BSGS算法
\(Baby Step Giant Step\)算法，即大步小步算法，缩写为\(BSGS\)
拔山盖世算法
它是用来解决这样一类问题
\(y^x = z (mod\ p)\)，给定\(y,z,p>=1\)求解\(x\)
普通的\(BSGS\)只能用来解决\(gcd(y,p)=1\)的情况
设\(x=a*m+b, m=\lceil \sqrt p \rceil, a\in[0,m), b\in[0,m)\)
那么\(y^{a*m}=z*y^{-b} (mod\ p)\)
怎么求解，为了方便，设\(x=a*m-b\)
那么\(y^{a*m}=z*y^b(mod \ p), a\in(0,m+1]\)
直接暴力辣，把右边的\(b\)枚举\([0,m)\)，算出\(z*y^b(mod \ p)\)，哈希存起来
然后左边\(a\)枚举\((0, m+1]\)，算出\(y^{a*m}(mod \ p)\)查表就行了
然后不知道为什么要用\(exgcd\)，只会\(map\)...
代码
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
template <class Int>
IL void Input(RG Int &x){
RG int z = 1; RG char c = getchar(); x = 0;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
x *= z;
}
IL void None(){
puts("Orz, I cannot find x!");
}
int p;
IL int Pow(RG ll x, RG ll y){
RG ll ret = 1;
for(; y; x = x * x % p, y >>= 1)
if(y & 1) ret = ret * x % p;
return ret;
}
map <int, int> pw;
IL void BSGS(RG int x, RG int y){
pw.clear();
if(y == 1){
puts("0");
return;
}
RG int ans = -1, m = sqrt(p) + 1, xx, s = y;
for(RG int i = 0; i < m; ++i){
pw[s] = i;
s = 1LL * s * x % p;
}
xx = Pow(x, m), s = 1;
for(RG int i = 1; i <= m + 1; ++i){
s = 1LL * s * xx % p;
if(pw.count(s)){
ans = i * m - pw[s];
break;
}
}
if(ans < 0) None();
else printf("%d\n", ans);
}
int T, k, y, z;
int main(RG int argc, RG char* argv[]){
for(Input(T), Input(k); T; --T){
Input(y), Input(z), Input(p);
if(k == 1) printf("%d\n", Pow(y, z));
else if(k == 2){
RG int d = (y % p) ? 1 : p;
if(z % d) None();
else printf("%lld\n", 1LL * Pow(y, p - 2) * z % p);
}
else{
if(y % p) BSGS(y % p, z % p);
else None();
}
}
return 0;
}
扩展BSGS
对于\(gcd(y, p)\ne1\)怎么办？
我们把它写成\(y*y^{x-1}+k*p=z, k\in Z\)的形式
根据\(exgcd\)的理论
那么如果\(y,p\)的\(gcd\)不是\(z\)的约数就不会有解
设\(d=gcd(y,p)\)
那么
\[\frac{y}{d}*y^{x-1}+k*\frac{p}{d}=\frac{z}{d}
\]
\]
递归到\(d=1\)
设之间的所有的\(d\)的乘积为\(g\)，递归\(c\)次
令\(x'=x-c, p'=\frac{p}{g},z'=\frac{z}{g}\)
那么
\[y^{x'}*\frac{y^c}{g}=z'(mod \ p')
\]
\]
那么\(BSGS\)求解就好了
代码
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
# define File(a) freopen(a".in", "r", stdin), freopen(a".out", "w", stdout)
using namespace std;
typedef long long ll;
template <class Int>
IL void Input(RG Int &x){
RG int z = 1; RG char c = getchar(); x = 0;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
x *= z;
}
map <int, int> pw;
IL int Gcd(RG int x, RG int y){
return !y ? x : Gcd(y, x % y);
}
IL int Pow(RG ll x, RG ll y, RG int p){
RG ll ret = 1;
for(; y; x = x * x % p, y >>= 1)
if(y & 1) ret = ret * x % p;
return ret;
}
int a, b, p;
IL int EX_BSGS(){
if(b == 1) return 0;
pw.clear();
RG int cnt = 0, t = 1, s, x, m;
for(RG int d = Gcd(a, p); d != 1; d = Gcd(a, p)){
if(b % d) return -1;
++cnt, b /= d, p /= d, t = 1LL * t * a / d % p;
if(b == t) return cnt;
}
s = b, m = sqrt(p) + 1;
for(RG int i = 0; i < m; ++i){
pw[s] = i;
s = 1LL * s * a % p;
}
x = Pow(a, m, p), s = t;
for(RG int i = 1; i <= m; ++i){
s = 1LL * s * x % p;
if(pw.count(s)) return i * m - pw[s] + cnt;
}
return -1;
}
int ans;
int main(RG int argc, RG char* argv[]){
for(Input(a), Input(p), Input(b); a + b + p;){
a %= p, b %= p, ans = EX_BSGS();
if(ans < 0) puts("No Solution");
else printf("%d\n", ans);
Input(a), Input(p), Input(b);
}
return 0;
}