2019牛客暑期多校训练营（第五场）

https://ac.nowcoder.com/acm/contest/885/C

这个跟平时那种求离散对数的bsgs不一样，虽然可以转化成离散对数来做不过会T掉。展开递推式然后合并具有a的项，发现其实是离散对数。

#include<cstdio>
#include<algorithm>
const int AA = 1000000;
std::pair<int, int> d[AA];
int mypow(long long x, int y, int p) {
    long long res = 1;
    while(y) {
        if(y & 1)
            res = res * x % p;
        x = x * x % p;
        y >>= 1;
    }
    return res;
}
int inv(int x, int p) {
    return mypow(x, p - 2, p);
}
int val[AA], pos[AA];
void solve() {
    long long n, x0, a, b, p;
    int Q;
    scanf("%lld%lld%lld%lld%lld%d", &n, &x0, &a, &b, &p, &Q);
    if(a == 0) {
        while(Q--) {
            int v;
            scanf("%d", &v);
            if(v == x0)
                puts("0");
            else if(v == b)
                puts("1");
            else
                puts("-1");
        }
        return;
    }
    long long now = x0;
    int m = std::min((long long)AA, n);
    for(int i = 0; i < m; i++) {
        d[i] = {now, i};
        now = (now * a + b) % p;
    }
    sort(d, d + m);
    {
        int new_m = 0;
        for(int i = 0; i < m; i++) {
            val[new_m] = d[i].first;
            pos[new_m++] = d[i].second;
            while(i + 1 < AA && d[i + 1].first == d[i].first)
                i++;
        }
        m = new_m;
    }
    int BB = p / AA + 3;
    int inv_a = inv(a, p);
    int inv_b = (p - b) % p * inv_a % p;
    long long aa = 1, bb = 0;
    for(int i = 0; i < AA; i++) {
        aa = aa * inv_a % p;
        bb = (bb * inv_a + inv_b) % p;
    }
    while(Q--) {
        int v;
        scanf("%d", &v);
        int it = std::lower_bound(val, val + m, v) - val;
        if(it < m && val[it] == v) {
            printf("%d
", pos[it]);
            continue;
        }
        if(n < AA) {
            puts("-1");
            continue;
        }
        bool suc = false;
        for(int i = 1; i <= BB; i++) {
            v = (aa * v + bb) % p;
            it = std::lower_bound(val, val + m, v) - val;
            if(it < m && val[it] == v) {
                suc = true;
                int res =  i * AA + pos[it];
                if(res >= n)
                    res = -1;
                printf("%d
", res);
                break;
            }
        }
        if(!suc)
            puts("-1");
    }
}
int main() {
    int T;
    scanf("%d", &T);
    while(T--)
        solve();
    return 0;
}

但这里要学的不是套这个模板，而是用BSGS算法的思路去改。好像在这里unorderedmap的表现不如二分好。

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAXBUF = 1 << 20;
char buf[MAXBUF], *fh, *ft;
inline char gc() {
    if(fh == ft) {
        int l = fread(buf, 1, MAXBUF, stdin);
        ft = (fh = buf) + l;
    }
    return *fh++;
}

inline int read() {
    int x = 0;
    char c = gc();
    for(; c < '0' || c > '9'; c = gc())
        ;
    for(; c >= '0' && c <= '9'; c = gc())
        x = (x << 3) + (x << 1) + c - '0';
    return x ;
}

inline ll readll() {
    ll x = 0;
    char c = gc();
    for(; c < '0' || c > '9'; c = gc())
        ;
    for(; c >= '0' && c <= '9'; c = gc())
        x = (x << 3) + (x << 1) + c - '0';
    return x ;
}

ll n, x0, a, b,  v, A, B;
int Q, p;

inline int qpow(ll a, int n, int p) {
    ll s = 1;
    while(n) {
        if(n & 1) {
            s = s * a;
            if(s >= p)
                s %= p;
        }
        a = a * a;
        if(a >= p)
            a %= p;
        n >>= 1;
    }
    return s;
}

inline int inv(ll a, int p) {
    return qpow(a % p, p - 2, p);
}

const int BSCEIL = 1000000;
int GSCEIL;

pair<int, int> pii[BSCEIL + 5];
int fi[BSCEIL + 5];
int se[BSCEIL + 5];
struct BinaryMap {
    int cnt,piitop;
    inline void clear() {
        piitop = 0;
    }
    inline void insert(int s, int i) {
        pii[piitop++] = {s, i};
    }
    inline void build() {
        sort(pii , pii  + piitop);
        cnt = 0;
        for(int i = 0; i < piitop; ++i) {
            if(i == 0 || pii[i].first != pii[i - 1].first) {
                fi[cnt] = pii[i].first;
                se[cnt++] = pii[i].second;
            }
        }
    }

    inline int query(int v) {
        int t = lower_bound(fi, fi+ cnt, v) - fi;
        if(t == cnt || fi[t] != v)
            return -1;
        else
            return se[t];
    }
} M;

void bs() {
    M.clear();
    ll s = x0;

    for(int i = 0; i < BSCEIL; ++i) {
        M.insert(s, i);
        s = s * a + b;
        if(s >= p)
            s %= p;
    }
    M.build();
    int inva = inv(a, p), invb = (p - b) % p * inva % p;
    A = 1, B = 0;
    for(int i = 1; i <= BSCEIL; ++i) {
        A *=  inva ;
        if(A >= p)
            A %= p;
        B = B * inva + invb;
        if(B >= p)
            B %= p;
    }
    GSCEIL = p / BSCEIL + 3;
}

ll gs() {
    ll s = v;
    int ans = M.query(s);
    if(ans >= n)
        return -1;
    if(ans!=-1)
        return ans;
    if(n < BSCEIL)
        return -1;
    for(int i = 1; i <= GSCEIL; ++i) {
        s = s*A+B;
        if(s >= p)
            s %= p;
        int tmp=M.query(s);
        if(tmp!=-1) {
            ll ans = 1ll * BSCEIL * i + tmp;
            if(ans >= n)
                ans = -1;
            return ans;
        }
    }
    return -1;
}

int main() {
#ifdef Yinku
    freopen("Yinku.in", "r", stdin);
#endif // Yinku
    int T = read();
    while(T--) {
        n = readll(), x0 = read(), a = read(), b = read(), p = read(), Q = read();
        if(a == 0) {
            while(Q--) {
                v = read();
                if(v == x0)
                    puts("0");
                else if(v == b)
                    puts("1");
                else
                    puts("-1");
            }
        } else {
            bs();
            while(Q--) {
                v = read();
                ll ans = gs();
                if(ans == -1) {
                    puts("-1");
                } else
                    printf("%lld
", ans);
            }
        }
    }
}