bzoj3122 [SDOI2013]随机数生成器

给定一个递推式， (X_i=(aX_{i-1}+b)mod P)

求满足 (X_k=t) 的最小整数解，无解输出 (-1)

(0leq a, b, t, Pleq10^9, P) 为质数

BSGS

首先化式子，推得

[X_k=a^{k-1}x+bdisplaystylesum_{i=0}^{k-2}a_i ]

因此

[egin{aligned}displaystylesum_{i=0}^{k-2}a_i&equivfrac{t-a^{k-1}x}{b}pmod P\frac{a^{k-1}-1}{a-1}&equivfrac{t-a^{k-1}x}{b}pmod P\a^{k-1}(b-x+ax)&equiv at-t+bpmod P\a^{k-1}&equivfrac{at-t+b}{b-x+ax}pmod Pend{aligned} ]

所以上 (BSGS)

然而这题特判很恶心，不加特判 (0 ext{pts})

特判如下：

(x=t:ans=1)
(a=1)
- (b=0:ans=-1)
- (b eq0:ans=frac{t-x}{b}+1)
(a=0)
- (b=t:ans=2)
- (b eq t:ans=-1)

时间复杂度 (O(Tsqrt P))

代码

#include <bits/stdc++.h>
using namespace std;

int P;

int qp(int a, int k) {
  int res = bool(a);
  for (; k; k >>= 1, a = 1ll * a * a % P) {
    if (k & 1) res = 1ll * res * a % P;
  }
  return res % P;
}

int bsgs(int a, int b) {
  if (!a && b) return -1;
  map <int, int> s;
  int sz = sqrt(P), inv_a = qp(a, P - 2), pw = qp(a, sz), cur = 1;
  for (int i = 0; i <= sz; i++) {
    s.insert(make_pair(1ll * b * cur % P, i)), cur = 1ll * cur * inv_a % P;
  }
  cur = 1;
  map <int, int> :: iterator it;
  for (int i = 0; i <= sz; i++, cur = 1ll * cur * pw % P) {
    if ((it = s.find(cur)) != s.end()) {
      return i * sz + (it -> second);
    }
  }
  return -1;
}

int main() {
  int Tests, a, b, x, t, A, B;
  scanf("%d", &Tests);
  while (Tests--) {
    scanf("%d %d %d %d %d", &P, &a, &b, &x, &t);
    a %= P, b %= P, x %= P, t %= P;
    if (x == t) {
      puts("1"); continue;
    } else if (a == 1) {
      if (!b) {
        puts("-1"); continue;
      }
      printf("%d
", 1ll * (t - x + P) * qp(b, P - 2) % P + 1);
      continue;
    } else if (!a) {
      puts(b == t ? "2" : "-1");
      continue;
    }
    A = a, B = 1ll * (1ll * a * t - t + b + P) % P * qp((b - x + 1ll * a * x + P) % P, P - 2) % P;
    int ans = bsgs(A, B);
    printf("%d
", ~ans ? ans + 1 : ans);
  }
  return 0;
}

bzoj3122 [SDOI2013]随机数生成器

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