Pohlig-Hellman algorithm 解决一类离散对数问题
原理：
https://blog.csdn.net/qq_26060285/article/details/90322054
适用范围大致为模数Ｐ为质数，且(P-1)的最大质因子 <= 1e12。Ｐ为合数的情况直接用EXBSGS就好了，因为P为合数的时候大部分情况下求不了原根, 若P有原根，则 P = 2,4,q^a,2q^a(q为质数)。模板如下：
洛谷3846
#include"iostream"
#include"cstdio"
#include"vector"
#include"unordered_map"
#include"map"
#include"cstring"
#include"algorithm"
#include"cmath"
using namespace std;
namespace DLP{
typedef long long LL;
typedef __int128 i128;
const int MX = 1e6+7;
LL qpow(LL a, LL n){
LL ret = 1;
while(n > 0){
if(n&1) ret = ret*a;
a = a*a;
n >>= 1;
}
return ret;
}
LL qpow(LL a, LL n, LL p){
LL ret = 1;
a %= p;
n %= p;
while(n > 0){
if(n&1) ret = (i128)ret*a%p;
a = (i128)a*a%p;
n >>= 1;
}
return ret;
}
int prime[MX],tt;
bool is_prime[MX];
struct node{
LL p;
int c;
};
void prime_init()
{
memset(is_prime,1,sizeof(is_prime));
int n = MX - 7;
for(int i = 2; i <= n; i++){
if(is_prime[i]) prime[++tt] = i;
for(int j = 1; j <= tt && prime[j] <= n/i; j++){
is_prime[i*prime[j]] = 0;
if(!i%prime[j]) break;
}
}
}
void find_fra(vector &v, LL num){
if(tt == 0) prime_init();
for(int i = 1; i <= tt && prime[i] <= num; i++){
if(num%prime[i] == 0){
int cnt = 0;
while(num%prime[i] == 0){
++cnt;
num /= prime[i];
}
v.push_back(node{prime[i],cnt});
}
}
if(num > 1) v.push_back(node{num,1});
}
int getroot(LL p, LL phi, vector &v) {
for(int k=2;; k++) {
int flag=1;
for(int i=0; i<(int)v.size(); i++)
if(qpow(k,phi/v[i].p,p)==1) {
flag=0;
break;
}
if(flag)return k;
}
}
LL BSGS(LL a, LL b, LL p, LL mod) {
a %= mod, b %= mod;
//特殊情况，具体看题目要怎么做
if(b == 1) return 0;
if(a==0){
if(b==0) return 1;
else return -1;
}
LL t = 1;
int m = int(sqrt(1.0 * p) + 1);
LL base = b;
unordered_map has;
for(int i = 0; i < m; ++i) {
has[base] = i;
base = (i128)base * a % mod;
}
base = qpow(a, m, mod);
LL now = t;
//对于A^x，从前往后，保证答案是从小到大的
for(int i = 1; i <= m + 1; ++i) {
now = (i128)now * base % mod;
if(has.count(now)) return i * m - has[now];//找到最小后，找到最小的x使得a^x=1 mod p,通解为 最小+k*x
}
return -1;
}
LL get_xi(LL g, LL h, LL p, LL c, LL N, LL mod){
vector pi;
LL tp = 1;
for(int i = 0; i <= c; i++){
pi.push_back(tp);
tp *= p;
}
LL gq = qpow(g,pi[c-1],mod);
LL inv = 0;
tp = 1;
for(int i = c-1; i >= 0; i--){
LL tx = tp*BSGS(gq, qpow( (i128)h*qpow(g,pi[c]-inv,mod)%mod, pi[i], mod), p, mod);
inv += tx;
tp *= p;
}
return inv;
}
LL exgcd(LL a,LL b,LL &x,LL &y) {
if(b==0) {
x=1,y=0;
return a;
}
LL d=exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
LL Inv(LL a,LL P){//求a在膜P下的逆
if(a==0)return 1;
LL x,y;
exgcd(a,P,x,y);
return (x%P+P)%P;
}
//ax+by=c
LL solve_exgcd(LL a,LL b,LL c,LL &x,LL &y) {
