莫比乌斯反演学习日记

@
如有错误,忘大佬不吝赐教,及时指出。
积性函数好难啊,求大佬教我积性函数,或者推荐点博客。。
如果感觉有点卡,请点这里。

莫比乌斯函数

参考:peng-ym

\(\sum_{d|n} \mu(d) = [n = 1]\) \((拆成二项式定理就很易证明了\)

\(\sum_{d|n}\phi(d) = n\) \((wiki上的证明挺易理解,就是化简\{\frac 1n,\frac 2n...\frac nn\}\)

\(\phi(n) = \sum_{d|n} \mu(d)\frac{n}{d}\)   \((容斥过程)\)

\(\sum_{d|gcd(i,j)} \mu(d) = [gcd(i,j) = 1]\;--式子C\)


莫比乌斯反演

\(F(n)=\sum_{d|n}f(d) => f(n)=\sum_{d|n}u(d)F(⌊\frac{n}{d}⌋)\;--式子B\)

\(F(n)=\sum_{n|d}f(d) => f(n)=\sum_{n|d}u(\frac{d}{n})F(d)\;--式子A\)

从莫比乌斯到欧拉:here


【CJOJ2512】gcd之和 -反演

求:\(ans = \sum_{i=1}^{n}\sum_{j=1}^{m}gcd(i,j)\;\;\;n\leq m\leq1e7\)

转换一下:
\[ans = \sum_{d=1}^{n}d \sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=d]=\sum_{d=1}^{n} d\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{m}{d}}[gcd(i,j)=1]\]

法1:

\[f(d)=\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=d]并且F(n)=\sum_{n|d}f(d)\]

\[F(n)=\sum_{i=1}^{n}\sum_{j=1}^{m}[d|gcd(i,j)]=⌊\frac{n}{d}⌋\times ⌊ \frac{m}{d}⌋\]

\[f(x)=\sum_{x|d}^{min(n,m)}u(\frac{d}{x})F(d)\]

\[f(1)=\sum_{d=1}^{min(n,m)}u(d)F(d)=\sum_{d=1}^{min(n,m)}u(d)⌊\frac{n}{d}⌋\times ⌊ \frac{m}{d}⌋\]

\[ans = \sum_{d=1}^{n}d\sum_{i=1}^{\frac{n}{d}}u(i)\frac{n}{d\times i}\frac{m}{d\times i}\]

参考:yyb jk_chen

法2:(无敌易理解)

\[\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=1]=\sum_{i=1}^{n}\sum_{j=1}^{m}\sum_{d|gcd(i,j)} u(d)\]

\[\sum_{d=1}^{n}u(d)\sum_{i=1}^{n}\sum_{j=1}^{m}[d|gcd(i,j)]=\sum_{d=1}^{n}u(d)⌊\frac{n}{d}⌋⌊\frac{m}{d}⌋\]

\[ans = \sum_{d=1}^{n}d\sum_{i=1}^{\frac{n}{d}}u(i)\frac{n}{d\times i}\frac{m}{d\times i}\]

参考:这是一个神奇的Blog
AC_CODE: 和下面类似,就不写了。


P2257 YY的GCD -反演

题目:传送门
分析:
\[ans=\sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=prime]\;\;n\leq m\leq 1e7\]
\(prime\)提到前面去:
\[ans=\sum_{d=2}^{n}\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}[gcd(i,j)=1]\;d\in prime\]
莫比乌斯反演一下:
\[ans=\sum_{d=2}^{n}\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}\sum_{k|gcd}\mu(k)\;d\in prime\]
\(k\)提到前面来:
\[ans=\sum_{d=2}^{n}\sum_{k=1}^{\frac{n}{d}}\mu(k)\frac{n}{kd}\frac{m}{kd}\;d\in prime\]

这样写会超时,因为多组数据,还得优化,就是交换求和顺序,令\(T=kd:\)

\[ans=\sum_{T=2}^{n}\frac{n}{T}\frac{m}{T}\sum_{d|T}\mu(\frac{T}{d})\;d\in prime\]

