[Baby steps giant steps] BSGS EXT-BSGS
BSGS
给定y,z,p，计算满足Y^x ≡ Z ( mod P)的最小非负整数。P为质数 或者更确切的 gcd(Y,P)|Z
还是引用黄学长的吧
![[Baby steps giant steps] BSGS EXT-BSGS_第1张图片](http://img.e-com-net.com/image/info5/3eff369f57f6491a84ca52aba570ddba.jpg)
用hash显然比map优
EXT_BSGS
当模不一定为质数时
我们要对算法进行一些修改，使得其可以进行上述的处理。具体证明去看AekdyCoin犇的题解，我就是简单讲一下我的理解。
一开始的方程等价于A^x+C*b=B(a,b是整数)，现在gcd(A，C)!=1,不妨令t=gcd(A,C)
我们考虑方程(A/t)A^x’%(C/t)=(B/t)的解x’与原来的解x有什么关系。
显然现在的方程等价于(A/t)A^x’+(C/t)*b’=B/t.
两端均乘以t得到：A^(x’+1)+C*b’=B
由系数相等有:x=x’+1,b=b’.
那么我们就有一种方法算出x了。先算出x’,再加上1就行了。
考虑如何求出x’,现在的方程如果A,C依旧不互质，就继续迭代将系数除以最大公约数，同时前面的常数增大。
然后求出解之后再加上总共除的次数就好了。
BZOJ3239
裸的BSGS 同poj2417
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;
typedef long long ll;
ll A,B,P;
inline ll Pow(ll a,ll b)
{
if (!b) return 1LL;
ll ret=Pow(a,b>>1);
if (b&1) return ret*ret%P*a%P;
return ret*ret%P;
}
map<ll,int> M;
inline void solve()
{
M.clear();
A%=P;
if (!A && !B) { printf("1\n"); return; }
if (!A) { printf("no solution\n"); return; }
ll m=ceil(sqrt((double)P)),tmp=1;
M[1]=m+1;
for (int i=1;i<m;i++)
{
(tmp*=A)%=P;
if (!M[tmp]) M[tmp]=i;
}
ll inv=1,T=Pow(A,P-1-m);
for (int k=0;k<m;k++)
{
int i=M[B*inv%P];
if (i)
{
if (i==m+1) i=0;
printf("%lld\n",k*m+i);
return;
}
(inv*=T)%=P;
}
printf("no solution\n");
}
int main()
{
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
while (~scanf("%lld%lld%lld",&P,&A,&B))
solve();
return 0;
}
BZOJ2242
一堆乱七八糟的操作放在一起
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;
typedef long long ll;
inline char nc()
{
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}
inline void read(ll &x)
{
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void write(ll x,char c)
{
if (x==0) { putchar('0'); putchar(c); return; }
if (x<0) putchar('-'),x=-x;
char s[55]={0},len=0;
while (x) s[++len]=x%10+'0',x/=10;
for (int i=len;i;i--) putchar(s[i]); putchar(c);
}
namespace Work1{
ll A,B,P;
inline ll Pow(ll a,ll b)
{
if (!b) return 1LL;
ll ret=Pow(a,b>>1);
if (b&1) return ret*ret%P*a%P;
return ret*ret%P;
}
inline void work()
{
read(A); read(B); read(P);
write(Pow(A,B),'\n');
}
}
namespace Work2{
#define X first
#define Y second
typedef pair<ll,ll> data;
ll A,B,P;
ll D;
inline data EXGCD(ll a,ll b)
{
if (!b) { D=a; return data(1,0); }
data ret=EXGCD(b,a%b);
return data(ret.Y,ret.X-a/b*ret.Y);
}
inline void solve()
{
data ret=EXGCD(A,P);
if (B%D==0)
{
ll ans=((ret.X*B/D)%P+P)%P;
write(ans,'\n');
}
else
printf("Orz, I cannot find x!\n");
}
inline void work()
{
read(A); read(B); read(P);
solve();
}
}
namespace Work3{
ll A,B,P;
inline ll Pow(ll a,ll b)
{
if (!b) return 1LL;
ll ret=Pow(a,b>>1);
if (b&1) return ret*ret%P*a%P;
return ret*ret%P;
}
map<ll,int> M;
inline void solve()
{
M.clear();
A%=P;
if (!A && !B) { printf("1\n"); return; }
if (!A) { printf("Orz, I cannot find x!\n"); return; }
ll m=ceil(sqrt((double)P)),tmp=1;
M[1]=m+1;
for (int i=1;i<m;i++)
{
(tmp*=A)%=P;
if (!M[tmp]) M[tmp]=i;
}
ll inv=1,T=Pow(A,P-1-m);
for (int k=0;k<m;k++)
{
int i=M[B*inv%P];
if (i)
{
if (i==m+1) i=0;
