[i] = mint(i);
cin >> y[i];
}
int T; cin >> T;
LagrangePolynomial<mint> lp(x, y);
cout << lp.interpolate(T) << endl;
}
int main() {
// ECR065_E();
ARC033_F_PART();
return 0;
}