\(O(nlog(n))\)预处理这个式子\(\sum_{d|T}\mu(\frac{T}{d})\):枚举每个素数,给它的\(x\)倍数加上\(\mu(x)\),然后前缀和。

这个式子也可以\(O(n)\)预处理:
在这里插入图片描述

AC_CODE:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

const int MXN = 1e7+ 7;
const int mod = 998244353;

LL n, m;
int noprime[MXN], pp[MXN], pcnt;
int phi[MXN], mu[MXN];
LL pre_mu[MXN];
void init_prime() {
    noprime[0] = noprime[1] = 1;
    mu[1] = 1; phi[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, phi[i] = i-1, mu[i] = -1;
        for(int j = 0; j < pcnt && pp[j] * i < MXN; ++j) {
            noprime[pp[j]*i] = 1;
            //phi[pp[j]*i] = (pp[j]-1)*phi[i];
            mu[pp[j]*i] = -mu[i];
            if(i % pp[j] == 0) {
                //phi[pp[j]*i] = pp[j]*phi[i];
                mu[pp[j]*i] = 0;
                break;
            }
        }
    }
}
void init_ans() {
    for(int i = 0; i < pcnt; ++i) {
        for(int j = 1; j*pp[i] < MXN; ++j) {
            pre_mu[j*pp[i]] += mu[j];
        }
    }
    for(int i = 2; i < MXN; ++i) pre_mu[i] += pre_mu[i-1];
}
int main() {
    init_prime();
    init_ans();
    int tim; scanf("%d", &tim);
    while(tim --) {
        scanf("%lld%lld", &n, &m);
        if(n > m) swap(n, m);
        LL ans = 0, tmp;
        for(LL L = 2, R; L <= n; L = R + 1) {
            R = min(n/(n/L),m/(m/L));
            tmp = pre_mu[R]-pre_mu[L-1];
            ans = (ans + tmp*(n/L)*(m/L));
        }
        printf("%lld\n", ans);
    }
    return 0;
}



小清新数论 -杜教筛

题目:wannafly winter camp day3 F
\[ans=\sum_{i=1}^{n}\sum_{j=1}^{n}\mu(gcd(i,j))\;\;n\leq1e11\]
这个式子与\(gcd\)之和那题非常相似,如果数据范围是\(1e7\)的话,这两题代码就基本一样了。可这题数据范围\(1e11\),就用杜教筛咯。
杜教筛学习博客:here
方法1:利用欧拉函数前缀和
\[ans=\sum_{i=1}^{n}\sum_{j=1}^{n}\mu(gcd(i,j))\;\;n\leq1e11\]

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

\[F(n)=\sum_{i=1}^{n}\sum_{j=1}^{n}[gcd(i,j)=1],\;P(n)=\sum_{i=1}^{n}\phi(i)=\sum_{i=1}^{n}\sum_{j=1}^{i}[gcd(i,j)=1]\]

\[F(n)=P(n)\times2 -1\]

\[ans = \sum_{d=1}^{n}\mu(d)\times(P(\frac{n}{d})\times 2-1)\]


方法2:利用莫比乌斯反演和迪利克雷卷积
\[ans=\sum_{i=1}^{n}\sum_{j=1}^{n}\mu(gcd(i,j))\;\;n\leq1e11\]
化简一下:\[ans = \sum_{d=1}^{n}\mu(d) \sum_{i=1}^{n}\sum_{j=1}^{m}[gcd(i,j)=d]=\sum_{d=1}^{n}\mu(d)\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}[gcd(i,j)=1]\]
反演一下:\[ans=\sum_{d=1}^{n}\mu(d)\sum_{i=1}^{\frac{n}{d}}\sum_{j=1}^{\frac{n}{d}}\sum_{k|gcd(i,j)}\mu(k)\]
\(k\)提出来:\[ans=\sum_{d=1}^{n}\mu(d)\sum_{k=1}^{\frac{n}{d}}\mu(k)(\frac n{kd})^2\]
又到了熟悉的步骤:令\(T=kd\)
\[ans=\sum_{T=1}^n(\frac nT)^2\sum_{d|T}\mu(d)\times \mu(\frac Td)=\sum_{T=1}^n(\frac nT)^2(\mu*\mu)(T)\]
接着问题变成求下面这个式子的前缀和:\[\sum_{i=1}^n (\mu*\mu)(i)\]

