BSGS

介紹

這是一個(gè)求解$a^x\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}p)$，的方法。并且$p$是質(zhì)數(shù)，$a,p$互質(zhì)，費(fèi)馬小定理可知，這個(gè)式子有周期性，

我們一般取$m=sqrt(p)$，假設(shè)$x=i?m+j，0<=i,j<=m$，則有

$$a^{i?m+j}\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}p)$$

$$a^{i+m}\equiv \frac{b}{a^j}(\mathrm{m}\mathrm{o}\mathrm{d}p)$$

為了方便我們?cè)O(shè)$x=i?m?j$，則有

$$a^{i?m}\equiv b?a^j(\mathrm{m}\mathrm{o}\mathrm{d}p)$$

所以我們只要通過一次枚舉$j$，記錄下$b?a^j$，再一次枚舉$i$去剛才枚舉過的$j$中尋找有沒有符合要求的答案即可，整體復(fù)雜度是$\sqrt{p}$的

P2485 [SDOI2011]計(jì)算器

模板題

/*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?？褪钇诙嘈Ｓ?xùn)練營(yíng)（第五場(chǎng)）C generator 2

思路

$$x_0=x_0$$

$$x_1=a?x_0?b$$

$$x_2=a?x_1+b=a^2?x_0+a?b+b$$

容易發(fā)現(xiàn)后項(xiàng)是一個(gè)等比數(shù)列求和

$$x_n=a^n{x}_0+\frac{b(1?a^n)}{1?a}$$

我們要求$x_n=v$，化簡(jiǎn)

$$a^n{x}_0+\frac{b(1?a^n)}{1?a}=v$$

$$a^n{x}_0(1?a)+b(1?a^n)=v(1?a)$$

$$a^n(x_0?a{x}_0?b)=v(1?a)?b$$

$$a^n=\frac{v(1?a)?b}{x_0?a{x}_0?b}$$

上面式子都是$\mathrm{m}\mathrm{o}\mathrm{d}p$下的同余等式，為了方便寫了$=$

看到這里就簡(jiǎn)單了，我們要求解的是$n$，顯然右邊這一坨都是已知的，我們假定為$B$，求解$a^n=B$，這不就是個(gè)裸題了嗎。

這道題目還要稍加分類討論一下：

• a == 0

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

• $a==1$

因?yàn)檫@種情況上面不能直接相除，所以我們需要特殊考慮$a_i=b+i?a$

也就是求解$i?a=v?b$，這個(gè)時(shí)候只要左右兩邊同時(shí)乘以$a$的逆元即可得到我們要的$i$

代碼

/*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

介紹

求解$a^x\equiv b(\mathrm{m}\mathrm{o}\mathrm{d}p)$，但是$a,b$不互質(zhì)，這里就要用到我們的$EXBSGS$了。

假定$$a^x+py=b$$

$$d_1=gcd(a,p)，如果有解則一定：gcd(a,p)\mid b$$

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

如果$gcd(a,\frac{p}{d_1})!=1$，繼續(xù)化簡(jiǎn)

$$d_2=gcd(a,\frac{p}{d_1})?>a^{x?2}\frac{a^2}{d_1{d}_2}+\frac{p}{d_1{d}_2}y=\frac{b}{d_1{d}_2}$$

如果$gcd(a,\frac{p^2}{d_1{d}_2})!=1$

重復(fù)上面操作，最后得到式子

$$a^{x?n}\frac{a^n}{d_1{d}_2\dots \dots d_n}+\frac{p}{d_1{d}_2\dots \dots d_n}y=\frac{b}{d_1{d}_2\dots \dots d_n}$$

記$A=ax?n,A^{\prime }=\frac{a^n}{d_1{d}_2\dots \dots d_n},P=\frac{p}{d_1{d}_2\dots \dots d_n},B=\frac{b}{d_1{d}_2\dots \dots d_n}$

則變成里求$$A^{x?n}\equiv B{A}^{\prime ?1}(\mathrm{m}\mathrm{o}\mathrm{d}P)$$

記錄進(jìn)行了多少次$gcd$求解，通過求$A^{\prime }$的逆元化簡(jiǎn)式子，再通過一次$BSGS$求得$x?n$，即可得到我們得答案$x$

P4195 【模板】擴(kuò)展BSGS

一定注意求逆元不能用費(fèi)馬小定理，我就入了這個(gè)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; }

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