2019年南京网络赛E题K Sum(莫比乌斯反演+杜教筛+欧拉降幂)

目录

• 题目链接
• 思路
• 代码

题目链接

传送门

思路

首先我们将原式化简：

\[\begin{aligned} &\sum\limits_{l_1=1}^{n}\sum\limits_{l_2=1}^{n}\dots\sum\limits_{l_k=1}^{n}gcd(l_1,l_2,\dots,l_k)^2&\\ =&\sum\limits_{d=1}^{n}d^2\sum\limits_{l_1=1}^{n}\sum\limits_{l_2=1}^{n}\dots\sum\limits_{l_k=1}^{n}[gcd(l_1,l_2,\dots,l_k)=d]&\\ =&\sum\limits_{d=1}^{n}d^2\sum\limits_{l_1=1}^{\frac{n}{d}}\sum\limits_{l_2=1}^{\frac{n}{d}}\dots\sum\limits_{l_k=1}^{\frac{n}{d}}[gcd(l_1,l_2,\dots,l_k)=1]& \end{aligned} \]

后面那一堆我们用经典反演套路进行反演得到：

\[\begin{aligned} \sum\limits_{d=1}^{n}d^2\sum\limits_{p=1}^{\lfloor\frac{n}{d}\rfloor}\mu(p)\lfloor\frac{n}{dp}\rfloor^k \end{aligned} \]

我们枚举\(T=dp\)得：

\[\begin{aligned} \sum\limits_{T=1}^{n}\lfloor\frac{n}{k}\rfloor^k\sum\limits_{t|T}\mu(t)\lfloor\frac{T}{t}\rfloor^2 \end{aligned} \]

因此题目要求的式子最后为：

\[\begin{aligned} &\sum\limits_{cnt=2}^{k}\sum\limits_{T=1}^{n}\lfloor\frac{n}{k}\rfloor^{cnt}\sum\limits_{t|T}\mu(t)\lfloor\frac{T}{t}\rfloor^2&\\ =&\sum\limits_{T=1}^{n}(\sum\limits_{cnt=2}^{k}\lfloor\frac{n}{k}\rfloor^{cnt})\sum\limits_{t|T}\mu(t)\lfloor\frac{T}{t}\rfloor^2&\\ =&\sum\limits_{T=1}^{n}(\frac{\lfloor\frac{n}{T}\rfloor\times(\lfloor\frac{n}{T}\rfloor^k-1)}{\lfloor\frac{n}{T}\rfloor-1}-\lfloor\frac{n}{T}\rfloor)\sum\limits_{t|T}\mu(t)\lfloor\frac{T}{t}\rfloor^2&\text{等比数列求和} \end{aligned}\\ \]

比赛的时候我写到这里就不会了，主要是后面那个不知道该怎么卷积进行杜教筛，赛后看了这篇博客才懂，下面思路基本上都是参考这位大佬的。

首先我们知道\(\mu\)为积性函数，\(id^2\)为积性函数，两者相乘还是积性函数，因此\(f(n)=\sum\limits_{d|n}\mu(d)\lfloor\frac{n}{d}\rfloor^2=\sum\limits_{d|n}\mu(d)id(\frac{n}{d})^2\)。

对于这个\(f\)的前缀和我们可以用杜教筛进行求解，设\(I*h=I*f=I*\mu *id^2\)，将其化简：

\[\begin{aligned} &\sum\limits_{t|n}I(\frac{n}{t})f(t)&\\ =&\sum\limits_{t|n}I(\frac{n}{t})\sum\limits_{d|t}\mu(t)id(\frac{t}{d})^2&\\ =&\sum\limits_{d|n}id(\frac{n}{d})^2\sum\limits_{x|\frac{n}{d}}I(\frac{n}{xd})\mu(\frac{xd}{d})&\\ =&\sum\limits_{d|n}id(\frac{n}{d})^2\sum\limits_{x|\frac{n}{d}}\mu(x)&\\ =&\sum\limits_{d|n}id(\frac{n}{d})^2[\frac{n}{d}=1]&\\ =&n^2& \end{aligned} \]

我当时看那篇博客时一直看不懂第二步到第三步是怎么来的，后面发现其实就是枚举\(x=\frac{t}{d}\)，然后那篇博客里面仍然用\(t\)来表示就导致看起来很迷。

然后进行杜教筛，设\(S(n)=\sum\limits_{i=1}^{n}f(i)=\sum\limits_{i=1}^{n}i^2-\sum\limits_{i=2}^{n}S(\frac{n}{i})=\frac{n(n+1)(2n+1)}{6}-\sum\limits_{i=2}^{n}S(\frac{n}{i})\)。

