BZOJ3239Discrete Logging

3239: Discrete Logging
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 372 Solved: 238
Description
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)
Input
Read several lines of input, each containing P,B,N separated by a space,
Output
for each line print the logarithm on a separate line. If there are several, print the smallest; if there is none, print “no solution”.
The solution to this problem requires a well known result in number theory that is probably expected of you for Putnam but not ACM competitions. It is Fermat’s theorem that states
B(P-1) == 1 (mod P)
for any prime P and some other (fairly rare) numbers known as base-B pseudoprimes. A rarer subset of the base-B pseudoprimes, known as Carmichael numbers, are pseudoprimes for every base between 2 and P-1. A corollary to Fermat’s theorem is that for any m
B(-m) == B(P-1-m) (mod P) .
Sample Input
5 2 1
5 2 2
5 2 3
5 2 4
5 3 1
5 3 2
5 3 3
5 3 4
5 4 1
5 4 2
5 4 3
5 4 4
12345701 2 1111111
1111111121 65537 1111111111
Sample Output
0
1
3
2
0
3
1
2
0
no solution
no solution
1
9584351
462803587
拓展BSGS。。
附上本蒟蒻的代码：

#include<cstdio>#include<map>#include<cmath>#include<iostream>using namespace std;long long gcd(long long x,long long y){    return y==0?x:gcd(y,x%y);}long long exgcd(long long a,long long b,long long &x,long long &y){    if (!b)      {        x=1,y=0;        return a;      }    else      {        long long g=exgcd(b,a%b,x,y),t=x;        x=y,y=t-a/b*y;        return g;      }}long long inv(long long a,long long p){    long long x,y,d=exgcd(a,p,x,y);    return d==1?(x+p)%p:-1;}long long pow(long long x,long long a,long long p){    long long t;    if (!a) return 1%p;    if (a==1) return x%p;    t=pow(x,a/2,p);    if (a%2) return ((t*t)%p*x)%p;    else return (t*t)%p;}int BSGS(long long a,long long b,long long p){    long long m=0;    for (;m*m<=p;m++);    b%=p;    map<long long,int>hash;    hash[b]=0;    long long e=b,v=inv(a,p),mul=pow(a,m,p);    for (int i=1;i<m;i++)      {        e=e*v%p;        if (!hash.count(e)) hash[e]=i;        else break;      }    e=1;    for (int i=0;i<=m;i++)      {        if (hash.count(e)) return hash[e]+i*m;        e=e*mul%p;      }    return -1;}void solve(long long a,long long b,long long p){    long long e=1;    b%=p,a%=p;    for (int i=0;i<100;i++)      {        if (e==b)           {            printf("%d\n",i);            return;          }        e=e*a%p;      }    int sum=0;    while (gcd(a,p)!=1)      {        long long d=gcd(a,p);        if (b%d)           {            printf("no solution\n");            return;          }        p/=d,sum++,b/=d;        b=b*inv(a/d,p)%p;      }    int ans=BSGS(a,b,p);    if (ans==-1)       {        printf("no solution\n");        return;      }    printf("%d\n",ans+sum);}int main(){    long long a,p,b;    while (cin>>p>>a>>b)      solve(a,b,p);    return 0;}