bzoj3122 [SDOI2013]随机数生成器

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 \(-1\)

\(0\leq a,\ b,\ t,\ P\leq10^9,\ P\) 为质数

BSGS

---

首先化式子，推得

\[X_k=a^{k-1}x+b\displaystyle\sum_{i=0}^{k-2}a_i
\]

因此

\[\begin{aligned}\displaystyle\sum_{i=0}^{k-2}a_i&\equiv\frac{t-a^{k-1}x}{b}\pmod P\\\frac{a^{k-1}-1}{a-1}&\equiv\frac{t-a^{k-1}x}{b}\pmod P\\a^{k-1}(b-x+ax)&\equiv at-t+b\pmod P\\a^{k-1}&\equiv\frac{at-t+b}{b-x+ax}\pmod P\end{aligned}
\]

所以上 \(BSGS\)

然而这题特判很恶心，不加特判 \(0\text{pts}\)

特判如下：

• \(x=t:ans=1\)
• \(a=1\)
  • \(b=0:ans=-1\)
  • \(b\neq0:ans=\frac{t-x}{b}+1\)
• \(a=0\)
  • \(b=t:ans=2\)
  • \(b\neq t:ans=-1\)

时间复杂度 \(O(T\sqrt P)\)

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\n", 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\n", ~ans ? ans + 1 : ans);
  }
  return 0;
}