在杜教筛之前我们需要先预处理出\(n\leq1000000\)的情况，我们可以发现\(f(1)=1,f(p^c)=p^{2c}-p^{2c-2},f(p_1^{c_1}p_2^{c_2}\dots p_n^{c_n})=f(p_1^{c_1})f(p_2^{c_2})\dots f(p_n^{c_n})\)，然后就可以在线性筛时一并进行预处理掉辣~

对于最终答案里面的\(\sum\limits_{T=1}^{n}(\frac{\lfloor\frac{n}{T}\rfloor\times(\lfloor\frac{n}{T}\rfloor^k-1)}{\lfloor\frac{n}{T}\rfloor-1}-\lfloor\frac{n}{T}\rfloor)\)我们可以用数论分块进行处理，注意处理公比为\(1\)的情况，然后这题就做完了。

代码

#include <set>
#include <map>
#include <deque>
#include <queue>
#include <stack>
#include <cmath>
#include <ctime>
#include <bitset>
#include <cstdio>
#include <string>
#include <vector>
#include <cassert>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <unordered_map>
using namespace std;

typedef long long LL;
typedef pair<LL, LL> pLL;
typedef pair<LL, int> pLi;
typedef pair<int, LL> pil;;
typedef pair<int, int> pii;
typedef unsigned long long uLL;

#define lson rt<<1
#define rson rt<<1|1
#define lowbit(x) x&(-x)
#define name2str(name) (#name)
#define bug printf("*********\n")
#define debug(x) cout<<#x"=["<<x<<"]" <<endl
#define FIN freopen("D://Code//in.txt","r",stdin)
#define IO ios::sync_with_stdio(false),cin.tie(0)

const double eps = 1e-8;
const int mod = 1000000007;
const int maxn = 1000000 + 7;
const double pi = acos(-1);
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;

bool v[maxn];
char s[100007];
int _, n, k0, k, cnt, inv;
int p[maxn], sum[maxn], mu[maxn];
unordered_map<int,int> dp;

int add(int x, int y) {
    x += y;
    if(x >= mod) x -= mod;
    if(x < 0) x += mod;
    return x;
}

void init() {
    sum[1] = 1;
    for(int i = 2; i < maxn; ++i) {
        if(!v[i]) {
            sum[i] = add(1LL * i * i % mod, -1);
            p[cnt++] = i;
        }
        for(int j = 0; j < cnt && i * p[j] < maxn; ++j) {
            v[i*p[j]] = 1;
            if(i % p[j] == 0) {
                sum[i*p[j]] = 1LL * sum[i] * p[j] % mod * p[j] % mod;
                break;
            }
            sum[i*p[j]] = 1LL * sum[i] * add(1LL * p[j] * p[j] % mod, -1) % mod;
        }
    }
    for(int i = 2; i < maxn; ++i) {
        sum[i] = add(sum[i], sum[i-1]);
    }
}

LL qpow(LL x, int n) {
    LL res = 1;
    while(n) {
        if(n & 1) res = res * x % mod;
        x = x * x % mod;
        n >>= 1;
    }
    return res;
}

LL get_sum(int n, int k) {
    LL ans = 1LL * n * add(qpow(n, k), -1) % mod;
    int tmp = qpow(add(n, -1), mod - 2);
    ans = ans * tmp % mod;
    ans = add(ans, -n);
    return ans;
}

int dfs(int x) {
    if(x < maxn) return sum[x];
    if(dp.count(x)) return dp[x];
    LL ans = 1LL * x * (x + 1) % mod * (2 * x + 1) % mod * inv % mod;
    for(int l = 2, r; l <= x; l = r + 1) {
        r = x / (x / l);
        ans = add(ans, -(1LL * (r - l + 1) * dfs(x / l) % mod));
    }
    return dp[x] = ans;
}

int main() {
#ifndef ONLINE_JUDGE
FIN;
#endif // ONLINE_JUDGE
    init();
    inv = qpow(6, mod - 2);
    for(scanf("%d", &_); _; _--) {
        scanf("%d%s", &n, s + 1);
        int len = strlen(s + 1);
        k0 = k = 0;
        for(int i = 1; i <= len; ++i) {
            k = (10LL * k + (s[i] - '0')) % (mod - 1);
            k0 = (10LL * k0 + (s[i] - '0')) % mod;
        }
        k0 = (k0 - 1 + mod) % mod;
        LL ans = 0;
        for(int l = 1, r; l <= n; l = r + 1) {
            r = n / (n / l);
            int x = n / l;
            LL tmp;
            if(x == 1) tmp = k0;
            else tmp = get_sum(x, k);
            ans = add(ans, tmp * add(dfs(r), -dfs(l - 1)) % mod);
        }
        printf("%lld\n", ans);
    }
    return 0;
}