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)))

 1 /**************************************************************
 2     Problem: 3239
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:280 ms
 7     Memory:2932 kb
 8 ****************************************************************/
 9
10 #include <cstdio>
11 #include <cmath>
12 #include <map>
13
14 using namespace std;
15 typedef long long ll;
16
17 ll p, y, z, B;
18 map <ll, int> mp;
19
20 ll pow(ll a, ll x) {
21   a %= p;
22   ll res = 1, base = a;
23   while (x) {
24     if (x & 1) (res *= base) %= p;
25     (base *= base) %= p;
26     x >>= 1;
27   }
28   return res;
29 }
30
31 int main() {
32   ll i, now, base, tmp;
33   while (scanf("%lld%lld%lld", &p, &y, &z) != EOF) {
34     mp.clear();
35     y %= p;
36     if (!y) {
37       if (!z) puts("1");
38       else puts("no solution");
39       goto end_of_work;
40     }
41     B = (int) ceil(sqrt(p)), now = 1;
42     mp[1] = B + 1;
43     for (i = 1; i < B; ++i) {
44       (now *= y) %= p;
45       if (!mp[now]) mp[now] = i;
46     }
47     now = 1, base = pow(y, p - B - 1);
48     for (i = 0; i < B; ++i) {
49       tmp = mp[z * now % p];
50       if (tmp) {
51     if (tmp == B + 1) tmp = 0;
52     printf("%lld\n", i * B + tmp);
53     goto end_of_work;
54       }
55       (now *= base) %= p;
56     }
57     puts("no solution");
58   end_of_work:;
59   }
60   return 0;
61 }

(p.s. bz上开了O2，所以还不是很慢恩！)