BZOJ 3122: [Sdoi2013]随机数生成器 (BSGS)

设目标状态为$X_n$.画一画柿子$egin{aligned}X_{i+1}&=a*X_i+b\X_{i+1}+frac b{a-1}&=a*left(X_i+frac b{a-1} ight)end{aligned}$

令$f_i=X_i+frac b{a-1}$.有$f_n=f_1*a^{n-1}$

所以题目转化为了求最小的$x$使得$f_1*a^{x-1}=f_n$

把$f_1$除过去就能$BSGS$做了,最后答案$+1$.

注意下面几个特判

• $X_1=X_n$ (这个也可以在几种情况内部判,但放在外面更方便)
• $a=0 o f_2=f_3=...=f_n=b$
• $a=1 o f_n=f_1+(n-1)*b$
• 要求逆元注意特判$0$的情况.
• 详见代码

CODE

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
inline LL qpow(LL a, LL b, LL c) {
    LL re = 1;
    while(b) {
        if(b & 1) re = re * a % c;
        a = a * a % c; b >>= 1;
    }
    return re;
}
int gcd(int a, int b) { return b ? gcd(b, a%b) : a; }
map<int, int>myhash;
inline int Baby_Step_Giant_Step(int a, int b, int p) {
    if(p == 1) return 0;
    a %= p, b %= p;
    if(b == 1) return 0;
    int cnt = 0; LL tmp = 1;
    for(int g = gcd(a, p); g != 1; g = gcd(a, p)) {
        if(b % g) return -1;
        b /= g, p /= g, tmp = tmp * (a/g) % p;
        ++cnt;
        if(b == tmp) return cnt;
    }
    myhash.clear();
    int m = int(sqrt(p) + 1);
    LL base = b;
    for(int i = 0; i < m; ++i) {
        myhash[base] = i;
        base = base * a % p;
    }
    base = qpow(a, m, p);
    for(int i = 1; i <= m+1; ++i) {
        tmp = tmp * base % p;
        if(myhash.count(tmp))
            return i*m - myhash[tmp] + cnt;
    }
    return -1;
}
int main() {
	int a, b, x1, xn, p, T;
	scanf("%d", &T);
	while(T--) {
        scanf("%d%d%d%d%d", &p, &a, &b, &x1, &xn);
        if(x1 == xn) puts("1");
        else if(a == 0) puts(xn == b ? "2" : "-1");
        else if(a == 1) {
            xn = (xn - x1 + p) % p;
            //b*x = Xn (mod p)  (t > 0)
            if(b == 0) puts("-1");
            else printf("%d
", int(1ll * xn * qpow(b, p-2, p) % p)+1);
        }
        else {
            int tmp = 1ll * b * qpow(a-1, p-2, p) % p;
            (xn += tmp) %= p, (x1 += tmp) %= p;
            //X1 * a^x = Xn (mod p) (x1 != tmp)
            if(x1 == 0) puts("-1");
            else {
                int ans = Baby_Step_Giant_Step(a, 1ll * xn * qpow(x1, p-2, p) % p, p);
                if(~ans) printf("%d
", ans+1);
                else puts("-1");
            }
        }
	}
}