BZOJ3122 推公式+逆元+BSGS

xn=axn−1+b

设 xn−k=a(xn−1−k)
解得 k=b1−a
即 xn−k=(x1−k)×an−1

将k代入，整理式子
问题转变为求最小的n满足等式
(1−a)xn−b=((1−a)x1−b)an−1

此时用BSGS求解即可：
令 m=⌈p√⌉
A=(1−a)xn−b
B=(1−a)x1−b
即求
A≡Ban−1(mod p)
Aaj≡Bamj
枚举左边和右边，判断相等的情况即可
特殊情况：a = 1（用扩展欧几里得算法求逆元） ， b = 1（特判）

一个地方没有取模调了一晚上gg

#include 
#include 
#include 
#include 
#include 

#define INF ( 1LL << 62 )
#define N 1000050
#define eps 0.0000001

using namespace std;

typedef long long LL;
struct Monster{ LL u,l; }lzk[N];
LL lhr[N],p,a,b,x,t,A,B,am;
LL m;

inline LL read() {
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&& ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

LL ex_gcd(LL a,LL b,LL &x,LL &y) {
    if (b == 0) { x = 1; y = 0; return a; }
    LL q = ex_gcd(b,a%b,y,x);
    y -= (a / b) * x;
    return q;
}

bool cmp(Monster p1,Monster p2) {
    return p1.u == p2.u ? p1.l > p2.l : p1.u < p2.u;
}

int main()
{
    int T = read();
    while (T--) {
        p = read() , a = read() , b = read() , x = read() , t = read();
        if (t >= p || x >= p)  assert(0);
        if (x == t) {
            printf("1\n"); continue;
        }

        if(a==0) {
            if(b == t) printf("2\n"); else printf("-1\n");
            continue;
        }

        if ( a == 1 ) {
            LL db,dp;
            LL lhr = ex_gcd(b,p,db,dp);
            LL C = (t - x + p) % p;
            if (C % lhr) {
                printf("-1\n"); continue;
            } 
            C /= lhr;
            db = db * C % p;
            while (db < 0) db += p;
            printf("%lld\n",db+1);
            continue;
        }   

        A = (1LL - a) * t - b;
        B = (1LL - a) * x - b;
        m = ceil( sqrt(p) ) + eps;
        A %= p; B %= p;
        while (A < 0) A += p;
        while (B < 0) B += p;

        lzk[0].u = A;
        for (int j=1;j<=m;j++) {
            lzk[j].u = ( lzk[j-1].u * a ) % p;
            lzk[j].l = j;
        }
        sort(lzk+1,lzk+m+1,cmp);

        lhr[0] = B; am = 1LL;
        for (int i=1;i<=m;i++) am = ( am * a ) % p;
        for (int i=1;i<=m;i++) lhr[i] = ( lhr[i-1] * am ) % p;

        LL ans = INF;
        for (LL i=1;i<=m;i++) {
            int l = 1 , r = m;
            while (l < r) {
                int mid = (l + r) >> 1;
                if (lzk[mid].u >= lhr[i]) r = mid; else l = mid + 1;
            }
            if (lzk[l].u == lhr[i]) 
                ans = min(ans,i * m - lzk[l].l);
        }
        if (ans == INF) printf("-1\n"); else printf("%lld\n",ans+1);
    }
    return 0;
}