BZOJ3239 Discrete Logging

一道裸的BSGS题目（叫baby step, giant step）

从爱酱的blog里学来的，是一个很神的根号算法。

如果我们有hash的工具的话，就是O(sqrt(p))的，这里又用了一个map所以是O(sqrt(p) * log(sqrt(p)))

/**************************************************************
    Problem: 3239
    User: rausen
    Language: C++
    Result: Accepted
    Time:280 ms
    Memory:2932 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
#include <map>
 
using namespace std;
typedef long long ll;
 
ll p, y, z, B;
map <ll, int> mp;
 
ll pow(ll a, ll x) {
  a %= p;
  ll res = 1, base = a;
  while (x) {
    if (x & 1) (res *= base) %= p;
    (base *= base) %= p;
    x >>= 1;
  }
  return res;
}
 
int main() {
  ll i, now, base, tmp;
  while (scanf("%lld%lld%lld", &p, &y, &z) != EOF) {
    mp.clear();
    y %= p;
    if (!y) {
      if (!z) puts("1");
      else puts("no solution");
      goto end_of_work;
    }
    B = (int) ceil(sqrt(p)), now = 1;
    mp[1] = B + 1;
    for (i = 1; i < B; ++i) {
      (now *= y) %= p;
      if (!mp[now]) mp[now] = i;
    }
    now = 1, base = pow(y, p - B - 1);
    for (i = 0; i < B; ++i) {
      tmp = mp[z * now % p];
      if (tmp) {
    if (tmp == B + 1) tmp = 0;
    printf("%lld
", i * B + tmp);
    goto end_of_work;
      }
      (now *= base) %= p;
    }
    puts("no solution");
  end_of_work:;
  }
  return 0;
}

(p.s. bz上开了O2，所以还不是很慢恩！)