LL d;
if(c%(d=__gcd(a,b)))return -1;//返回-1是无解
a/=d;
b/=d;
c/=d;
exgcd(a,b,x,y);
x=(i128)x*c%b;
while(x<=0) x+=b;
return x;
}
LL CRT(vector xi, vector v){
LL sz = v.size();
LL M = 1, ans = 0;
vector m;
for(int i = 0; i &v){
vector xi;
for(node tp : v){
LL tg = qpow(g,N/qpow(tp.p,tp.c),mod);
LL th = qpow(h,N/qpow(tp.p,tp.c),mod);
xi.push_back(get_xi(tg, th, tp.p, tp.c, N, mod));
}
return CRT(xi,v);
}
//p为质数且(p-1)分解质因子较大的情况时，算法比较优秀，如果ｐ=2*q+1,q为质数, q<=1e12皆可做，则算法与BSGS差不多，不为质数的话可以考虑EXBSGS
LL solve(LL g, LL h, LL p){
if(h == 1) return 0;
LL phi_p = p-1;
vector v;
find_fra(v,phi_p);
int rt = getroot(p,phi_p,v);
LL x = get_x(g,rt,phi_p,p,v);
LL y = get_x(h,rt,phi_p,p,v);
LL a = 0, b = 0;
if(x==0){
if(y==0) return 1;
else if(y == 1) return 0;
else return -1; //不满足返回-1
}
else return solve_exgcd(x,phi_p,y,a,b);
}
}
typedef long long LL;
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
LL g,h,p;
while(~scanf("%lld%lld%lld",&p,&g,&h)){
LL ans = DLP::solve(g,h,p);
if(ans != -1) printf("%lld\n",ans);
else printf("no solution\n");
}
return 0;
} hdu6632
#include"bits/stdc++.h"
using namespace std;
namespace DLP{
typedef long long LL;
typedef __int128 i128;
const int MX = 1e6+7;
LL qpow(LL a, LL n){
LL ret = 1;
while(n > 0){
if(n&1) ret = ret*a;
a = a*a;
n >>= 1;
}
return ret;
}
LL qpow(LL a, LL n, LL p){
LL ret = 1;
a %= p;
n %= p;
while(n > 0){
if(n&1) ret = (i128)ret*a%p;
a = (i128)a*a%p;
n >>= 1;
}
return ret;
}
int prime[MX],tt;
bool is_prime[MX];
struct node{
LL p;
int c;
};
void prime_init()
{
memset(is_prime,1,sizeof(is_prime));
int n = MX - 7;
for(int i = 2; i <= n; i++){
if(is_prime[i]) prime[++tt] = i;
for(int j = 1; j <= tt && prime[j] <= n/i; j++){
is_prime[i*prime[j]] = 0;
if(!i%prime[j]) break;
}
}
}
void find_fra(vector &v, LL num){
if(tt == 0) prime_init();
for(int i = 1; i <= tt && prime[i] <= num; i++){
if(num%prime[i] == 0){
int cnt = 0;
while(num%prime[i] == 0){
++cnt;
num /= prime[i];
}
v.push_back(node{prime[i],cnt});
}
}
if(num > 1) v.push_back(node{num,1});
}
int getroot(LL p, LL phi, vector &v) {
for(int k=2;; k++) {
int flag=1;
for(int i=0; i<(int)v.size(); i++)
if(qpow(k,phi/v[i].p,p)==1) {
flag=0;
break;
}
if(flag)return k;
}
}
LL BSGS(LL a, LL b, LL p, LL mod) {
a %= mod, b %= mod;
//特殊情况，具体看题目要怎么做
if(b == 1) return 0;
if(a==0){
if(b==0) return 1;
else return -1;
}
LL t = 1;
int m = int(sqrt(1.0 * p) + 1);
LL base = b;
unordered_map has;
for(int i = 0; i < m; ++i) {
has[base] = i;
base = (i128)base * a % mod;
}
base = qpow(a, m, mod);
LL now = t;
//对于A^x，从前往后，保证答案是从小到大的
for(int i = 1; i <= m + 1; ++i) {
now = (i128)now * base % mod;