AC_CODE1:

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;

const int MXN = 1e7+ 7;
const LL mod = 998244353;
const LL inf = 1e18;
LL n, m, inv2;
int noprime[MXN], pp[MXN], pcnt;
int phi[MXN], mu[MXN];
LL pre_mu[MXN], pre_phi[MXN];
//unordered_map<LL, LL> mp1, mp2;
void init_prime() {
    noprime[0] = noprime[1] = 1;
    mu[1] = 1; phi[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, phi[i] = i-1, mu[i] = -1;
        for(int j = 0; j < pcnt && pp[j] * i < MXN; ++j) {
            noprime[pp[j]*i] = 1;
            phi[pp[j]*i] = (pp[j]-1)*phi[i];
            mu[pp[j]*i] = -mu[i];
            if(i % pp[j] == 0) {
                phi[pp[j]*i] = pp[j]*phi[i];
                mu[pp[j]*i] = 0;
                break;
            }
        }
    }
    for(int i = 1; i < MXN; ++i) {
        pre_mu[i] = pre_mu[i-1] + mu[i];
        pre_phi[i] = (pre_phi[i-1] + phi[i])%mod;
    }
}
struct Hash_map{
    static const int mask=0x7fffff;
    LL p[mask+1],q[mask+1];
    void clear(){
        memset(q,0,sizeof(q));
    }
    LL& operator [](LL k){
        LL i;
        for(i=k&mask;q[i]&&p[i]!=k;i=(i+1)&mask);
        p[i]=k;
        return q[i];
    }
}mp1, mp2;
LL solve_mu(LL n) {
    if(n < MXN) return pre_mu[n];
    if(mp1[n] == inf) return 0;
    if(mp1[n]) return mp1[n];
    LL ans = 1;
    for(LL L = 2, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans -= (R-L+1LL)%mod*solve_mu(n/L);
    }
    mp1[n] = ans;
    if(mp1[n] == 0) mp1[n] = inf;
    return ans;
}
LL solve_phi(LL n) {
    if(n < MXN) return pre_phi[n];
    if(mp2[n]) return mp2[n];
    LL ans = n%mod*(n+1)%mod*inv2%mod;
    for(LL L = 2, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans - (R-L+1LL)%mod*solve_phi(n/L)%mod)%mod;
    }
    ans = (ans + mod) % mod;
    mp2[n] = ans;
    return ans;
}
int main() {
    inv2 = 499122177;
    init_prime();
    scanf("%lld", &n);
    LL ans = 0;
    for(LL L = 1, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans + (solve_mu(R)-solve_mu(L-1))%mod*(solve_phi(n/L)*2LL%mod-1LL)%mod+mod)%mod;
    }
    printf("%lld\n", (ans+mod)%mod);
    return 0;
}

AC_CODE2:

#include<bits/stdc++.h>
#define fi first
#define se second
#define iis std::ios::sync_with_stdio(false);cin.tie(0)
#define pb push_back
#define o2(x) (x)*(x)
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int INF = 0x3f3f3f3f;
const LL MOD = 998244353;
const int MXN = 5e6 + 6;

LL n, q, m;
int noprime[MXN], pp[MXN/2], pcnt;
int mu[MXN], phi[MXN];
int pre_mu[MXN];//莫比乌斯函数的前缀和
LL mumu[MXN];//莫比乌斯函数卷积莫比乌斯函数的前缀和
unordered_map<LL,LL>mp1,mp2;

