[BZOJ3434][Wc2014]时空穿梭（莫比乌斯反演）

Address

https://www.luogu.org/problemnew/show/P4152
https://www.lydsy.com/JudgeOnline/problem.php?id=3434
http://uoj.ac/problem/54

Solution

考虑枚举第一个点的坐标 $(x_{1,1},x_{1,2},...,x_{1,n})$ 和
最后一个点的坐标 $(x_{c,1},x_{c,2},...,x_{c,n})$ ，
那么容易得出，这两个点连成的线段上的整点（不包括端点）个数为：

$$\gcd(x_{c,1}-x_{1,1},x_{c,2}-x_{1,2},...,x_{c,n}-x_{1,n})-1$$
对答案的贡献：
$$C_{\gcd(x_{c,1}-x_{1,1},x_{c,2}-x_{1,2},...,x_{c,n}-x_{1,n})-1}^{c-2}$$
考虑枚举 $n$ 维向量 $z$ 满足：
$$z_i=x_{c,1}-x_{1,1}$$
那么 $x_{1,i}$ 就有 $m_i-z_i$ 种合法取值。
得到答案为：
$$\sum_{z_1=1}^{m_1}\sum_{z_2=1}^{m_2}...\sum_{z_n=1}^{m_n}C_{\gcd(z_1,z_2,...,z_n)-1}^{c-2}\prod_{i=1}^n(m_i-z_i)$$
这是一个显然的 Mobius 反演。考虑把 $\gcd$ 解决掉：
$$\sum_d\sum_{z_1=1}^{m_1}\sum_{z_2=1}^{m_2}...\sum_{z_n=1}^{m_n}C_{d-1}^{c-2}\prod_{i=1}^n(m_i-z_i)[\gcd(z_1,z_2,...,z_n)=d]$$
$$=\sum_d\sum_{z_1=1}^{\lfloor\frac{m_1}d\rfloor}\sum_{z_2=1}^{\lfloor\frac{m_2}d\rfloor}...\sum_{z_n=1}^{\lfloor\frac{m_n}d\rfloor}C_{d-1}^{c-2}\prod_{i=1}^n(m_i-z_id)\sum_{k|\gcd(z_1,z_2,...,z_n)}\mu(k)$$
$$=\sum_d\sum_k\mu(k)\sum_{z_1=1,k|z_1}^{\lfloor\frac{m_1}d\rfloor}\sum_{z_2=1,k|z_2}^{\lfloor\frac{m_2}d\rfloor}...\sum_{z_n=1,k|z_n}^{\lfloor\frac{m_n}d\rfloor}C_{d-1}^{c-2}\prod_{i=1}^n(m_i-z_id)$$
$$=\sum_d\sum_k\mu(k)C_{d-1}^{c-2}\prod_{i=1}^n\sum_{j=1}^{\lfloor\frac{m_i}{dk}\rfloor}(m_i-jdk)$$
$$=\sum_u\prod_{i=1}^n\sum_{j=1}^{\lfloor\frac{m_i}u\rfloor}(m_i-uj)\sum_{d|u}C_{d-1}^{c-2}\mu(\frac ud)$$
$$=\sum_u\prod_{i=1}^n(m_i\lfloor\frac{m_i}u\rfloor-\frac{\lfloor\frac{m_i}u\rfloor(\lfloor\frac{m_i}u\rfloor+1)}2u)\sum_{d|u}C_{d-1}^{c-2}\mu(\frac ud)$$
$\prod_{i=1}^n(m_i\lfloor\frac{m_i}u\rfloor-\frac{\lfloor\frac{m_i}u\rfloor(\lfloor\frac{m_i}u+1)}2u)$ 是一个 $n$ 次多项式，可以用 $O(n^2)$ 的时间求出，设 $i$ 次项为 $a_i$ 。
$$=\sum_u\prod_{i=0}^na_iu^i\sum_{d|u}C_{d-1}^{c-2}\mu(\frac ud)$$
预处理 $f(u,c,i)=u^i\sum_{d|u}C_{d-1}^{c-2}\mu(\frac ud)$ 后：
$$=\sum_u\prod_{i=0}^na_if(u,c,i)$$
可以得出，由于多项式 $a$ 的系数只有对 $u$ 整除（下取整）的结果，故 $a$ 的取值不超过 $O(n\sqrt m)$ 种。
对下界分块之后使用 $f$ 的前缀和计算。
复杂度 $O(mnc+Tn^3\sqrt m)$ （这能过？）
~~辣鸡 BZOJ 评测机卡常数~~

