51Nod 最大公因数之和+最小公倍数之和(杜教筛)

1238 最小公倍数之和

http://www.51nod.com/Challenge/Problem.html#!#problemId=1238

1237 最大公因数之和

http://www.51nod.com/Challenge/Problem.html#!#problemId=1237

先来说最大公因数之和：

1237 最大公因数之和

$\sum _{i=1}^n\sum _{j=1}^n\gcd (i,j)$
$=\sum _{i=1}^n\sum _{j=1}^{n}d[\gcd (i,j)=d]$
$=\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }[\gcd (i,j)=1]$
$而\gcd (i,j)=1等同于\sum _{k|\gcd (i,j)}\mu (k)$
$所以原式=\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{k|\gcd (i,j)}\mu (k)$
$把k提到前面$
$$\begin{gathered} 原式=\sum _{d=1}^{n}d\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)\sum _{i=1}^{\lfloor \frac{n}{kd}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{kd}\rfloor }1 \\ 设函数S(n)=\sum _{i=1}^n\sum _{j=1}^{n}1 \\ 原式=\sum _{d=1}^{n}d\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)S(\frac{n}{kd}) \\ 令T=kd,原式=\sum _{T=1}^{n}S(\frac{n}{T})\sum _{k|T}\mu (k)\frac{T}{k} \\ 因为\mu \ast id=\phi (\ast 为狄利克雷卷积) \\ 所以原式=\sum _{T=1}^{n}S(\frac{n}{T})\phi (T) \end{gathered}$$
那么结果就显而易见了，结果就是欧拉函数的和，但是我们发现这个n是1e10的大小，线性筛会超时，所以欧拉函数的前缀和我们需要用杜教筛来求得，
我们知道$\varphi \ast I=id$,所以我们可以直接套杜教筛的公式
$$\begin{gathered} S(n)=\sum _{i=1}^n\phi (i) \\ g(1)S(n)=\sum _{i=1}^n(g\ast S)(n)-\sum _{i=2}^{n}g(i)S(\lfloor \frac{n}{i}\rfloor ) \end{gathered}$$
g函数为$I$
那么原式=$$\begin{gathered} \sum _{i=1}^{n}i-\sum _{i=2}^{n}S(\lfloor \frac{n}{i}\rfloor ) \\ =\frac{n(n+1)}{2}-\sum _{i=2}^{n}S(\lfloor \frac{n}{i}\rfloor ) \end{gathered}$$
对于杜教筛我们可以预先处理$n^{\frac{2}{3}}$的结果，然后后面的用杜教筛公式递归求得

AC代码：

#include 

using namespace std;
#define LL long long
const int M = 5e6;
const int Mod = 1e9 +7;
const double eps = 0.000000001;
LL phi[M + 30], prim[M + 30], tot;
bool mark[M + 30];
map<LL, LL> m;
#define inv_2 (Mod+1)/2

LL Add(LL a, LL b) {
    LL c = (a + b) % Mod;
    if(c >= Mod) return c - Mod;
    if(c < 0) return c + Mod;
    return c;
}

void init() {
    phi[1] = 1;
    for(LL i = 2; i <= M; i ++) {
        if(!mark[i]) {
            prim[++tot] = i;
            phi[i] = i - 1;
        }
        for (LL j = 1; j <= tot; j ++) {
            if(i * prim[j] > M) break;
            mark[i * prim[j]] = 1;
            if(i % prim[j] == 0) {
                phi[i * prim[j]] = phi[i] * prim[j];
                break;
            }
            phi[i *prim[j]] = phi[i] * phi[prim[j]];
        }
    }
    for (int i = 1; i <= M; i ++) phi[i] = Add(phi[i-1], phi[i]);
}

LL getPhi(LL n) {
    if(n <= M) return phi[n];
    if(m[n]) return m[n];
    LL ans;
    ans = 1LL * n % Mod * (n % Mod + 1) % Mod * inv_2 % Mod;
    for (LL l = 2, r; l <= n; l = r + 1) {
        r = n/(n/l);
        ans = (ans - (r - l + 1) % Mod * getPhi(n/l) % Mod + Mod) % Mod;
    }
    return m[n] = ans;
}
LL solve(LL n) {
    LL ans = 0;
    for (LL l = 1, r; l <= n; l = r + 1) {
        r = n/(n/l);
        ans = Add(ans, 1ULL * (n/l) % Mod * (n/l) % Mod * (Add(getPhi(r) % Mod, -getPhi(l-1) % Mod)  ) % Mod);
    }
    return ans;
}

int main()
{
    init();
    LL n;
    cin >> n;
    m.clear();
    cout << solve(n) << endl;
    return 0;
}