void init_rime() {
    noprime[0] = noprime[1] = 1;
    mu[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, mu[i]=-1;
        for(int j = 0; j < pcnt && i*pp[j] < MXN; ++j) {
            noprime[i*pp[j]] = 1;
            mu[i*pp[j]] = -mu[i];
            if(i % pp[j] == 0) {
                mu[i*pp[j]] = 0;
                break;
            }
        }
    }
    for(int i = 1; i < MXN; ++i) pre_mu[i] = pre_mu[i-1] + mu[i], mumu[i] = mu[i];
    for(int i = 2; i < MXN; ++i) {//预处理莫比乌斯卷积莫比乌斯
        for(int j = i; j < MXN; j += i) {
            mumu[j] += mu[i]*mu[j/i];
            if(mumu[j] >= MOD) mumu[j] %= MOD;
        }
    }
    for(int i = 2; i < MXN; ++i) mumu[i] = (mumu[i]+mumu[i-1]+MOD)%MOD;
}

LL solve_u(LL n) {//求莫比乌斯前缀和
    if(n < MXN) return pre_mu[n];
    if(mp1.count(n)) return mp1[n];
    LL ans = 1;
    for(LL L = 2, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans - (R-L+1)%MOD*solve_u(n/L)%MOD+MOD)%MOD;
    }
    mp1[n] = ans;
    return ans;
}
LL solve_uu(LL n) {//求莫比乌斯卷积莫比乌斯前缀和
    if(n < MXN) return mumu[n];
    if(mp2.count(n)) return mp2[n];
    LL ans = 0;
    for(LL L = 1, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans + (solve_u(R)-solve_u(L-1)+MOD)%MOD*solve_u(n/L)%MOD);
        if(ans >= MOD) ans %= MOD;
    }
    mp2[n] = (ans+MOD)%MOD;
    return ans;
}
int main(int argc, char const *argv[]) {
    init_rime();
    scanf("%lld", &n);
    LL ans = 0;
    for(LL L = 1, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans + (n/L)%MOD*(n/L)%MOD*(solve_uu(R)-solve_uu(L-1)+MOD)%MOD)%MOD;
    }
    printf("%lld\n", (ans+MOD)%MOD);
    return 0;
}
/*
void init_rime() {//因为莫比乌斯卷积莫比乌斯是积性函数,所以可以On预处理
    noprime[0] = noprime[1] = 1;
    mu[1] = 1;mumu[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, mu[i]=-1, mumu[i]=-2;
        for(int j = 0; j < pcnt && i*pp[j] < MXN; ++j) {
            noprime[i*pp[j]] = 1;
            mu[i*pp[j]] = -mu[i];
            mumu[i*pp[j]] = mumu[i]*mumu[pp[j]];
            if(i % pp[j] == 0) {
                mu[i*pp[j]] = 0;
                if((i/pp[j])%pp[j]) mumu[i*pp[j]] = mumu[i/pp[j]];
                else mumu[i*pp[j]] = 0;
                break;
            }
        }
    }
    for(int i = 1; i < MXN; ++i) pre_mu[i] = pre_mu[i-1] + mu[i];
    for(int i = 2; i < MXN; ++i) mumu[i] = (mumu[i]+mumu[i-1]+MOD)%MOD;
}
*/



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

2154 化简一下:
\[ans=\sum_{i=1}^n\sum_{j=1}^{m}lcm(i,j)=\sum_{i=1}^n\sum_{j=1}^{m}\frac{ij}{gcd(i,j)}\]

经过一系列化简可得:

\[ans=\sum_{d=1}^nd\sum_{k=1}^{min(\frac nd,\frac md)}\mu(k)\times k^2\sum_{i=1}^{\frac n{dk}}i\sum_{j=1}^{\frac m{kd}}j,O(n)\]

这个式子是\(O(n)\),但是通过改变枚举方法可以\(O(\sqrt n)\)求解。

\(T=kd,h(n)=\sum_{i=1}^ni\)可得:

\[ans=\sum_{d=1}^nd\sum_{k=1}^{\frac nd}\mu(k)\times k^2\times h(\frac nT)\times h(\frac mT)\]
\[ans=\sum_{T=1}^nh(\frac nT)h(\frac mT)\sum_{d|T}d\times \mu(\frac Td)\times (\frac Td)^2\]
\[ans=\sum_{T=1}^nh(\frac nT)h(\frac mT)\sum_{d|T}\mu(d)\times d\times T\]
然后,一些神仙发现\(f(n)=\sum_{d|n}d\times \mu(d)\)是积性函数。为什么呢?

