BZOJ_3239_Discrete Logging_BSGS
题意：Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 2 <= N < P, compute the discrete logarithm of N, base B, modulo P. That is, find an integer L such that BL == N (mod P)
分析：BSGS裸题
数据很水，少了很多特判依然能过
BSGS思想核心思想：根号预处理+哈希/map
代码：
// luogu-judger-enable-o2
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <map>
#include <math.h>
using namespace std;
#define LL long long
map<LL,int> mp;
LL qp(LL x,LL y,LL mod){
LL re=1;
while(y){
if(y&1ll)re=re*x%mod;
x=x*x%mod;
y>>=1ll;
}
return re;
}
void exgcd(LL a,LL b,LL &x,LL &y,LL &p){
if(!b){x=1;y=0;p=a;return ;}
exgcd(b,a%b,y,x,p);
y-=(a/b)*x;
}
LL BSGS(LL n,LL a,LL b){
if(n==1)if(!b)return a!=1; else return -1;
if(b==1)if(a)return 0; else return -1;
if(a%n==0)if(!b)return 1; else return -1;
LL m=ceil(sqrt(n)),d=1,base=1;
mp.clear();
for(int i=0;i<m;i++)
{
if(!mp.count(base))mp[base]=i;
base=(base*a)%n;
}
for(int i=0;i<m;i++)
{
LL x,y,s;
exgcd(d,n,x,y,s);
x=(x*b%n+n)%n;
if(mp.count(x))return i*m+mp[x];
d=(d*base)%n;
}
return -1;
}
int main()
{
LL n,a,b;
while(scanf("%lld%lld%lld",&n,&a,&b)!=EOF)
{
LL x=BSGS(n,a,b);
if(x==-1)puts("no solution");
else printf("%lld\n",x);
}
}
