[SNOI2017]遗失的答案

题目

首先\(G,L\)肯定会满足\(G|L\)，否则直接全部输出\(0\)

之后我们考虑一下能用到的质因数最多只有\(8\)个

同时我们能选择的数\(x\)肯定是\(L\)的约数，还得是\(G\)的倍数，还不能超过\(N\)，感性理解一下这样的\(x\)显然不多，我们直接\(dfs\)出来

对于每一个数我们把它压成一个\(2\times 8\)的二进制数，第\(2i-1,2i\)位分别表示第\(i\)个质因子是否到上界/下界，我们直接一个\(dp[i][j]\)表示前\(i\)个数选择状态为\(j\)的方案数，同时我们还能处理出一个后缀的\(dp\)值来

对于一个必须选择的\(x\)，我们用前缀和后缀做一下或\(FWT\)，求所有和这个数的状态或起来为全集的位置的和就好了

复杂度看起来不是很科学，但是常数比较小，就跑过去了

代码

#include <bits/stdc++.h>
#define re register
const int mod = 1e9 + 7;
inline int read() {
    char c = getchar();
    int x = 0;
    while (c < '0' || c > '9') c = getchar();
    while (c >= '0' && c <= '9') x = (x << 3) + (x << 1) + c - 48, c = getchar();
    return x;
}
std::map<int, int> ma;
int N, G, L, Q, M, n, len;
int f[50005], p[20005];
int cnt[2], b[10], a[10][2], l[10], r[10], beg[10];
int c[850], st[850], ans[850], vis[850];
int dp[850][65537], tp[850][65537], d[65537];
void dfs(int x, int now, int S) {
    if (x == cnt[0] + 1) {
        c[++n] = now, st[n] = S;
        return;
    }
    int t = beg[x];
    for (re int i = l[x]; i <= r[x]; ++i)
        if (now <= N / t) {
            int s = S;
            if (i == l[x])
                s |= (1 << (2 * x - 2));
            if (i == r[x])
                s |= (1 << (2 * x - 1));
            dfs(x + 1, now * t, s);
            t = t * a[x][0];
        } else
            break;
}
inline int ksm(int a, int b) {
    int S = 1;
    for (; b; b >>= 1, a = a * a)
        if (b & 1)
            S = S * a;
    return S;
}
inline int qm(int x) { return x < 0 ? x + mod : x % mod; }
inline void Fwt(int *f, int o) {
    for (re int i = 2; i <= len; i <<= 1)
        for (re int ln = i >> 1, l = 0; l < len; l += i)
            for (re int x = l; x < l + ln; ++x) f[x + ln] = qm(f[x + ln] + o * f[x]);
}
int main() {
    N = read(), G = read(), L = read(), Q = read();
    M = std::ceil(std::sqrt(L));
    if (L % G) {
        for (re int i = 1; i <= Q; i++) puts("0");
        return 0;
    }
    for (re int i = 2; i <= M; i++) {
        if (!f[i])
            p[++p[0]] = i;
        for (re int j = 1; j <= p[0] && p[j] * i <= M; j++) {
            f[p[j] * i] = 1;
            if (i % p[j] == 0)
                break;
        }
    }
    std::swap(G, L);
    for (re int i = 1; i <= p[0]; i++) {
        if (G % p[i])
            continue;
        a[++cnt[0]][0] = p[i];
        while (G % p[i] == 0) G /= p[i], ++r[cnt[0]];
    }
    if (G > 1)
        a[++cnt[0]][0] = G, r[cnt[0]] = 1;
    for (re int i = 1; i <= p[0]; i++) {
        if (L % p[i])
            continue;
        a[++cnt[1]][1] = p[i];
        while (L % p[i] == 0) L /= p[i], ++b[cnt[1]];
    }
    if (L > 1)
        a[++cnt[1]][1] = L, b[cnt[1]] = 1;
    for (re int i = 1; i <= cnt[0]; i++)
        for (re int j = 1; j <= cnt[1]; j++)
            if (a[j][1] == a[i][0]) {
                l[i] = b[j];
                break;
            }
    for (re int i = 1; i <= cnt[0]; ++i) beg[i] = ksm(a[i][0], l[i]);
    dfs(1, 1, 0);
    for (re int i = 1; i <= n; i++) ma[c[i]] = i;
    dp[0][0] = 1;
    len = 1 << (2 * cnt[0]);
    for (re int i = 1; i <= n; i++)
        for (re int j = 0; j < len; j++) {
            if (!dp[i - 1][j])
                continue;
            dp[i][j] = (dp[i][j] + dp[i - 1][j]) % mod;
            dp[i][j | st[i]] = (dp[i][j | st[i]] + dp[i - 1][j]) % mod;
        }
    tp[n + 1][0] = 1;
    for (re int i = n; i; --i)
        for (re int j = 0; j < len; j++) {
            if (!tp[i + 1][j])
                continue;
            tp[i][j] = (tp[i][j] + tp[i + 1][j]) % mod;
            tp[i][j | st[i]] = (tp[i][j | st[i]] + tp[i + 1][j]) % mod;
        }
    while (Q--) {
        int x = read(), pos = ma[x];
        if (!pos) {
            puts("0");
            continue;
        }
        if (vis[pos]) {
            printf("%d\n", ans[pos]);
            continue;
        }
        Fwt(dp[pos - 1], 1), Fwt(tp[pos + 1], 1);
        for (re int i = 0; i < len; i++) d[i] = 1ll * dp[pos - 1][i] * tp[pos + 1][i] % mod;
        Fwt(d, -1);
        for (re int i = 0; i < len; i++)
            if ((i | st[pos]) + 1 == len)
                ans[pos] = (ans[pos] + d[i]) % mod;
        vis[pos] = 1;
        printf("%d\n", ans[pos]);
    }
    return 0;
}