1238 最小公倍数之和

$$\begin{gathered} \sum _{i=1}^n\sum _{j=1}^{n}lcm(i,j) \\ =\sum _{i=1}^n\sum _{j=1}^n\frac{ij}{d}[\gcd (i,j)=d] \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }ij[\gcd (i,j)=1] \\ =\sum _{d=1}^{n}d\sum _{i=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{d}\rfloor }\sum _{k|\gcd (i,j)}\mu (k)ij \\ =\sum _{d=1}^{n}d\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)k^2\sum _{i=1}^{\lfloor \frac{n}{kd}\rfloor }\sum _{j=1}^{\lfloor \frac{n}{kd}\rfloor }ij \end{gathered}$$
$$\begin{gathered} 令S(n)=\sum _{i=1}^n\sum _{j=1}^{n}ij \\ 原式=\sum _{d=1}^{n}d\sum _{k=1}^{\lfloor \frac{n}{d}\rfloor }\mu (k)k^{2}S(\lfloor \frac{n}{kd}\rfloor ) \\ 令T=kd \\ 原式=\sum _{T=1}^{n}S(\lfloor \frac{n}{T}\rfloor )\sum _{k|T}\mu (k)k^2(\frac{T}{k}) \\ =\sum _{T=1}^{n}S(\lfloor \frac{n}{T}\rfloor )T\sum _{k|T}\mu (k)k \\ 令f(n)=n\sum _{k|n}\mu (k)k \\ 原式化简为=\sum _{T=1}^{n}S(\lfloor \frac{n}{T}\rfloor )f(T) \\ 令f\ast i{d}^2(\ast 为狄利克雷卷积) \\ f\ast i{d}^2=\sum _{d|n}f(d)\cdot (\frac{n}{d}{)}^2 \\ =\sum _{d|n}d\cdot \sum _{k|d}\mu (k)\cdot k\cdot (\frac{n}{d}{)}^2 \\ =\sum _{d|n}n\sum _{k|d}\mu (k)\cdot k\cdot (\frac{n}{d}) \\ 令t(n)=\sum _{k|d}\mu (d)\cdot d \\ t(n)=[(\mu \cdot id)\ast 1] \\ 则f\ast i{d}^2=n\sum _{d|n}\cdot t(d)\cdot (\frac{n}{d}) \\ =n[(\mu \cdot id)\ast 1\ast id] \\ =n[(\mu \cdot id)\ast id\ast 1] \\ =n[(\sum _{d|n}(\mu (d)\cdot d)\cdot (\frac{n}{d}))\ast 1] \\ =n[(n\sum _{d|n}\mu (d))\ast 1] \\ =n[e\ast 1] \\ =n \\ 带入杜教筛的公式 \\ g(1)S(n)=\sum _{i=1}^n(g\ast S)(n)-\sum _{i=1}^{n}g(i)S(\lfloor \frac{n}{i}\rfloor ) \\ g(n)=i{d}^2,S(n)=n\sum _{k|n}\mu (k)k \\ 所以g(1)S(n)=\sum _{i=1}^{n}i-\sum _{i=1}^n{i}^{2}S(\lfloor \frac{n}{i}\rfloor ) \\ =\frac{n\cdot (n+1)}{2}-\sum _{i=1}^n{i}^{2}S(\lfloor \frac{n}{i}\rfloor ) \end{gathered}$$
这样f的前做个就可以算出，而结果$\sum _{T=1}^{n}S(\lfloor \frac{n}{T}\rfloor )f(T)$
分块处理一下即可

AC代码：

#include 

using namespace std;
#define LL long long
const int M = 5e6 + 5;
const int Mod = 1e9 + 7;
const double eps = 0.000000001;
const int inf = 0x3f3f3f3f;
#define inv_2 (Mod+1)/2
#define inv_6 (Mod+1)/6

LL mu[M + 30], prim[M + 30];
bool vis[M + 30];
LL F[M + 30];
map<LL, LL> m;


void Add(LL &a, LL b) {
    a += b;
    if(a >= Mod) a -= Mod;
    if(a <= - Mod) a += Mod;
}


void Mobius(){
    mu[1] = 1;
    int tot = 0;
    for (int i = 2; i <= M; i ++) {
        if(!vis[i]) {
            prim[tot ++] = i;
            mu[i] = -1;
        }
        for (int j = 0; j < tot; j ++) {
            if(i * prim[j] > M) break;
            vis[i * prim[j]] = true;
            if(i % prim[j] == 0) {
                mu[i * prim[j]] = 0;
                break;
            }
            mu[i * prim[j]] = -mu[i];
        }
    }
    for (int i = 1; i <= M; i ++)
        for (int j = 1; j * i <= M; j ++)
            Add(F[i*j], mu[i] * i % Mod);
    for (int i = 1; i <= M; i ++) F[i] = F[i] * i % Mod;
    for (int i = 1; i <= M; i ++) Add(F[i], F[i-1]);
}

LL getX(LL X) {
    X %= Mod;
    return 1ULL * X * (X + 1) % Mod * (2 * X + 1) % Mod * inv_6 % Mod;
}

LL getX_2(LL l, LL r) {
    return 1ULL * (r - l + 1) % Mod * ((r + 1) % Mod) % Mod * inv_2 % Mod;
}

LL getS(LL n) {
    if(n <= M) return F[n];
    if(m[n]) return m[n];
    LL ans = 1ULL * n % Mod * (n+1) % Mod * inv_2 % Mod;
    for (LL l = 2, r; l <= n; l = r + 1) {
        r = n/(n/l);
        Add(ans, -((getX(r) - getX(l-1)) % Mod * getS(n/l) % Mod));
    }
    return m[n] = ans;
}

LL solve(LL n) {
    LL ans = 0;
    for (LL l = 1, r; l <= n; l = r + 1) {
        r = n/(n/l);
        LL k = getX_2(1, n/l);
        Add(ans, (k % Mod * k % Mod * ((getS(r) - getS(l-1)) % Mod) % Mod));
    }
    return (ans + Mod) % Mod;
}

int main()
{
    Mobius();
    LL n;
    scanf("%lld", &n);
    printf("%lld\n", solve(n));
    return 0;
}