poj2417 bzoj3239 Discrete Logging（bsgs）

Description

Given a prime P, 2 <= P < 231, an integer B, 2 <= B < P, and an integer N, 1 <= 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”.

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

Hint
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) .

分析：
bsgs

tip

板子打对，万事ok

费马定理的推论
X^(-m)≡X^(p-1-m) (mod p)
p为素数

终于知道为什么tmp是这么计算的了

这里写代码片
#include<cstdio>
#include<iostream>
#include<cmath>
#include<map>
#define ll long long

using namespace std;

ll x,z,p;
map<ll,ll> mp;

ll KSM(ll a,ll b,ll p)
{
    ll t=1;
    a%=p;
    while (b)
    {
        if (b&1) 
           t=(t%p*a%p)%p;
        b>>=1;
        a=(a%p*a%p)%p;
    }
    return t%p;
}

ll bsgs(ll x,ll z,ll p)
{
    mp.clear();
    x%=p; z%=p;
    if (x==0&&z==0) return 0;
    if (x==0) return -1;
    ll m=(ll)ceil(sqrt((double)p)),now=1;
    mp[1]=m+1;
    for (ll i=1;i<m;i++)
    {
        now=(now%p*x%p)%p;
        if (!mp[now]) mp[now]=i;
    }
    ll inv=1,tmp=KSM(x,p-m-1,p);
    for (ll k=0;k<m;k++)
    {
        ll x1=z%p*inv%p;
        ll i=mp[(z%p*inv%p)%p];
        if (i)
        {
            if (i==m+1) i=0;
            return k*m+i;
        }
        inv=(inv%p*tmp%p)%p;
    }
    return -1;  
}

int main()
{
    while (scanf("%lld%lld%lld",&p,&x,&z)!=EOF)
    {
        ll ans=bsgs(x,z,p);
        if (ans!=-1) printf("%lld
",ans);
        else printf("no solution
");
    }
    return 0;
}