数论板子——自己用的
- 从“自己用的板子”中搬出
文章目录
-
-
- 1. gcd与lcm
- 2. ex_gcd
- 3. 素数筛
-
- 埃式筛
- 线性筛
- 4. 逆元
-
- 线性版
- 扩欧版
- 费马小定理版
- 5. 快速幂
- 6. 矩阵快速幂
-
- 结构体版
- vector重载运算符版
- 7. 高斯消元
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- 普通浮点数高斯消元，洛谷模板题
- 浮点数高斯约旦消元法， 洛谷模板题
- 模意义下的高斯消元法，POJ - 2065 SETI
- 异或的高斯消元法（带解决自由变元的），POJ1681 Painter's Problem
- 8. 卢卡斯定理
-
- lucas
- exlucas
- 9. 线性基
- 10. 中国剩余定理
-
- Crt
- ExCrt
- 11. 欧拉函数
-
- 求单个数的欧拉函数
- 线性复杂度O(n)
- 12. 求组合数
-
- 递推版，针对取余p为质数（如果有的话），且m较小的情况,O(m)
- 素因子版，针对取余p不为质数，其n较小的情况，O(nlogn)
- 13. FFT
- 14. FNTT
-
-
1. gcd与lcm
typedef long long ll;
//最大公因数，公约数
ll gcd(ll a, ll b, ll m = 1) { while(b) m = a % b, a = b, b = m; return a; }
//最小公倍数
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
2. ex_gcd
/*
通解为
x' = x * c / gcd + (b / gcd) * k
y' = y * c / gcd - (a / gcd) * k
*/
typedef long long ll;
// 扩展欧几里得算法核心函数
void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
if (!b) {
g = a, x = 1, y = 0;
return;
}
exgcd(b, a % b, g, y, x);
y -= x * (a / b);
}
/**
* 此函数为 求解 ax + by = c
* 返回 x的最小正整数解
* 返回 -1 说明无解
*/
ll minx(ll a, ll b, ll c) {
ll x, y, g;
exgcd(a, b, g, x, y);
if (c % g != 0) return -1;
ll t = abs(b / g);
x = (x * c / g % t + t) % t;
return x == 0 ? x + t : x; // 返回最小正整数解
}
3. 素数筛
埃式筛
const int M = 1e5 + 5;
int pri[M], cnt = 0;
bool isp[M];
// 复杂度O(nloglogn)
// true 为非素数， false 为素数
void table_prime() {
isp[0] = isp[1] = true;
for (int i = 2; i < M; i++) {
if (isp[i]) continue;
pri[cnt++] = i;
for (int j = i + i; j < M; j += i) isp[j] = true;
}
}
线性筛
// 复杂度O(n)
const int M = 1e5 + 10;
vector<int> pri;
int minp[M]; // 存放x的最小素因子
void table(int n = 1e5) { // if x >= 2 && minp[x] == x 则为素数
for (int i = 1; i <= n; ++i) minp[i] = i;
for (int i = 2; i <= n; ++i) {
if (minp[i] == i) pri.push_back(i);
for (int p : pri) {
if (p * i > n) break;
minp[p * i] = p;
if (i % p == 0) break;
}
}
}
4. 逆元
线性版
typedef long long ll;
// 时间复杂度O(n)
void fny(const int &n, ll *inv, const ll mod) {
inv[0] = inv[1] = 1;
for (ll i = 2; i <= n; i++) {
inv[i] =((mod - mod / i) * inv[mod % i]) % mod;
}
}
扩欧版
typedef long long ll;
void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
if (!b) {
g = a, x = 1, y = 0;
return;
}
exgcd(b, a % b, g, y, x);
y -= x * (a / b);
}
// ax=1(mod m)
ll inverse(ll a, ll m) {//扩展欧几里得法求逆元，返回-1代表没有逆元
ll g, x, y;
exgcd(a, m, g, x, y);
return g == 1 ? (x % m + m) % m : -1;
}
费马小定理版
//a ^ (p - 1) = 1 (mod p), p为素数
//a ^ (p - 2) = a ^ (-1) (mod p)
//a 的逆元为 a ^ (p - 2)
typedef long long ll;
ll ksm(ll a, ll b, const ll mod = 1e9 + 7) {// 返回a^b % mod
if (a == 0 && b != 0) return 0;
ll res = 1;
