BSGS及其拓展

BSGS

介绍

这是一个求解 $\equiv b \pmod p$ ，的方法。并且 $p$ 是质数， $a, p$ 互质，费马小定理可知，这个式子有周期性，

我们一般取 $m = s q r t (p)$ ，假设 $x = i * m + j ， 0 < = i, j < = m$ ，则有

$\equiv b \pmod p$

$\equiv \frac {b} {a ^ j} \pmod p$

为了方便我们设 $x = i * m - j$ ，则有

$\equiv b * a ^ {j} \pmod p$

所以我们只要通过一次枚举 $j$ ，记录下 $b * a ^ j$ ，再一次枚举 $i$ 去刚才枚举过的 $j$ 中寻找有没有符合要求的答案即可，整体复杂度是 $p\sqrt {p}$ 的

P2485 [SDOI2011]计算器

模板题

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>// #include <cstdio>
// #include <iostream>
// #include <stdlib.h>
// #include <algorithm>
// #include <cmath>#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-')    f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = (ans * a) % mod;n >>= 1;a = (a * a) % mod;}return ans;
}ll quick_mult(ll a, ll b, ll mod) {ll ans = 0;while(b) {if(b & 1) ans = (ans + a) % mod;b >>= 1;a = (a + a) % mod;}return ans;
}ll ex_gcd(ll a, ll b, ll & x, ll & y) {if(!b) {x = 1, y = 0;return a;}ll gcd = ex_gcd(b, a % b, x, y);ll temp = x;x = y;y = temp - a / b * y;return gcd;
}void BSGC(ll a, ll b, ll p) {map<ll, ll> MP;int m = sqrt(p) + 1;ll x = b, nex = quick_pow(a, m, p);for(int i = 0; i <= m; i++) {MP[x] = i;x = (x * a) % p;}x = 1;if(nex == 0) {if(b == 0) {puts("0");}else {puts("Orz, I cannot find x!");}return ;}for(int i = 0; i <= m; i++) {if(MP.count(x)) {printf("%d\n", i * m - MP[x]);return ;}x = (x * nex) % p;}puts("Orz, I cannot find x!");
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int n = read(), k = read();for(int i = 1; i <= n; i++) {ll a = read(), b = read(), p = read();if(k == 1) {printf("%lld\n", quick_pow(a, b, p));}else if(k == 2) {ll x, y;ll gcd = ex_gcd(a, p, x, y);if(b % gcd) {puts("Orz, I cannot find x!");}else {b /= gcd;x = (((x % p + p) % p) * b) % p;printf("%lld\n", x);}}   else {BSGC(a, b % p, p);}}return 0;
}

