POJ 2417 Discrete Logging（离散对数-小步大步算法）

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

    B

^L

 == 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".

题目大意：求离散对数，B^L = N (mod P)，其中P是素数，求L。

思路：大步小步算法的模板题。

PS：白书上的模板少了个ceil向上取整……

PS：用map来哈希TLE了。速度和手写哈希原来差距这么大……

代码（47MS）：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #include <cmath>
 using namespace std;
 typedef long long LL;

 const int SIZEH = ;

 struct hash_map {
     int head[SIZEH], size;
     int next[SIZEH];
     LL state[SIZEH], val[SIZEH];

     void init() {
         memset(head, -, sizeof(head));
         size = ;
     }

     void insert(LL st, LL sv) {
         LL h = st % SIZEH;
         for(int p = head[h]; ~p; p = next[p])
             if(state[p] == st) return ;
         state[size] = st; val[size] = sv;
         next[size] = head[h]; head[h] = size++;
     }

     LL find(LL st) {
         LL h = st % SIZEH;
         for(int p = head[h]; ~p; p = next[p])
             if(state[p] == st) return val[p];
         return -;
     }
 } hashmap;

 LL inv(LL a, LL n) {
     if(a == ) return ;
     return ((n - n / a) * inv(n % a, n)) % n;
 }

 LL pow_mod(LL x, LL p, LL n) {
     LL ret = ;
     while(p) {
         if(p & ) ret = (ret * x) % n;
         x = (x * x) % n;
         p >>= ;
     }
     return ret;
 }

 LL BabyStep(LL a, LL b, LL n) {
     LL e = , m = LL(ceil(sqrt(n + 0.5)));
     hashmap.init();
     hashmap.insert(, );
     for(int i = ; i < m; ++i) {
         e = (e * a) % n;
         hashmap.insert(e, i);
     }
     LL v = inv(pow_mod(a, m, n), n);
     for(int i = ; i < m; ++i) {
         LL t = hashmap.find(b);
         if(t != -) return i * m + t;
         b = (b * v) % n;
     }
     return -;
 }

 int main() {
     LL p, b, n;
     while(cin>>p>>b>>n) {
         LL ans = BabyStep(b, n, p);
         if(ans == -) puts("no solution");
         else cout<<ans<<endl;
     }
 }