for (a %= mod; b; b >>= 1, a = a * a % mod)
if (b & 1) res = res * a % mod;
return res;
}
ll inverse(ll a, ll m) {
return ksm(a, m - 2, m);
}
5. 快速幂
typedef long long ll;
ll ksm(ll a, ll b, const ll mod = 1e9 + 7) {// 返回a^b % mod
if (a == 0 && b != 0) return 0;
ll res = 1;
for (a %= mod; b; b >>= 1, a = a * a % mod)
if (b & 1) res = res * a % mod;
return res;
}
6. 矩阵快速幂
结构体版
const ll MOD = 1e9 + 7;
#define MO(x) ((x) % MOD)
struct Mat {
ll mat[10][10];
int n; // n * n 阶矩阵
Mat(int n = 2) : n(n) { memset(mat, 0, sizeof mat); }//记得修改
void to_one() {
for (int i = 0; i < n; i++)
mat[i][i] = 1;
}
Mat operator*(const Mat a) const {
Mat res;
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
ll sum = 0;
for (int k = 0; k < n; k++) {
sum += MO(this->mat[i][k] * a.mat[k][j]);
}
res.mat[i][j] = MO(sum);
}
}
return res;
}
ll *operator[](int x) { return mat[x]; }
};
Mat pow_f(Mat a, ll b) { //a ^ b次幂
Mat ans;
ans.to_one();
while (b) {
if (b & 1) ans = ans * a;
a = a * a;
b >>= 1;
}
return ans;
}
vector重载运算符版
#define ll long long
#define vi vector<ll>
#define Mat vector<vi>
const int X = 6;
const ll mod = 1e9 + 7;
Mat operator* (Mat a, Mat b) {
Mat res(X, vi(X, 0));
for (int i = 0; i < X; ++i)
for (int j = 0; j < X; ++j)
for (int k = 0; k < X; ++k)
res[i][j] = (res[i][j] + a[i][k] * b[k][j] % mod) % mod;
return res;
}
// 要保证矩阵A不为0
Mat operator^ (Mat a, long long b) {
Mat res(X, vi(X, 0));
for (int i = 0; i < X; ++i) res[i][i] = 1;
while (b) {
if (b & 1) res = res * a;
a = a * a, b >>= 1;
}
return res;
}
7. 高斯消元
普通浮点数高斯消元，洛谷模板题
// 洛谷模板题
#include 浮点数高斯约旦消元法， 洛谷模板题
// 洛谷模板题
#include 模意义下的高斯消元法，POJ - 2065 SETI
// POJ - 2065 SETI
#include 异或的高斯消元法（带解决自由变元的），POJ1681 Painter’s Problem
//POJ1681 Painter's Problem
#include 8. 卢卡斯定理
lucas
const ll mod = 10007;// 注意lucas算法，mod必须为质数
ll fac[mod + 10], inv[mod + 10];
void fny(const int &n, ll *inv, const ll mod) {
fac[0] = fac[1] = inv[0] = inv[1] = 1;
for (ll i = 2; i <= n; i++) {
inv[i] =((mod - mod / i) * inv[mod % i]) % mod;
fac[i] = fac[i - 1] * i % mod;
}
}
ll comb(ll n, ll m) {
if (m > n) return 0;
return fac[n] * inv[fac[n - m] * fac[m] % mod] % mod;
}
ll lucas(ll n, ll m) { // 本函数不考虑comb的时间复杂度的话就是O(logmod)
if (m == 0) return 1;
if (n < mod) return comb(n, m);
return comb(n % mod, m % mod) * lucas(n / mod, m / mod) % mod;
}
exlucas
ll ksm(ll a, ll b, const ll mod = 1e9 + 7) {// 返回a^b % mod
if (a == 0 && b != 0) return 0;
ll res = 1;
for (a %= mod; b; b >>= 1, a = a * a % mod)
if (b & 1) res = res * a % mod;
return res;
}
void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
if (!b) {
g = a, x = 1, y = 0;
