HDU 6537 Neko and function（min_25筛，容斥，组合数学）

参考博客：https://www.90yang.com/2019ccpc-xiangtan-f-neko-and-function/

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思维比较巧妙，首先从贡献的角度计算 $f(n,k)$，枚举 n 的每一个质因子，设其幂次为 $e$，问题等价于将这 $e$ 个 $p$ 放到 $k$ 个盒子里。

由于盒子不允许为空，这个贡献很难计算，考虑盒子可以为空，用容斥得到最后的答案：令 $g(n,k)$ 为 $f(n,k)$ 去掉 $a_i > 1$ 的限制。

没有这个限制，每一个素数的贡献独立，即 $g(n,k)$ 是一个积性函数

考虑容斥得到 $f(n,k)$：枚举有几个盒子为空，容斥一下即可：
$\displaystyle f(n,k) = \sum_{x = 0}^k(-1)^{k - x}*C(k,x)*g(n,x)$
$\displaystyle \sum_{i = 1}^nf(i,k) = \sum_{x = 0}^k(-1)^{k - x}*C(k,x)*\sum_{i = 1}^ng(i,x)$
由于 $g(n,k)$ 是积性函数，考虑用min_25筛得到 $\displaystyle\sum_{i = 1}^ng(i,x)$

由于这题卡常，我最终也没有卡过去，待update

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#include<bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 10;
typedef long long ll;
const int mod = 1e9 + 7;
int num,tot,k;
bool ispri[maxn];
int sum[maxn],pri[maxn],w[maxn],g[maxn],id1[maxn],id2[maxn];
int n,sqr,fact[maxn],ifact[maxn],ans[maxn];
inline void sieve(int x) {
    num = 0; ispri[0] = ispri[1] = true;
    for (int i = 2; i <= x; i++) {
        if (!ispri[i]) {
            pri[++num] = i;
            sum[num] = (sum[num - 1] + 1) % mod;
        }
        for (int j = 1; j <= num && i * pri[j] <= x; j++) {
            ispri[i * pri[j]] = true;
            if (i % pri[j] == 0) break;
        }
    }
}
int fpow(int a,int b) {
    int r = 1;
    while (b) {
        if (b & 1) r = 1ll * r * a % mod;
        b >>= 1;
        a = 1ll * a * a % mod;
    }
    return r;
}
inline int C(int x,int y) {
    if (y > x || y < 0) return 0;
    return 1ll * fact[x] * ifact[y] % mod * ifact[x - y] % mod;
}
inline int S(ll x,int y,int z) {
    if (pri[y] >= x) return 0;
    int k = x <= sqr ? id1[x] : id2[n / x];
    int ans = 1ll * z * (g[k] - sum[y] + mod) % mod;
    for (int i = y + 1; i <= num && pri[i] <= x / pri[i]; i++)
        for (ll e = 1, pe = pri[i]; pe <= x; e++, pe = pe * pri[i]) 
            ans = (ans + 1ll * (S(x / pe,i,z) + (e != 1)) % mod * C(e + z - 1,z - 1) % mod) % mod;
    return ans;
}
int main() {
    int N = maxn - 10;
    sieve(N);
    fact[0] = 1;
    for (int i = 1; i <= N; i++) 
        fact[i] = 1ll * fact[i - 1] * i % mod;
    ifact[N] = fpow(fact[N],mod - 2);
    for (int i = N - 1; i >= 0; i--)
        ifact[i] = 1ll * ifact[i + 1] * (i + 1) % mod;
    while(scanf("%d%d",&n,&k) != EOF) {
    	if (k == 1) {
    		printf("%d\n",(n - 1) % mod);
    		continue;
		}
    	else if (n < (1<<k)) {
    		puts("0");
    		continue;
		}
    	else if (n < (1 << k) + (1 << (k - 1))) {
    		puts("1");
			continue;	
		}
        tot = 0;
        sqr = sqrt(n);
        for (int i = 1,j; i <= n; i = j + 1) {
            j = n / (n / i);
            w[++tot] = n / i;
            int p = n / i % mod;
            g[tot] = (p - 1) % mod;
            if (n / i <= sqr) id1[n / i] = tot;
            else id2[j] = tot;
        }
        for (int i = 1; i <= num; i++) {
            for (int j = 1; j <= tot && pri[i] <= w[j] / pri[i]; j++) {
                int x = w[j] / pri[i];
                x = x <= sqr ? id1[x] : id2[n / x];
                g[j] -= (g[x] - sum[i - 1] + mod) % mod;
                if (g[j] < 0) g[j] += mod;
            }
        }
        int res = 0, t = 1;
        for (int i = k; i >= 0; i--) {
        	if (k == 0) res = (res + C(k,i) * t % mod) % mod;
        	else if (k == 1) res = (res + 1ll * n * C(k,i) % mod * t % mod) % mod;
            else res = (res + 1ll * S(n,0,i) * C(k,i) % mod * t % mod) % mod;
            if (res < 0) res += mod;
            t *= -1;
        }
        printf("%d\n",res);
    }
    return 0;
}