if(has.count(now)) return i * m - has[now];//找到最小后，找到最小的x使得a^x=1 mod p,通解为 最小+k*x
}
return -1;
}
LL get_xi(LL g, LL h, LL p, LL c, LL N, LL mod){
vector pi;
LL tp = 1;
for(int i = 0; i <= c; i++){
pi.push_back(tp);
tp *= p;
}
LL gq = qpow(g,pi[c-1],mod);
LL inv = 0;
tp = 1;
for(int i = c-1; i >= 0; i--){
LL tx = tp*BSGS(gq, qpow( (i128)h*qpow(g,pi[c]-inv,mod)%mod, pi[i], mod), p, mod);
inv += tx;
tp *= p;
}
return inv;
}
LL exgcd(LL a,LL b,LL &x,LL &y) {
if(b==0) {
x=1,y=0;
return a;
}
LL d=exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
LL Inv(LL a,LL P){//求a在膜P下的逆
if(a==0)return 1;
LL x,y;
exgcd(a,P,x,y);
return (x%P+P)%P;
}
//ax+by=c
LL solve_exgcd(LL a,LL b,LL c,LL &x,LL &y) {
LL d;
if(c%(d=__gcd(a,b)))return -1;//返回-1是无解
a/=d;
b/=d;
c/=d;
exgcd(a,b,x,y);
x=(i128)x*c%b;
while(x<=0) x+=b;
return x;
}
LL CRT(vector xi, vector v){
LL sz = v.size();
LL M = 1, ans = 0;
vector m;
for(int i = 0; i &v){
vector xi;
for(node tp : v){
LL tg = qpow(g,N/qpow(tp.p,tp.c),mod);
LL th = qpow(h,N/qpow(tp.p,tp.c),mod);
xi.push_back(get_xi(tg, th, tp.p, tp.c, N, mod));
}
return CRT(xi,v);
}
//p为质数且(p-1)分解质因子较大的情况时，算法比较优秀，如果ｐ=2*q+1,q为质数, q<=1e12皆可做，则算法与BSGS差不多，不为质数的话可以考虑EXBSGS
LL solve(LL g, LL h, LL p){
if(h == 1) return 0;
LL phi_p = p-1;
vector v;
find_fra(v,phi_p);
int rt = getroot(p,phi_p,v);
LL x = get_x(g,rt,phi_p,p,v);
LL y = get_x(h,rt,phi_p,p,v);
LL a = 0, b = 0;
if(x==0){
if(y==0) return 1;
else if(y == 1) return 0;
else return -1; //不满足返回-1
}
else return solve_exgcd(x,phi_p,y,a,b);
}
}
typedef long long LL;
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
int T;
scanf("%d",&T);
while(T--){
LL g,h,p;
scanf("%lld%lld%lld",&p,&g,&h);
printf("%lld\n",DLP::solve(g,h,p));
}
return 0;
} poj2417
#include"iostream"
#include"cstdio"
#include"vector"
#include"map"
#include"map"
#include"cstring"
#include"algorithm"
#include"cmath"
using namespace std;
namespace DLP{
typedef long long LL;
typedef long long i128;
const int MX = 1e6+7;
LL qpow(LL a, LL n){
LL ret = 1;
while(n > 0){
if(n&1) ret = ret*a;
a = a*a;
n >>= 1;
}
return ret;
}
LL qpow(LL a, LL n, LL p){
LL ret = 1;
a %= p;
n %= p;
while(n > 0){
if(n&1) ret = (i128)ret*a%p;
a = (i128)a*a%p;
n >>= 1;
}
return ret;
}
int prime[MX],tt;
bool is_prime[MX];
struct node{
LL p;
int c;
};
void prime_init()
{
memset(is_prime,1,sizeof(is_prime));
int n = MX - 7;
for(int i = 2; i <= n; i++){
if(is_prime[i]) prime[++tt] = i;
for(int j = 1; j <= tt && prime[j] <= n/i; j++){
is_prime[i*prime[j]] = 0;
if(!i%prime[j]) break;
}
}
}
void find_fra(vector &v, LL num){
if(tt == 0) prime_init();
