BSGS板子

1

Discrete Logging

Given a prime P, 2 <= P < 2 31, 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) .

本题是一个典型并且裸的BSGS 

所谓BSGS

就是求B^l == N (mod P)

常用的一种方式 当 p较小是可以通过快速幂进行解决当是这种题目时即为 p有的值极大 无法用快速幂求解b^

b^l == n mod p;'

设l==i*m-j

其中m=ceil(sqrt(p)) ceil意思是向上取整  即 2.1=3   ；3.9=4  ；-1.5=-1；  -1.2=-1;

所以b^(i+m-j)=n mod p;

所以b^(i+m)=(n * b^j) mod p

然后暴力枚举 右边存入哈希表中 其中使用一个 map 函数 

相当于 扩展数组大小和一把钥匙 直接就可以判断当前数组中是否含有 当前数

然后枚举左边 同时进行哈希表的判断 如果有 

那么 i+m-j 即是答案

下面 是BSGS 的模板

ll bsgs(ll a,ll b,ll p)
{
    ll m=ceil(sqrt(p));
    map<ll,int>flag;
    ll temp;
    for(ll j=0;j<=m;j++){
        if(j==0) temp=b;
        else temp*=a;
        temp%=p;
        flag[temp]=j;
    }
    ll z=1;
    temp=qpow(a,m,p);
    for(ll i=1;i<m;i++){
            z*=temp;
            z%=p;
            if(flag[z])
                return i*m-flag[z];
    }
    return -1;
}