2019牛客暑期多校训练营（第五场）C generator 2

思路

$x_0 = x_0$

$x_1 = a * x_0 * b$

$x_2 = a * x_1 + b = a ^{2} * x_0 + a * b + b$

容易发现后项是一个等比数列求和

$xn=anx0+b(1−an)1−ax_n = a ^ {n} x_0 + \frac {b (1 - a ^ n)} {1 - a}$

我们要求 $x_n = v$ ，化简

$x_0 + \frac {b (1 - a ^ n)} {1 - a} = v$

$a ^{n} x_0(1 - a) + b (1 - a ^ n) = v(1 - a)$

$a ^ n (x_0 - ax_0 - b) = v(1 - a) - b$

$an=v(1−a)−bx0−ax0−ba^n =\frac {v (1 - a) - b} {x_0 - a x_0 - b}$

上面式子都是 $modp\mod p$ 下的同余等式，为了方便写了 $=$

看到这里就简单了，我们要求解的是 $n$ ，显然右边这一坨都是已知的，我们假定为 $B$ ，求解 $a ^ n = B$ ，这不就是个裸题了吗。

这道题目还要稍加分类讨论一下：

a == 0

$x_0 = x_0, x_i = b(i >= 1)$

$a = = 1$

因为这种情况上面不能直接相除，所以我们需要特殊考虑 $a_i = b + i * a$

也就是求解 $i * a = v - b$ ，这个时候只要左右两边同时乘以 $a$ 的逆元即可得到我们要的 $i$

代码

/*Author : lifehappy
*/
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include <bits/stdc++.h>
#define mp make_pair
#define pb push_back
#define endl '\n'using namespace std;typedef long long ll;
typedef unsigned long long ull;
typedef pair<int, int> pii;const double pi = acos(-1.0);
const double eps = 1e-7;
const int inf = 0x3f3f3f3f;inline ll read() {ll f = 1, x = 0;char c = getchar();while(c < '0' || c > '9') {if(c == '-') f = -1;c = getchar();}while(c >= '0' && c <= '9') {x = (x << 1) + (x << 3) + (c ^ 48);c = getchar();}return f * x;
}void print(ll x) {if(x < 10) {putchar(x + 48);return ;}print(x / 10);putchar(x % 10 + 48);
}ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = (ans * a) % mod;a = (a * a) % mod;n >>= 1;}return ans;
}int main() {// freopen("in.txt", "r", stdin);// freopen("out.txt", "w", stdout);// ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);int T = read();while(T--) {ll n = read(), x0 = read(), a = read(), b = read(), p = read();ll x = 1, Unit = quick_pow(a, 1000, p);unordered_map<ll, int> MP;for(int i = 1; i <= 1000000; i++) {x = (x * Unit) % p;if(!MP.count(x)) {MP[x] = i * 1000;}}ll inv = quick_pow(((x0 - a * x0 - b) % p + p) % p, p - 2, p);int t = read();while(t--) {ll v = read();if(a == 0) {if(v % p == x0 % p) {puts("0");}else if(v % p == b % p) {puts("1");}else {puts("-1");}continue;}if(a == 1) {ll ans = (((((v - x0) % p + p) % p) * quick_pow(b, p - 2, p))) % p;if(ans < n) {printf("%lld\n", ans);}else {puts("-1");}continue;}v = (((v * (1 - a) - b) % p + p) % p * inv) % p;if(Unit == 0) {if(v == 0) {puts("0");}else {puts("-1");}continue;}x = v;int ans = p + 1;for(int i = 0; i <= 1000; i++) {if(MP.count(x)) {ans = min(ans, MP[x] - i);}x = (x * a) % p;}if(ans != p + 1 && ans < n) {printf("%d\n", ans);}else {puts("-1");}}}return 0;
}

BSGS拓展

介绍

求解 $\equiv b \pmod p$ ，但是 $a, b$ 不互质，这里就要用到我们的 $E X B S G S$ 了。

假定 $a ^ x + py = b$

$d1=gcd(a,p)，如果有解则一定：gcd(a,p)∣bd_1 = gcd(a, p)，如果有解则一定：gcd(a, p) \mid b$

得到 $^{x - 1}\frac{a} {d_1} + \frac{p}{d_1}y = \frac {b} {d_1}$

如果 $\frac{p} {d_1}) != 1$ ，继续化简

$d2=gcd(a,pd1)−>ax−2a2d1d2+pd1d2y=bd1d2d_2 = gcd(a, \frac{p} {d_1})->a ^{x - 2} \frac{a ^ 2} {d_1d_2} + \frac{p} {d_1d_2}y = \frac{b}{d_1d_2}$

如果 $\frac{p ^ 2} {d_1d_2}) != 1$

重复上面操作，最后得到式子

$^{x - n} \frac {a ^ n} {d_1d_2……d_n} + \frac{p}{d_1d_2……d_n}y = \frac{b}{d_1d_2……d_n}$

记 $\frac{a^n}{d_1d_2……d_n}, P = \frac{p}{d_1d_2……d_n}, B = \frac{b} {d_1d_2……d_n}$

则变成里求 $Ax−n≡BA′−1(modP)A^{x - n} \equiv B A'^{-1} \pmod{P}$

记录进行了多少次 $g c d$ 求解，通过求 $A^{'}$ 的逆元化简式子，再通过一次 $B S G S$ 求得 $x - n$ ，即可得到我们得答案 $x$

P4195 【模板】扩展BSGS

一定注意求逆元不能用费马小定理，我就入了这个坑。

/*Author : lifehappy
*/
#include <bits/stdc++.h>using namespace std;typedef long long ll;ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;
}ll quick_pow(ll a, ll n, ll mod) {ll ans = 1;while(n) {if(n & 1) ans = (ans * a) % mod;a = (a * a) % mod;n >>= 1;}return ans;
}int exgcd(int a, int b, int & x, int & y) {if(!b) {x = 1, y = 0;return a;}int gcd = exgcd(b, a % b, x, y);int temp = x;x = y;y = temp - a / b * y;return gcd;
}int inv(int a, int b) {int x, y;exgcd(a, b, x, y);return (x % b + b) % b;
}int BSGS(ll a, ll b, ll p) {int m = sqrt(p) + 1;unordered_map<int, int> mp;int x = b;for(int i = 0; i <= m; i++) {mp[x] = i;x = (1ll * x * a) % p;}x = 1;int Unit = quick_pow(a, m, p);if(Unit == 0) {if(b == 0) {return 0;}else {return -1;}}for(int i = 1; i <= m; i++) {x = (1ll * x * Unit) % p;if(mp.count(x)) {return i * m - mp[x];}}return -1;
}void EXBSGS(ll a, ll b, ll p) {a %= p, b %= p;if(b == 1 || p == 1) {puts("0");return ;}int cnt = 0, d, ad = 1;while((d = gcd(a, p)) != 1) {if(b % d) {puts("No Solution");return ;}cnt++;b /= d, p /= d;ad = (1ll * ad * a / d) % p;if(ad == b) {printf("%d\n", cnt);return ;}}int ans = BSGS(a % p, (1ll * b * inv(ad, p)) % p, p);if(ans == -1) {puts("No Solution");}else {printf("%d\n", ans + cnt);}
}int main() {// freopen("in.txt", "r", stdin);ll a, b, p;while(scanf("%lld %lld %lld", &a, &p, &b) && (a || b || p)) {EXBSGS(a, b, p);}return 0;
}