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 = , base = a;
   while (x) {
     if (x & ) (res *= base) %= p;
     (base *= base) %= p;
     x >>= ;
   }
   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("");
       else puts("no solution");
       goto end_of_work;
     }
     B = (int) ceil(sqrt(p)), now = ;
     mp[] = B + ;
     for (i = ; i < B; ++i) {
       (now *= y) %= p;
       if (!mp[now]) mp[now] = i;
     }
     now = , base = pow(y, p - B - );
     for (i = ; i < B; ++i) {
       tmp = mp[z * now % p];
       if (tmp) {
     if (tmp == B + ) tmp = ;
     printf("%lld\n", i * B + tmp);
     goto end_of_work;
       }
       (now *= base) %= p;
     }
     puts("no solution");
   end_of_work:;
   }
   return ;
 }

(p.s. bz上开了O2，所以还不是很慢恩！)