bzoj3122 [SDOI2013]隨機數生成器
阿新 • • 發佈:2019-04-05
滿足 ret 時間復雜度 pmod class include scan ++ ont
\(\verb|bzoj3122 [SDOI2013]隨機數生成器|\)
給定一個遞推式, \(X_i=(aX_{i-1}+b)\mod P\)
求滿足 \(X_k=t\) 的最小整數解,無解輸出 \(-1\)
\(0\leq a,\ b,\ t,\ P\leq10^9,\ P\) 為質數
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
\[\displaystyle\sum_{i=0}^{k-2}a_i\equiv\frac{t-a^{k-1}x}{b}\pmod P\]
\[\frac{a^{k-1}-1}{a-1}\equiv\frac{t-a^{k-1}x}{b}\pmod P\]
\[b(a^{k-1}-1)\equiv(a-1)(t-a^{k-1}x)\pmod P\]
\[ba^{k-1}-b\equiv at-a^kx-t+a^{k-1}x\pmod P\]
\[a^{k-1}(b-x+ax)\equiv at-t+b\pmod P\]
\[a^{k-1}\equiv\frac{at-t+b}{b-x+ax}\pmod P\]
所以上 \(BSGS\)
然而這題特判很惡心,不加特判 \(0\text{pts}\)
特判如下:
- \(x=t:ans=1\)
- \(a=1\)
- \(b=0:ans=-1\)
- \(b\neq0:ans=\frac{t-x}{b}+1\)
- \(a=0\)
- \(b=t:ans=2\)
- \(b\neq t:ans=-1\)
時間復雜度 \(O(T\sqrt P)\)
代碼
#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; }
bzoj3122 [SDOI2013]隨機數生成器