write(k*m+i,'\n');
return;
}
(inv*=T)%=P;
}
printf("Orz, I cannot find x!\n");
}
inline void work()
{
read(A); read(B); read(P);
solve();
}
}
ll Q,K;
int main()
{
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(Q); read(K);
while (Q--)
{
if (K==1)
Work1::work();
else if (K==2)
Work2::work();
else if (K==3)
Work3::work();
}
return 0;
}
求数列通项 再用BSGS解
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
typedef long long ll;
inline char nc()
{
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}
inline void read(ll &x)
{
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void write(ll x,char c)
{
if (x==0) { putchar('0'); putchar(c); return; }
if (x<0) putchar('-'),x=-x;
char s[55]={0},len=0;
while (x) s[++len]=x%10+'0',x/=10;
for (int i=len;i;i--) putchar(s[i]); putchar(c);
}
namespace MOD{
#define X first
#define Y second
typedef pair<ll,ll> data;
ll D;
inline data EXGCD(ll a,ll b)
{
if (a<b) { data ret=EXGCD(b,a); return data(ret.Y,ret.X); }
if (!b) { D=a; return data(1,0); }
data ret=EXGCD(b,a%b);
return data(ret.Y,ret.X-a/b*ret.Y);
}
inline ll Inv(ll A,ll P)
{
data ret=EXGCD(A,P);
return (ret.X%P+P)%P;
}
inline ll Solve(ll A,ll B,ll P)
{
data ret=EXGCD(A,P);
if (B%D==0)
{
ll ans=((ret.X*B/D)%P+P)%P;
return ans;
}
else
return -1;
}
}
namespace BSGS{
inline ll Pow(ll a,ll b,ll p)
{
ll ret=1;
for (;b;b>>=1,a=a*a%p)
if (b&1)
(ret*=a)%=p;
return ret;
}
map<ll,int> M;
inline ll Solve(ll A,ll B,ll P)
{
M.clear();
A%=P;
if (!A && !B) return 1LL;
if (!A) return -1;
ll m=sqrt((double)P)+1,tmp=1;
M[1]=m+1;
for (int i=1;i<m;i++)
{
(tmp*=A)%=P;
if (!M[tmp]) M[tmp]=i;
}
ll inv=1,T=Pow(A,P-1-m,P);
for (int k=0;k<m;k++)
{
int i=M[B*inv%P];
if (i)
{
if (i==m+1) i=0;
return k*m+i;
}
(inv*=T)%=P;
}
return -1;
}
}
ll P,A,B,X,T;
ll N;
int main()
{
ll Q;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(Q);
while (Q--)
{
read(P); read(A); read(B); read(X); read(T);
if (X==T) { printf("1\n"); continue; }
if (A==0)
{
if (B==T) { printf("2\n"); continue; }
else { printf("-1\n"); continue; }
}
if (A==1)
{
N=MOD::Solve(B,((T-X+B)%P+P)%P,P);
if (N==0) N=P;
}
else
{
ll C=MOD::Inv(A-1,P);
ll AN=MOD::Solve((X+B*C)%P,(B*C+T)%P,P);
if (AN==-1)
N=-1;
else
{
N=BSGS::Solve(A,AN,P);
if (N!=-1) N++;
}
}
write(N,'\n');
}
return 0;
}
扩展BSGS
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<map>
#include<algorithm>
#include<cstring>
using namespace std;
typedef long long ll;
inline char nc()
{
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}
inline void read(ll &x)
{
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void write(ll x,char c)
{
if (x==0) { putchar('0'); putchar(c); return; }
if (x<0) putchar('-'),x=-x;
char s[55]={0},len=0;
while (x) s[++len]=x%10+'0',x/=10;
for (int i=len;i;i--) putchar(s[i]); putchar(c);
}
#define mod 1313131
struct Hashset {
int head[mod], next[35010], f[35010], v[35010], ind;
void reset() {
ind = 0;
memset(head, -1, sizeof head);
}
void insert(int x, int _v) {
int ins = x % mod;
for(int j = head[ins]; j != -1; j = next[j])