return;
}
exgcd(b, a % b, g, y, x);
y -= x * (a / b);
}
// ax=1(mod m)
ll inverse(ll a, ll m) {//扩展欧几里得法求逆元，返回-1代表没有逆元
ll g, x, y;
exgcd(a, m, g, x, y);
return g == 1 ? (x % m + m) % m : -1;
}
ll crt(ll *a, ll *b, int n) {// x % b[i] = a[i], 返回最小的x， b[i]中互质， O(nlogn)
ll mul = 1, ret = 0;
for (int i = 0; i < n; ++i) mul *= b[i];
for (int i = 0; i < n; ++i) {
ll minlcm = mul / b[i];
ll inv = inverse(minlcm, b[i]); // 求逆元
ret = (ret + minlcm * inv * a[i]) % mul;
}
return (ret + mul) % mul;
}
ll cal(ll n, ll p, ll pk) { // 计算n!中取出所有p因子后mod pk的结果
if (n == 0) return 1;
ll res = 1;
for (int i = 1; i < pk; ++i) {
if (i % p) res = res * i % pk;
}
res = ksm(res, n / pk, pk);
int len = n % pk;
for (int i = 1; i <= len; ++i) {
if (i % p) res = res * i % pk;
}
return res * cal(n / p, p, pk) % pk;
}
ll comb(ll n, ll m, ll p, ll pk) {// 计算C(n, m) % pk的结果，其中p^k = pk
if (n < m) return 0;
ll up = cal(n, p, pk), down = cal(m, p, pk) * cal(n - m, p, pk) % pk, cnt = 0;
for (ll i = n; i; i /= p) cnt += i / p;
for (ll i = m; i; i /= p) cnt -= i / p;
for (ll i = n - m; i; i /= p) cnt -= i / p;
return up * inverse(down, pk) % pk * ksm(p, cnt, pk) % pk;
}
//直接调用此函数即可
ll exlucas(ll n, ll m, ll p) { // 计算C(n, m) % p，其中p不为质数
ll b[50], a[50], len = 0, tmp = p;
for (ll i = 2; i * i <= tmp; ++i) {
if (tmp % i) continue;
b[len] = 1;
while (tmp % i == 0) b[len] *= i, tmp /= i;
a[len] = comb(n, m, i, b[len]);
len += 1;
}
if (tmp > 1) b[len] = tmp, a[len] = comb(n, m, tmp, tmp), len += 1;
return crt(a, b, len);
}
9. 线性基
struct LB {
using ll = long long;
ll d[65], cnt, num; // cnt 原序列的个数， num 基的个数
bool re;
LB () : cnt(0), num(0), re(false) { memset(d, 0, sizeof d); }
// 添加一个数x
void add(ll x) {
++cnt;
for (int i = 60; ~i && x; --i) {
if ((x >> i) & 1) {
if (d[i]) x ^= d[i];
else d[i] = x, x = 0, ++num, re = 0;
}
}
}
// 询问是否能异或出x
bool check(ll x) {
for (int i = 60; ~i && x; --i) {
if ((x >> i) & 1) {
if (d[i]) x ^= d[i];
else return true;
}
}
return false;
}
ll get_max() {
ll res = 0;
for (int i = 60; ~i; --i) {
if ((d[i] ^ res) > res) res ^= d[i];
}
return res;
}
// 求的是线性基的异或最小值，不是原序列，否则要特判是否为0
ll get_min() {
for (int i = 0; i <= 60; ++i) {
if (d[i]) return d[i];
}
}
void rebuild() {
for (int i = 0; i <= 60; ++i) {
for (int j = 0; j < i; ++j) {
if ((d[i] >> j) & 1) {
d[i] ^= d[j];
}
}
}
re = true;
}
ll k_th(ll k) {
if (!re) rebuild();
if (k == 1 && num < cnt) return 0; // 如果异或得到0
if (num < cnt) --k;
ll res = 0;
for (int i = 0; i <= 60; ++i) {
if (d[i]) {
if (k & 1) res ^= d[i];
k >>= 1;
}
}
return res;
}
};
10. 中国剩余定理
Crt
ll crt(ll *a, ll *b, int n) {// x % b[i] = a[i], 返回最小的x， b[i]中互质， O(nlogn)