Code

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define For(i, a, b) for (i = a; i <= b; i++)
#define Mobius(i, n) for (i = 1; i <= n;)
using namespace std;

inline int read()
{
    int res = 0; bool bo = 0; char c;
    while (((c = getchar()) < '0' || c > '9') && c != '-');
    if (c == '-') bo = 1; else res = c - 48;
    while ((c = getchar()) >= '0' && c <= '9')
        res = (res << 3) + (res << 1) + (c - 48);
    return bo ? ~res + 1 : res;
}

const int MaxN = 1e5, E = 23, P = 14, N = MaxN + 5, MOD = 10007;
int tot, pri[N], miu[N], C[N][E], pw[N][P], f[N][E], sum[N][E][P],
n, c, m[P], a[P], b[P];
bool mark[N];

int Min(int a, int b) {return a < b ? a : b;}

void sieve()
{
    int i, j, k;
    For (i, 0, MaxN) C[i][0] = pw[i][0] = 1;
    For (i, 1, MaxN)
    {
        For (j, 1, Min(18, i))
            C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
        For (j, 1, 11) pw[i][j] = 1ll * pw[i][j - 1] * i % MOD;
    }
    mark[0] = mark[miu[1] = 1] = 1;
    For (i, 2, MaxN)
    {
        if (!mark[i]) miu[pri[++tot] = i] = -1;
        For (j, 1, tot)
        {
            if (1ll * i * pri[j] > MaxN) break;
            mark[i * pri[j]] = 1;
            if (i % pri[j] == 0) break;
            else miu[i * pri[j]] = -miu[i];
        }
    }
    For (k, 2, 20)
        For (i, 1, MaxN)
        {
            int dl = MaxN / i;
            For (j, 1, dl)
                f[i * j][k] = (f[i * j][k] + MOD +
                    C[i - 1][k - 2] * miu[j]) % MOD;
            For (j, 0, 11)
                sum[i][k][j] = (sum[i - 1][k][j] +
                    f[i][k] * pw[i][j]) % MOD;
        }
}

void prod(int x, int y, int n)
{
    int i;
    a[n + 1] = b[0] = 0;
    For (i, 1, n + 1) b[i] = a[i - 1] * y % MOD;
    For (i, 0, n + 1) a[i] = (a[i] * x - b[i] + MOD) % MOD;
}

void work()
{
    int i, j, _min, ans = 0;
    n = read(); c = read();
    For (i, 1, n) m[i] = _min = read();
    For (i, 1, n) _min = Min(_min, m[i]);
    Mobius (i, _min)
    {
        int nxt = m[1] / (m[1] / i);
        For (j, 2, n) nxt = Min(nxt, m[j] / (m[j] / i)), a[j] = 0;
        a[a[0] = 1] = 0;
        For (j, 1, n) prod(1ll * m[j] * (m[j] / i) % MOD,
            (1ll * (m[j] / i) * (m[j] / i + 1) >> 1) % MOD, j - 1);
        For (j, 0, n) ans = (ans + a[j] * (sum[nxt][c][j] -
            sum[i - 1][c][j] + MOD)) % MOD;
        i = nxt + 1;
    }
    printf("%d\n", ans);
}

int main()
{
    int T = read();
    sieve();
    while (T--) work();
    return 0;
}