if (f[j] == x) {
v[j] = min(v[j], _v);
return;
}
f[ind] = x, v[ind] = _v;
next[ind] = head[ins], head[ins] = ind++;
}
int operator [] (const int &x) const {
int ins = x % mod;
for(int j = head[ins]; j != -1; j = next[j])
if (f[j] == x)
return v[j];
return -1;
}
}M;
#define X first
#define Y second
typedef pair<ll,ll> data;
inline data EXGCD(ll a,ll b)
{
if (a<b) { data ret=EXGCD(b,a); return data(ret.Y,ret.X); }
if (!b) { return data(1,0); }
data ret=EXGCD(b,a%b);
return data(ret.Y,ret.X-a/b*ret.Y);
}
inline ll Inv(ll A,ll P)
{
data ret=EXGCD(A,P);
return (ret.X%P+P)%P;
}
inline ll Pow(ll a,ll b,ll p)
{
ll ret=1;
for (;b;b>>=1,a=a*a%p)
if (b&1)
(ret*=a)%=p;
return ret;
}
//map<ll,int> M;
inline ll BSGS(ll A,ll B,ll P)
{
// M.clear();
M.reset();
A%=P;
if (!A && !B) return 1LL;
if (!A) return -1;
ll m=sqrt((double)P)+1,tmp=1;
M.insert(1,m+1);
for (int i=1;i<m;i++)
{
(tmp*=A)%=P;
if (M[tmp]==-1) M.insert(tmp,i);
}
ll inv=1,T=Inv(Pow(A,m,P),P);
for (int k=0;k<m;k++)
{
int i=M[B*inv%P];
if (i!=-1)
{
if (i==m+1) i=0;
return k*m+i;
}
(inv*=T)%=P;
}
return -1;
}
ll A,B,P,ans;
inline ll Calc(ll A,ll B,ll P)
{
ll tmp=1;
for (int i=0;i<100;(tmp*=A)%=P,i++)
if (tmp==B)
return i;
ll C=1,D,num=0,ret;
data G;
while (1)
{
G=EXGCD(A,P);
D=A*G.first+P*G.second;
if (D==1)
{
(B*=Inv(C,P))%=P;
ret=BSGS(A,B,P);
if (ret==-1) return -1;
return ret+num;
}
if (B%D)
return -1;
B/=D; P/=D;
(C*=A/D)%=P;
num++;
}
}
int main()
{
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
while (1)
{
read(A); read(P); read(B);
if (!A && !B && !P) break;
ans=Calc(A,B,P);
if (ans==-1)
printf("No Solution\n");
else
write(ans,'\n');
}
return 0;
}
又是一堆乱七八糟的，组合数取模参见另一篇博文
#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<map>
#include<cstring>
#define cl(x) memset(x,0,sizeof(x))
using namespace std;
typedef long long ll;
inline char nc()
{
static char buf[100000],*p1=buf,*p2=buf;
if (p1==p2) { p2=(p1=buf)+fread(buf,1,100000,stdin); if (p1==p2) return EOF; }
return *p1++;
}
inline void read(ll &x)
{
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void write(ll x,char c)
{
if (x==0) { putchar('0'); putchar(c); return; }
if (x<0) putchar('-'),x=-x;
char s[55]={0},len=0;
while (x) s[++len]=x%10+'0',x/=10;
for (int i=len;i;i--) putchar(s[i]); putchar(c);
}
inline ll Mul(ll A,ll B,ll P)
{
A%=P,B%=P;
ll ret=A*B-(ll)((double)A/P*B+0.5)*P;
return ret<0?ret+P:ret;
}
inline ll Pow(ll A,ll B,ll P)
{
ll ret=1;
for (;B;B>>=1,A=Mul(A,A,P))
if (B&1)
ret=Mul(ret,A,P);
return ret;
}
#define X first
#define Y second
typedef pair<ll,ll> data;
inline data EXGCD(ll a,ll b)
{
if (a<b) { data ret=EXGCD(b,a); return data(ret.Y,ret.X); }
if (!b) { return data(1,0); }
data ret=EXGCD(b,a%b);
return data(ret.Y,ret.X-a/b*ret.Y);
}
inline ll Inv(ll A,ll P)
{
data ret=EXGCD(A,P);
return (ret.X%P+P)%P;
}
namespace Work1{
ll A,B,P;
inline void Solve()
{
read(A); read(B); read(P);
write(Pow(A,B,P),'\n');
}
}
namespace Work2{
#define mod 1313131
struct Hashset {