ll mul = 1, ret = 0;
for (int i = 0; i < n; ++i) mul *= b[i];
for (int i = 0; i < n; ++i) {
ll minlcm = mul / b[i];
ll inv = inverse(minlcm, b[i]); // 求逆元
ret = (ret + minlcm * inv * a[i]) % mul;
}
return (ret + mul) % mul;
}
ExCrt
ll mul(ll a, ll b, ll p) { // 快速乘
if (a >= p) a %= p;
if (b >= p) b %= p;
if (p <= 1000000000) return a * b % p;
if (p <= 1000000000000ll)
return (((a * (b >> 20) % p) << 20)
+ (a * (b & ((1 << 20) - 1)))) % p;
ll d = (ll)floor(a * (long double)b / p + 0.5);
ll ret = (a * b - d * p) % p;
if (ret < 0) ret += p;
return ret;
}
void exgcd(ll a, ll b, ll &g, ll &x, ll &y) {
if (!b) {
g = a, x = 1, y = 0;
return;
}
exgcd(b, a % b, g, y, x);
y -= x * (a / b);
}
ll excrt(ll *a, ll *b, int n) {// x % b[i] = a[i], 返回最小的x，下标从0开始
ll res = a[0], minlcm = b[0], x, y, g;
for (int i = 1; i < n; ++i) {
ll c = (a[i] - res % b[i] + b[i]) % b[i];
exgcd(minlcm, b[i], g, x, y);
if (c % g != 0) return -1; // 无解
x = mul(x, c / g, b[i] / g);// 快速乘
res += x * minlcm;
minlcm *= b[i] / g;
res = (res % minlcm + minlcm) % minlcm;
}
return (res % minlcm + minlcm) % minlcm;
}
11. 欧拉函数
求单个数的欧拉函数
int euler_phi(int n) {
int ans = n;
for (int i = 2; i * i <= n; ++i) {
if (n % i == 0) {
ans = ans / i * (i - 1);
while (n % i == 0) n /= i;
}
}
if (n > 1) ans = ans / n * (n - 1);
return ans;
}
线性复杂度O(n)
const int M = 1e6 + 5;
int phi[M], minp[M];
vector<int> pri;
void table_phi(int n = 1e5) {
for (int i = 1; i <= n; ++i) minp[i] = i;
phi[1] = 1;
for (int i = 2; i <= n; ++i) {
if (minp[i] == i) pri.push_back(i), phi[i] = i - 1;
for (int p : pri) {
if (i * p > n) break;
minp[i * p] = p;
if (i % p == 0) {
phi[i * p] = p * phi[i];
break;
}
phi[i * p] = (p - 1) * phi[i];
}
}
}
12. 求组合数
递推版，针对取余p为质数（如果有的话），且m较小的情况,O(m)
ll comb(ll n, ll m) {
if (n < m) return 0;
ll ret = 1;
for (int i = 1; i <= m; ++i) ret = ret * (n - i + 1) / i;
return ret;
}
素因子版，针对取余p不为质数，其n较小的情况，O(nlogn)
// 计算分子分母的素数个数差来计算结果O(nlogn),需要用到质数表，和快速幂
ll comb(int n, int m, int p) {// 求C(n, m) % p的结果，其中n, m <= 1e6, p <= 1e9
if (n < m) return 0;
ll res = 1;
for (int i = 0; i < cnt && pri[i] <= n; ++i) {
int num = 0;
for (int j = n; j; j /= pri[i]) num += j / pri[i];
for (int j = m; j; j /= pri[i]) num -= j / pri[i];
for (int j = n - m; j; j /= pri[i]) num -= j / pri[i];
res = res * powf(pri[i], num, p) % p;
}
return res;
}
13. FFT
template <class T>
class FFT {// 时间复杂度O(nlogn), 若被卡常，可尝试手写复数
#define cpd complex<double>
constexpr static double PI = acos(-1.0);
vector<int> rev;
void fft(vector<cpd>& vc, int opt, int n) { // opt=1:FFT(系数式->点值式), opt=-1:IFFT
