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。

 inline void solve3() {
     Hash::clear();
     //mp.clear();
     scanf("%lld%lld%lld", &Y, &Z, &MO);
     Z %= MO;
     Y %= MO;
     /// BSGS
     if(Y == ) {
         if(!Z) printf("1\n");
         else printf("Orz, I cannot find x!\n");
         return;
     }
     int T = sqrt(MO);
     LL x = , y = ;
     for(int i = ; i < T; i++) {
         Hash::insert(x, i);
         //mp[x] = i;
         x = x * Y % MO;
     }
     x = qpow(x, MO - ); /// x = 1 / (Y ^ T)
     for(int i = ; i <= MO / T; i++) {
         int t = Hash::find(Z * y % MO);
         if(t != -) {
             printf("%lld\n", 1ll * i * T + t);
             return;
         }
         y = y * x % MO;
     }
     printf("Orz, I cannot find x!\n");
     return;
 }

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