[Luogu 3768]简单的数学题

Description

输入一个整数n和一个整数p，你需要求出$(sum_{i=1}^nsum_{j=1}^n ijgcd(i,j))~mod~p$，其中gcd(a,b)表示a与b的最大公约数。

Input

一行两个整数p、n。

Output

一行一个整数$(sum_{i=1}^nsum_{j=1}^n ijgcd(i,j))~mod~p$。

Sample Input

998244353 2000

Sample Output

883968974

HINT

对于20%的数据，$n leq 1000$。

对于30%的数据，$n leq 5000$。

对于60%的数据，$n leq 10^6$，时限1s。

对于另外20%的数据，$n leq 10^9$，时限3s。

对于最后20%的数据，$n leq 10^{10}$，时限6s。

对于100%的数据，$5 imes 10^8 leq p leq 1.1 imes 10^9$且p为质数。

题解

题目要求 $$sum_{i=1}^nsum_{j=1}^n ijcdot gcd(i,j)$$

提出 $gcd$ egin{aligned}&sum_{d=1}^ndsum_{i=1}^{n}sum_{j=1}^nij[gcd(i,j)=d]\=&sum_{d=1}^nd^3sum_{i=1}^{leftlfloorfrac{n}{d} ight floor}sum_{j=1}^{leftlfloorfrac{n}{d} ight floor}ij[gcd(i,j)=1]\=&sum_{d=1}^nd^3sum_{i=1}^{leftlfloorfrac{n}{d} ight floor}sum_{j=1}^{leftlfloorfrac{n}{d} ight floor}ijsum_{kmid gcd(i,j)}mu(k)\=&sum_{d=1}^nd^3sum_{k=1}^{leftlfloorfrac{n}{d} ight floor}mu(k)k^2sum_{i=1}^{leftlfloorfrac{n}{kd} ight floor}sum_{j=1}^{leftlfloorfrac{n}{kd} ight floor}ijend{aligned}

令 $F(x)=sumlimits_{i=1}^xi=frac{x(x+1)}{2}$ ， $T=kd$ egin{aligned}Rightarrow&sum_{d=1}^nd^3sum_{k=1}^{leftlfloorfrac{n}{d} ight floor}mu(k)k^2F^2left(frac{n}{kd} ight)\=&sum_{T=1}^nF^2left(frac{n}{T} ight)sum_{dmid T}d^3left(frac{T}{d} ight)^2muleft(frac{T}{d} ight)\=&sum_{T=1}^nF^2left(frac{n}{T} ight)T^2sum_{dmid T}dcdotmuleft(frac{T}{d} ight)end{aligned}

前面的很好处理，但那个狄利克雷卷积是什么鬼...我们把它拎出来调教一下: $$sum_{dmid T}dcdotmuleft(frac{T}{d} ight)$$

这个形式似乎不好看，我们让它女装变个样子： $$sum_{dmid T}frac{T}{d}cdotmu(d)=Tsum_{dmid T}frac{mu(d)}{d}$$

这玩意不就是 $varphi(T)$ 么。带入原柿 $$sum_{T=1}^nF^2left(frac{n}{T} ight)T^2varphi(T)$$

现在就好搞♂了，美滋滋。记 $f(T)=T^2varphi(T)$ 显然他是个积性函数，可以杜教筛了。

考虑求 $S(n)=sumlimits_{i=1}^nf(i)$

上述式子 $$g(1)S(n)=sum_{i=1}^n(g*f)(i)-sum_{i=2}^ng(i)Sleft(leftlfloorfrac{n}{i} ight floor ight)$$

考虑到 $sumlimits_{dmid n}varphi(d)=n$ ，又由于 $(g*f)(n)=sumlimits_{dmid n}varphi(d)d^2cdot gleft(frac{n}{d} ight)$ 。我们考虑让 $g(n)=id^2(n)$ ，那么 $(id*f)(n)=sumlimits_{dmid n}n^2cdotvarphi(d)=n^3$ 。由于 $sumlimits_{i=1}^ni^3=frac{n^2(n+1)^2}{4}=left(frac{n(n+1)}{2} ight)^2=F^2(n)$ 。显然这个卷积的前缀为 $sumlimits_{i=1}^n(g*f)(i)=F^2(n)$ 。

故对于 $f$ $$S(n)=F^2(n)-sum_{i=2}^ni^2cdot Sleft(leftlfloorfrac{n}{i} ight floor ight)$$

由公式 $sumlimits_{i=1}^ni^2=frac{n(n+1)(2n+1)}{6}$ 现在整个柿子都好算了。

//It is made by Awson on 2018.1.25
#include <set>
#include <map>
#include <cmath>
#include <ctime>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
#define LL long long
#define Abs(a) ((a) < 0 ? (-(a)) : (a))
#define Max(a, b) ((a) > (b) ? (a) : (b))
#define Min(a, b) ((a) < (b) ? (a) : (b))
#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))
#define writeln(x) (write(x), putchar('
'))
#define lowbit(x) ((x)&(-(x)))
using namespace std;
const int N = 2333333;
void read(LL &x) {
    char ch; bool flag = 0;
    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());
    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());
    x *= 1-2*flag;
}
void write(LL x) {
    if (x > 9) write(x/10);
    putchar(x%10+48);
}

LL n, p, phi[N+5], inv2, inv6;
LL prime[N+5], isprime[N+5], tot;
map<LL,LL>mp;

LL quick_pow(LL a, LL b) {
    LL ans = 1; a %= p;
    while (b) {
    if (b&1) ans = ans*a%p;
    a = a*a%p; b >>= 1;
    }
    return ans;
}
void get_pre() {
    memset(isprime, 1, sizeof(isprime)); isprime[1] = 0, phi[1] = 1;
    for (int i = 2; i <= N; i++) {
    if (isprime[i]) prime[++tot] = i, phi[i] = 1ll*(i-1)*i%p*i%p;
    for (int j = 1; j <= tot && i*prime[j] <= N; j++) {
        isprime[i*prime[j]] = 0;
        if (i%prime[j]) phi[i*prime[j]] = phi[i]*(prime[j]-1)%p*prime[j]%p*prime[j]%p;
        else {phi[i*prime[j]] = phi[i]*prime[j]%p*prime[j]%p*prime[j]%p; break; }
    }
    (phi[i] += phi[i-1]) %= p;
    }
}
LL sum(LL n) {n %= p; return n*(n+1)%p*inv2%p; }
LL sum2(LL n) {n %= p; return n*(n+1)%p*(n*2+1)%p*inv6%p; }
LL get_phi(LL n) {
    if (n <= N) return phi[n];
    if (mp.count(n)) return mp[n];
    LL ans = sum(n)*sum(n)%p;
    for (LL i = 2, last; i <= n; i = last+1) {
    last = n/(n/i); (ans -= get_phi(n/i)*((sum2(last)-sum2(i-1))%p)%p) %= p;
    }
    return mp[n] = ans;
}
void work() {
    read(p), read(n); get_pre(); inv2 = quick_pow(2, p-2), inv6 = quick_pow(6, p-2); LL ans = 0;
    for (LL i = 1, last; i <= n; i = last+1) {
    last = n/(n/i); LL s = sum(n/i);
    (ans += s*s%p*((get_phi(last)-get_phi(i-1))%p)%p) %= p;
    }
    writeln((ans+p)%p);
}
int main() {
    work();
    return 0;
}