for(int i = 1; i <= tt && prime[i] <= num; i++){
if(num%prime[i] == 0){
int cnt = 0;
while(num%prime[i] == 0){
++cnt;
num /= prime[i];
}
v.push_back(node{prime[i],cnt});
}
}
if(num > 1) v.push_back(node{num,1});
}
int getroot(LL p, LL phi, vector &v) {
for(int k=2;; k++) {
int flag=1;
for(int i=0; i<(int)v.size(); i++)
if(qpow(k,phi/v[i].p,p)==1) {
flag=0;
break;
}
if(flag)return k;
}
}
LL BSGS(LL a, LL b, LL p, LL mod) {
a %= mod, b %= mod;
//特殊情况，具体看题目要怎么做
if(b == 1) return 0;
if(a==0){
if(b==0) return 1;
else return -1;
}
LL t = 1;
int m = int(sqrt(1.0 * p) + 1);
LL base = b;
map has;
for(int i = 0; i < m; ++i) {
has[base] = i;
base = (i128)base * a % mod;
}
base = qpow(a, m, mod);
LL now = t;
//对于A^x，从前往后，保证答案是从小到大的
for(int i = 1; i <= m + 1; ++i) {
now = (i128)now * base % mod;
if(has.count(now)) return i * m - has[now];//找到最小后，找到最小的x使得a^x=1 mod p,通解为 最小+k*x
}
return -1;
}
LL get_xi(LL g, LL h, LL p, LL c, LL N, LL mod){
vector pi;
LL tp = 1;
for(int i = 0; i <= c; i++){
pi.push_back(tp);
tp *= p;
}
LL gq = qpow(g,pi[c-1],mod);
LL inv = 0;
tp = 1;
for(int i = c-1; i >= 0; i--){
LL tx = tp*BSGS(gq, qpow( (i128)h*qpow(g,pi[c]-inv,mod)%mod, pi[i], mod), p, mod);
inv += tx;
tp *= p;
}
return inv;
}
LL exgcd(LL a,LL b,LL &x,LL &y) {
if(b==0) {
x=1,y=0;
return a;
}
LL d=exgcd(b,a%b,y,x);
y-=a/b*x;
return d;
}
LL Inv(LL a,LL P){//求a在膜P下的逆
if(a==0)return 1;
LL x,y;
exgcd(a,P,x,y);
return (x%P+P)%P;
}
//ax+by=c
LL solve_exgcd(LL a,LL b,LL c,LL &x,LL &y) {
LL d;
if(c%(d=__gcd(a,b)))return -1;//返回-1是无解
a/=d;
b/=d;
c/=d;
exgcd(a,b,x,y);
x=(i128)x*c%b;
while(x<=0) x+=b;
return x;
}
LL CRT(vector xi, vector v){
LL sz = v.size();
LL M = 1, ans = 0;
vector m;
for(int i = 0; i &v){
vector xi;
for(int i = 0; i < v.size(); i++){
node tp = v[i];
LL tg = qpow(g,N/qpow(tp.p,tp.c),mod);
LL th = qpow(h,N/qpow(tp.p,tp.c),mod);
xi.push_back(get_xi(tg, th, tp.p, tp.c, N, mod));
}
return CRT(xi,v);
}
//p为质数且(p-1)分解质因子较大的情况时，算法比较优秀，如果ｐ=2*q+1,q为质数, q<=1e12皆可做，则算法与BSGS差不多，不为质数的话可以考虑EXBSGS
LL solve(LL g, LL h, LL p){
if(h == 1) return 0;
LL phi_p = p-1;
vector v;
find_fra(v,phi_p);
int rt = getroot(p,phi_p,v);
LL x = get_x(g,rt,phi_p,p,v);
LL y = get_x(h,rt,phi_p,p,v);
LL a = 0, b = 0;
if(x==0){
if(y==0) return 1;
else if(y == 1) return 0;
else return -1; //不满足返回-1
}
else return solve_exgcd(x,phi_p,y,a,b);
}
}
typedef long long LL;
int main(){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
#endif
LL g,h,p;
while(~scanf("%lld%lld%lld",&p,&g,&h)){
LL ans = DLP::solve(g,h,p);
if(ans != -1) printf("%lld\n",ans);
else printf("no solution\n");
}
return 0;
}