for (int i = 0; i < n; ++i) if (i < rev[i]) swap(vc[i], vc[rev[i]]); // 位逆序置换
for (int h = 2; h <= n; h <<= 1) { // 蝴蝶操作
cpd wn(cos(PI * 2 / h), sin(PI * 2 / h) * opt); // 主n次单位根
for (int j = 0; j < n; j += h) {
cpd w(1, 0);
for (int k = j; k < j + h / 2; ++k, w *= wn) {
cpd tmp = w * vc[k + h / 2], u = vc[k];
vc[k] = u +tmp, vc[k + h / 2] = u - tmp;
}
}
}
if (opt == 1) return;
for (int i = 0; i < n; ++i) vc[i].real(vc[i].real() / n);
}
public:
vector<T> result; // 用来存最后的答案
void convolution(vector<T>& a, vector<T> &b) { // 传入a和b两个多项式，是int就int
int Alen = a.size(), Blen = b.size(), m = 1, n = 2;
while (n < Alen + Blen) n <<= 1, m += 1;
vector<cpd> A(n, cpd(0, 0)), B(n, cpd(0, 0));
rev.resize(1, 0), result.clear();
for (int i = 1; i < n; ++i) rev.push_back((rev[i >> 1] >> 1) | (i & 1) << (m - 1));
for (int i = 0; i < n; ++i) {
if (i < Alen) A[i].real(a[i]);
if (i < Blen) B[i].real(b[i]);
}
fft(A, 1, n), fft(B, 1, n);
for (int i = 0; i < n; ++i) A[i] *= B[i];
fft(A, -1, n);
// 根据题意自己改最后要的是什么
for (int i = 0; i < Alen + Blen - 1; ++i) result.push_back(round(A[i].real()));
}
#undef cpd
};
14. FNTT
typedef long long ll;
ll ksm(ll a, ll b, const ll mod = 1e9 + 7) {// 返回a^b % mod
if (a == 0 && b != 0) return 0;
ll res = 1;
for (a %= mod; b; b >>= 1, a = a * a % mod)
if (b & 1) res = res * a % mod;
return res;
}
template <class T>
class NTT {
vector<int> rev;
constexpr static ll g = 3, gi = 332748118, mod = 998244353; // g * gi % mod = 1
void ntt(vector<ll>& vt, int opt, int n) { // opt=1:FFT, opt=-1:IFFT
for (int i = 0; i < n; ++i) if (i < rev[i]) swap(vt[i], vt[rev[i]]); // 位逆序置换
for (int i = 1; i < n; i <<= 1) {
ll gn = ksm(opt ? g : gi, (mod - 1) / (i << 1), mod); // 单位原根
for (int j = 0; j < n; j += (i << 1)) {
ll g0 = 1;
for (int k = 0; k < i; ++k, g0 = g0 * gn % mod) {
ll u = vt[j + k], tmp = g0 * vt[i + j + k] % mod;
vt[j + k] = (u + tmp) % mod;
vt[i + j + k] = (u - tmp + mod) % mod;
}
}
}
}
public:
vector<T> result; // 用来存最后的答案
void convolution(vector<T>& a, vector<T> &b) { // 传入a和b两个多项式，是int就int
int Alen = a.size(), Blen = b.size(), m = 1, n = 2;
while (n < Alen + Blen - 2) n <<= 1, m += 1;
vector<ll> A(n, 0), B(n, 0);
rev.resize(1, 0), result.clear();
for (int i = 1; i < n; ++i) rev.push_back((rev[i >> 1] >> 1) | (i & 1) << (m - 1));
for (int i = 0; i < n; ++i) {
if (i < Alen) A[i] = (a[i] + mod) % mod;
if (i < Blen) B[i] = (b[i] + mod) % mod;
}
ntt(A, 1, n), ntt(B, 1, n);
for (int i = 0; i < n; ++i) A[i] = A[i] * B[i] % mod;
ntt(A, 0, n);
ll inv = ksm(n, mod - 2, mod);
for (int i = 0; i < Alen + Blen - 1; ++i) result.push_back(A[i] * inv % mod);
}
};