\((a,b)=1\)时,\(a\)的因子\(j\)\(b\)的因子\(k\)当然也是互质的,然后\(\mu\)是积性函数。

所以\[j\times \mu(j)\times k\times \mu(k)=jk\times \mu(jk)\]

也就是说\(f(a)\)的某一项乘上\(f(b)\)的某一项一定是\(f(ab)\)的某一项且是完备的。

很自然的可以得到:\(f(n)\)是一个积性函数

\[f(1)=1,f(a)=1-a,\;\;a\in prime\]
\(gcd(a,b)=1\)时:\[f(ab)=f(a)\times f(b)\]
\(gcd(a,b)!=1\)\(b\in prime\)\(b\)\(a\)的最小素因子时:\[f(ab)=f(a)\]

我们知道\(f(ab)=\sum_{d|(ab)}d\times \mu(d)\)

如果\(d\)素因数分解后\(b\)的系数小于等于\(1\),那么这一项一定在\(f(a)\)中也会出现;
如果\(d\)素因数分解后\(b\)的系数大于\(1\),那么\(\mu(d)=0\)。故上式成立。

证明完\(f(n)=\sum_{d|n}d\times \mu(d)\)是积性函数和上述式子,\(f(n)和f(n)\times n\)都可以线性筛的同时预处理出来:

void init_rime() {
    noprime[0] = noprime[1] = 1;
    F[1] = sum_F[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, F[i] = 1 - i;
        for(int j = 0; j < pcnt && i*pp[j] < MXN; ++j) {
            noprime[i*pp[j]] = 1;
            F[i*pp[j]] = F[pp[j]]*F[i];
            if(i % pp[j] == 0) {
                F[i*pp[j]] = F[i];
                break;
            }
        }
        sum_F[i] = sum_F[i-1] + F[i]*i;
    }
}

所以\(O(\sqrt n)\)即可求解:\[ans=\sum_{T=1}^nh(\frac nT)h(\frac mT)\times f(T)\times T,O(\sqrt n)\]


\(求\sum_{i=1}^n\sum_{j=1}^{m}d(i\times j)\)

3994 有一个公式:\[d(i\times j)=\sum_{x|i}\sum_{y|j}[gcd(x,y)=1]\]
\[ans=\sum_{i=1}^n\sum_{x|i}\sum_{j=1}^m\sum_{y|j}[gcd(x,y)=1]\]
\[ans=\sum_{k=1}^n\mu(k)\times \sum_{x=1}^{\frac nk}\frac n{kx}\times \sum_{y=1}^{\frac mk}\frac m{kx},\;O(\sqrt n)\]
\[\sum_{i=1}^n \frac ni=\sum_{i=1}^nd(i)\]
因为\(\frac ni\)表示的是\(1,..,n\)有几个数是\(i\)的倍数。当然也可以\(n\sqrt n\)预处理。
\(O(n)\)筛法:

bool noprime[MXN];
int pp[MXN/2], pcnt, mu[MXN], pre_mu[MXN];
int a1[MXN];
LL g[MXN];
void init_prime(int MXN) {
    noprime[0] = noprime[1] = mu[1] = pre_mu[1] = 1;
    g[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, mu[i] = -1, g[i] = 2, a1[i] = 1;
        for(int j = 0; j < pcnt && i * pp[j] < MXN; ++j) {
            noprime[i*pp[j]] = 1; mu[i*pp[j]] = -mu[i];
            g[i*pp[j]] = g[i] * 2;
            a1[i*pp[j]] = 1;
            if(i % pp[j] == 0) {
                mu[i*pp[j]] = 0;
                a1[i*pp[j]] = a1[i] + 1;
                g[i*pp[j]] = g[i]*(a1[i*pp[j]]+1)/a1[i*pp[j]];
                break;
            }
        }
        pre_mu[i] = pre_mu[i-1] + mu[i];
    }
    for(int i = 2; i < MXN; ++i) g[i] += g[i-1];
}

