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

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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 
    
     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 
     
       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 
      
        :: 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;}