int head[mod], next[35010], f[35010], v[35010], ind;
void reset() {
ind = 0;
memset(head, -1, sizeof head);
}
void insert(int x, int _v) {
int ins = x % mod;
for(int j = head[ins]; j != -1; j = next[j])
if (f[j] == x) {
v[j] = min(v[j], _v);
return;
}
f[ind] = x, v[ind] = _v;
next[ind] = head[ins], head[ins] = ind++;
}
int operator [] (const int &x) const {
int ins = x % mod;
for(int j = head[ins]; j != -1; j = next[j])
if (f[j] == x)
return v[j];
return -1;
}
}M;
inline ll BSGS(ll A,ll B,ll P)
{
M.reset();
A%=P;
if (!A && !B) return 1LL;
if (!A) return -1;
ll m=sqrt((double)P)+1,tmp=1;
M.insert(1,0);
for (int i=1;i<m;i++)
{
(tmp*=A)%=P;
if (M[tmp]==-1) M.insert(tmp,i);
}
ll inv=1,T=Inv(Pow(A,m,P),P);
for (int k=0;k<m;k++)
{
int i=M[B*inv%P];
if (i!=-1)
return k*m+i;
(inv*=T)%=P;
}
return -1;
}
ll A,B,P,ans;
inline ll Calc(ll A,ll B,ll P)
{
for (ll tmp=1,i=0;i<100;(tmp*=A)%=P,i++)
if (tmp==B)
return i;
ll C=1,D,num=0,ret;
data G;
while (1)
{
G=EXGCD(A,P);
D=A*G.first+P*G.second;
if (D==1)
{
(B*=Inv(C,P))%=P;
ret=BSGS(A,B,P);
if (ret==-1) return -1;
return ret+num;
}
if (B%D)
return -1;
B/=D; P/=D;
(C*=A/D)%=P;
num++;
}
}
inline void Solve()
{
read(A); read(B); read(P);
ans=Calc(A,B,P);
if (ans==-1)
printf("Math Error\n");
else
write(ans,'\n');
}
}
namespace Work3{
namespace China{
ll A[100005],N[100005],M[100005];
int tot=0;
ll P=1;
inline void clear(){
P=1; tot=0;
}
inline void add(ll a,ll n){
A[++tot]=a; N[tot]=n; P*=n;
}
inline ll Solve(){
ll ret=0,inv;
for (int i=1;i<=tot;i++)
{
M[i]=P/N[i];
inv=Inv(M[i],N[i]);
(ret+=Mul(A[i],Mul(M[i],inv,P),P))%=P;
}
return ret;
}
}
int vst[100005],prime[100005],num;
inline void Pre()
{
const int maxn=100000;
for (int i=2; i<=maxn; i++)
if (!vst[i])
{
prime[++num]=i;
for (int j=i+i; j<=maxn; j+=i) vst[j]=1;
}
}
ll M,ans;
ll p[50005],a[50005],p_a[50005],cnt;
ll n,m;
inline void Machine(ll M)
{
for (int i=1; i<=num; i++)
if (M%prime[i]==0) {
p[++cnt]=prime[i];
while (M%prime[i]==0) a[cnt]++,M/=prime[i];
p_a[cnt]=Pow(p[cnt],a[cnt],1000000);
}
}
inline data Calc(ll x,ll p,ll a,ll p_a)
{
ll A=1,B=1;
for (ll i=1;i<p_a;i++)
if (i%p)
(A*=i)%=p_a;
A=Pow(A,x/p_a,p_a);
for (ll i=x-x%p_a+1;i<=x;i++)
if (i%p)
(A*=i)%=p_a;
B=x/p;
if (B)
{
data tmp=Calc(B,p,a,p_a);
(A*=tmp.first)%=p_a;
B+=tmp.second;
}
return data(A,B);
}
inline void Solve()
{
China::clear();
cnt=0; cl(p); cl(a); cl(p_a);
data ret,tmp;
read(m); read(n); read(M);
Machine(M);
for (int i=1;i<=cnt;i++)
{
ret=Calc(n,p[i],a[i],p_a[i]);
tmp=Calc(m,p[i],a[i],p_a[i]);
ret.second-=tmp.second;
(ret.first*=Inv(tmp.first,p_a[i]))%=p_a[i];
tmp=Calc(n-m,p[i],a[i],p_a[i]);
ret.second-=tmp.second;
(ret.first*=Inv(tmp.first,p_a[i]))%=p_a[i];
(ret.first*=Pow(p[i],ret.second,p_a[i]))%=p_a[i];
China::add(ret.first,p_a[i]);
}
ans=China::Solve();
write(ans,'\n');
}
}
int main()
{
ll Q,order;
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(Q);
Work3::Pre();
while (Q--)
{
read(order);
if (order==1) Work1::Solve();
if (order==2) Work2::Solve();
if (order==3) Work3::Solve();
}
return 0;
}