「BZOJ 2154」Crash的数字表格「莫比乌斯反演」

版权声明：本文为hzy原创文章，未经博主允许不可随意转载。 https://blog.csdn.net/Binary_Heap/article/details/82223303

题目传送门

题意

求$\sum_{i=1}^{n} \sum_{j=1}^{m} lcm(i,j)$，对$20101009$取模

题解

不妨设$n \leq m$.

$\sum_{i=1}^{n} \sum_{j=1}^{m} lcm(i,j)$

$=\sum_{i=1}^{n} \sum_{j=1}^{m} \frac{ij}{gcd(i,j)}$

$=\sum_{d=1}^{n}\sum_{i=1}^{n} \sum_{j=1}^{m} \frac{ij}{d} [gcd(i,j)=d]$

$=\sum_{d=1}^{n}\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} ijd [gcd(i,j)=1]$

$=\sum_{d=1}^{n}d\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} ij[gcd(i,j)=1]$

$=\sum_{d=1}^{n}d\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} ije(gcd(i,j))$

$=\sum_{d=1}^{n}d\sum_{i=1}^{\lfloor\frac{n}{d}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{d}\rfloor} ij \sum_{k|gcd(i,j)} \mu(k)$

$=\sum_{d=1}^{n}d \sum_{k=1}^{\lfloor\frac{n}{d}\rfloor} \mu(k) \sum_{i=1}^{\lfloor\frac{n}{dk}\rfloor} \sum_{j=1}^{\lfloor\frac{m}{dk}\rfloor} ijk^2$

$=\sum_{d=1}^{n}d \sum_{k=1}^{\lfloor\frac{n}{d}\rfloor} k^2\mu(k) \sum_{i=1}^{\lfloor\frac{n}{dk}\rfloor} i\sum_{j=1}^{\lfloor\frac{m}{dk}\rfloor} j$

记$sum(k)$表示$\sum_{i=1}^{k}i$，即$\frac{k(k+1)}{2}$

$=\sum_{d=1}^{n}d \sum_{k=1}^{\lfloor\frac{n}{d}\rfloor} k^2\mu(k) sum(\lfloor\frac{n}{dk}\rfloor)sum(\lfloor\frac{m}{dk}\rfloor)$

枚举$T=dk$：

$=\sum_{T=1}^{n}sum(\lfloor\frac{n}{T}\rfloor)sum(\lfloor\frac{m}{T}\rfloor)\sum_{d|T}d (\frac{T}{d})^2\mu(\frac{T}{d})$

考虑到枚举$d$相当于枚举$T/d$：

$=\sum_{T=1}^{n}sum(\lfloor\frac{n}{T}\rfloor)sum(\lfloor\frac{m}{T}\rfloor)\sum_{d|T}\frac{T}{d} (d)^2\mu(d)$

$=\sum_{T=1}^{n}sum(\lfloor\frac{n}{T}\rfloor)sum(\lfloor\frac{m}{T}\rfloor)\sum_{d|T} Td\mu(d)$

$=\sum_{T=1}^{n}sum(\lfloor\frac{n}{T}\rfloor)sum(\lfloor\frac{m}{T}\rfloor)T\sum_{d|T} d\mu(d)$

记$f(T)=\sum_{d|T} d\mu(d)$，这可以线性筛

如果$p$是$i$的一个质因子，那么$f(ip)=f(i)$（因为$p$加入不会产生新的有贡献的约数$d$）。否则这个$p$乘上之前的每一个约数$d$，使得$\mu(pd)=-\mu(d)$，加入答案里，即$f(ip)=f(i)-pf(i)=(1-p)f(i)$，注意到$f(p)=1-p$（根据定义），即$f(ip)=f(i)f(p)$.

于是就可以愉快的数论分块+线性筛了

#include <iostream>
#include <cstdio>
using namespace std;

const int MOD = 20101009;
const int MAXN = 1e7 + 10;

int n, m;
int pr[MAXN], tot;
int f[MAXN], sum[MAXN];
bool tag[MAXN];

void init(int n) {
    tag[1] = true;
    f[1] = sum[1] = 1;
    for(int i = 2; i <= n; i ++) {
        if(!tag[i]) {
            pr[++ tot] = i;
            f[i] = ((1 - i) % MOD + MOD) % MOD;
        }
        for(int j = 1; j <= tot && i * 1ll * pr[j] <= n; j ++) {
            tag[i * pr[j]] = true;
            if(i % pr[j] == 0) {
                f[i * pr[j]] = f[i];
                break ;
            }
            f[i * pr[j]] = f[i] * 1ll * f[pr[j]] % MOD;
        }
    }
    for(int i = 2; i <= n; i ++) {
        f[i] = (f[i - 1] + 1ll * f[i] * i % MOD) % MOD;
        sum[i] = (sum[i - 1] + 1ll * i) % MOD;
    }
}

int main() {
    scanf("%d%d", &n, &m);
    if(n > m) swap(n, m);
    init(m);
    int ans = 0;
    for(int i = 1, j; i <= n; i = j + 1) {
        j = min(n / (n / i), m / (m / i));
        ans = (ans + ((( (f[j] - f[i-1]) % MOD + MOD ) % MOD) * 1ll \
        * sum[n / i] % MOD * 1ll * sum[m / i] % MOD)) % MOD;
    }
    printf("%d\n", ans);
    return 0;
}