预处理\(2^{63}\)范围内\(d(n)\)的前缀和:洛谷SP26073


小D的demo

牛客练习赛40:\[x=\prod_{i=1}^qp_i^{k_i},g(x)=\sum_{i=1}^qk_i,g(1)=1\]
\[g(prime)=1,g(a\times prime)=g(a)+1\]
所以线性筛的时候,可以顺便把\(g(n)\)给筛出来。

待求解式子:\[\prod_{i=1}^n\prod_{j=1}^m g(gcd(i,j))\]
提取\(gcd\)\[\prod_{d=1}^n g(d)^{\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=d]}\]
反演一下指数那个式子:\[\sum_{i=1}^n\sum_{j=1}^m[gcd(i,j)=d]=\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}[gcd(i,j)=1]=\sum_{i=1}^{\frac nd}\sum_{j=1}^{\frac md}\sum_{k|gcd(i,j)}\mu(k)\]
得到这个式子:\[\prod_{d=1}^n g(d)^{\sum_{k=1}^{\frac nd}\mu(k)\frac{n}{kd}\frac{m}{kd}}\]
上面这个式子可以\(O(n)\)求值了,但是还不够,因为这题有多组数据。
想到一个常用的优化方式:
先把指数提下来:\[\prod_{d=1}^n \prod_{k=1}^{\frac nd} g(d)^{\mu(k)\frac{n}{kd}\frac{m}{kd}}\]
\(T=kd\),换元求积:\[\prod_{T=1}^n\prod_{d|T}g(d)^{\mu(\frac Td)\frac nT \frac mT}\]
再变换一下:\[\prod_{T=1}^n(\prod_{d|T}g(d)^{\mu(\frac Td)})^{\frac nT \frac mT}\]
发现这个式子\(\prod_{d|T}g(d)^{\mu(\frac Td)}\)可以\(O(nlog(n))\)预处理出来,\(O(1)\)求值。

所以最后总的复杂度是\(O(nlog(n)+T\sqrt n log(MOD))\)

代码:小D的demo


2440: [中山市选2011]完全平方数

\(T=50\),求第\(k(1e9)\)个非完全平方数倍数的数字。
显然二分是一个不错的选择,当你列出求\(n\)以内完全平方数倍数的个数的式子后,你会发现这东西不是可以用莫比乌斯函数求吗?都不需要反演的。复杂度就是枚举因数的复杂度:\(O(\sqrt n)\)


HDU 6134

\(n\leq 1e6\),求这个式子的值:\[f(n)=\sum_{d=1}^n\mu(d)\sum_{x=1}^{\frac nd}\sum_{y=1}^x ⌈\frac xy⌉\]
牛批网友说:\[g(n)=\sum_{i=1}^n⌈\frac ni⌉,\phi(n)\sum_{i=1}^n g(i)\]


BZOJ 3529 [Sdoi2014]数表

\(T(2e4) n,m(1e5),A(1e9)\),求:\[\sum_{i=1}^n\sum_{j=1}^m\sigma(gcd(i,j))\]
\[\sigma(n)=\sum_{d|n}d;if(\sigma(n)>A)\;\sigma(n)=0;\]
若不考虑\(A\),可以很容易得到:\[\sum_{d=1}^n\sigma(d)\sum_{k=1}^{\frac nd}\mu(k)\frac n{kd}\frac n{kd}\]
再优化一下:\[\sum_{T=1}^n\frac nT\frac mT\sum_{d|T}\sigma(d)\times \mu(\frac Td)\]

