题目大意
给定递推序列：
F[i] = a*F[i-1] + b (mod c)
求一个最小的i使得F[i] == t
题解
我们首先要化简这个数列，作为一个学渣，我查阅了一些资料：
http://d.g.wanfangdata.com.cn/Periodical_cczl200924107.aspx
http://wenku.baidu.com/view/7162471b650e52ea5518982d.html
推一下，就有：
[a_{n+1}=ba_n+c\
a_{n+1}+frac{c}{b-1}=ba_n+c+frac{c }{ b-1}=b(a_n+frac{c}{b-1})\
a_{n+1}+frac c{b-1}=b^{n-1}(a_1+frac c{b-1})
]
[F[i] = (F[1] + frac{b}{a-1}) * a^{i-1} - frac{b}{a-1}
]
令F[i] = t;
可以知道:
a^(i-1) = (t+b/(a-1)) / (x1+b/(a-1))
对于这个式子，我们直接调用BSGS算法求解即可。
特别的，某些情况需要特判。
代码
#include <bits/stdc++.h>
#define ll long long
using namespace std;
ll p, a, b, X1, t, T;
ll pow(ll a, ll b, ll p) {
ll ans = 1;
while(b) {
if(b & 1) ans = ans * a % p;
b >>= 1;
a = a * a % p;
}
return ans;
}
ll inv(ll a, ll p) {
return pow(a, p-2, p);
}
map<ll, ll> mp;
ll BSGS(ll A, ll B, ll C) {
mp.clear();
if(A % C == 0) return -2;
ll m = ceil(sqrt(C));
ll ans;
for(int i = 0; i <= m; i++) {
if(i == 0) {
ans = B % C;
mp[ans] = i;
continue;
}
ans = (ans * A) % C;
mp[ans] = i;
}
ll t = pow(A, m, C);
ans = t;
for(int i = 1; i <= m; i++) {
if(i != 1)ans = ans * t % C;
if(mp.count(ans)) {
int ret = i * m % C - mp[ans] % C;
return (ret % C + C)%C;
}
}
return -2;
}
int main() {
// freopen("input", "r", stdin);
scanf("%lld", &T);
while(T--) {
scanf("%lld %lld %lld %lld %lld", &p, &a, &b, &X1, &t);
if(X1 == t) {
printf("%d
", 1);
continue;
}
if(a == 0) {
if(t == b) {
printf("%d
", 2);
}
else printf("%d
", -1);
continue;
}
if(a == 1) {
if(b == 0) {
printf("%d
", -1);
continue;
}
ll ans = (((t-X1)%p + p)%p * inv(b, p)) % p;
printf("%lld
", ans+1);
continue;
}
X1 %= p, a %= p, b %= p, t%= p;
ll tmp = (b%p * inv(a-1, p))%p;
ll B = ((t+tmp)%p * inv((X1+tmp) % p, p)) % p;
ll A = a;
ll ans = BSGS(A, B, p);
printf("%lld
", ans+1);
}
return 0;
}