BSGS

北上广深/拔山盖世算法。

y^aT+b = z mod p

p为质数，Hash表存b，枚举a，复杂度p^0.5

记得特判y = 0的情况。

 1 inline void solve3() {
 2     Hash::clear();
 3     //mp.clear();
 4     scanf("%lld%lld%lld", &Y, &Z, &MO);
 5     Z %= MO;
 6     Y %= MO;
 7     /// BSGS
 8     if(Y == 0) {
 9         if(!Z) printf("1\n");
10         else printf("Orz, I cannot find x!\n");
11         return;
12     }
13     int T = sqrt(MO);
14     LL x = 1, y = 1;
15     for(int i = 0; i < T; i++) {
16         Hash::insert(x, i);
17         //mp[x] = i;
18         x = x * Y % MO;
19     }
20     x = qpow(x, MO - 2); /// x = 1 / (Y ^ T)
21     for(int i = 0; i <= MO / T; i++) {
22         int t = Hash::find(Z * y % MO);
23         if(t != -1) {
24             printf("%lld\n", 1ll * i * T + t);
25             return;
26         }
27         y = y * x % MO;
28     }
29     printf("Orz, I cannot find x!\n");
30     return;
31 }