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MXN = 1e5 + 20;
const LL mod = 1LL<<31;
/*
T(2e4) n,m(1e5)
不考虑A的话,式子很好化简,可以sqrt(n)算答案
这题不离线下来,处理巨奇怪,也不清楚正确性
离线下来的话,主要复杂度是O(n*sqrt(n)*log(n))吧?
先把莫比乌斯函数和因数和函数给线性筛出来,离线询问按A排序
不考虑A的话,我们正常做法就是枚举gcd,用的是g(i)算答案
考虑了A,我们还要按g(i)从小到大的顺序来枚举gcd,算出卷积的贡献
因为要求卷积的前缀和,所以要用一个数据结构维护,树状数组就够了,所以复杂度多了log。
取模也可以不用取模,因为模数是1LL<<31,利用int的自然溢出就行了
*/
bool noprime[MXN];
int pp[MXN/2], pcnt;
int mu[MXN], pre_mu[MXN];
int num1[MXN];//p1^a1
int g[MXN];//因数和函数
int p[MXN];
struct lp {
    int n, m, A, id;    
}cw[MXN];
int ANS[MXN];
int bit[MXN];
int lowbit(int x) {return x&(-x);}
void add_bit(int x, int val, int N) {for(;x <= N; x += lowbit(x)) bit[x] = (bit[x]+val);}
int query_bit(int x) {int ans = 0; for(; x; x -= lowbit(x)) ans = (ans+bit[x]); return ans;}
 
void init_prime(int MXN) {
    noprime[0] = noprime[1] = mu[1] = pre_mu[1] = 1;
    g[1] = 1; p[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, mu[i] = -1, g[i] = 1+i, num1[i] = i;
        for(int j = 0; j < pcnt && i * pp[j] < MXN; ++j) {
            noprime[i*pp[j]] = 1;
            mu[i*pp[j]] = -mu[i];
            g[i*pp[j]] = g[i] * (1+pp[j]);
            num1[i*pp[j]] = pp[j];
            if(i % pp[j] == 0) {
                mu[i*pp[j]] = 0;
                num1[i*pp[j]] = num1[i] * pp[j];
                g[i*pp[j]] = (g[i]/g[num1[i]])*(g[num1[i]]*pp[j]+1);
                break;
            }
        }
        pre_mu[i] = pre_mu[i-1] + mu[i]; p[i] = i;
    }
}
bool cmp(const lp &a, const lp &b) {
    return a.A < b.A;
}
bool cmp1(const int &a, const int &b) {
    return g[a] < g[b];
}
int solve(int n, int m) {
    int ans = 0, lst = 0, now;//除法分块的一点点优化,还是可以快一点哒
    for(int L = 1, R; L <= n; L = R + 1) {
        R = min(n/(n/L), m/(m/L));
        now = query_bit(R);
        ans = (ans+(n/L)*(m/L)*(now-lst));
        lst = now;
    }
    return (ans);
}
int main(){
    int tim; scanf("%d", &tim);
    int ret = 0;
    for(int i = 1; i <= tim; ++i) {
        scanf("%d%d%d", &cw[i].n, &cw[i].m, &cw[i].A);
        if(cw[i].n > cw[i].m) swap(cw[i].n, cw[i].m);
        cw[i].id = i;
        ret = max(ret, cw[i].n);
    }
    init_prime(ret+10);
    sort(cw + 1, cw + 1 + tim, cmp);
    int j = 1;
    sort(p + 1, p + 1 + ret, cmp1);
    for(int i = 1; i <= tim; ++i) {
        for(; j <= ret && g[p[j]] <= cw[i].A; ++j) {
            for(int k = p[j]; k <= ret; k += p[j]) {
                if(mu[k/p[j]]) add_bit(k, g[p[j]]*mu[k/p[j]], ret);
            }
        }
        ANS[cw[i].id] = solve(cw[i].n, cw[i].m);
    }
    for(int i = 1; i <= tim; ++i) {
        if(ANS[i] < 0) ANS[i] += 2147483647, ++ ANS[i];
        printf("%d\n", ANS[i]);
    }
    return 0;
}

在这里插入图片描述


2019西安邀请赛B. Product

在这里插入图片描述
显然是一个欧拉降幂套上一个莫比乌斯反演,看这数据范围还得杜教筛。
由:\[\prod_{i=1}^{n}\prod_{j=1}^{n}\prod_{k=1}^{n}m^{gcd(i,j)[k|gcd(i,j)]}\]
可化简为:\[m^{\sum_{i=1}^n\sum_{j=1}^ngcd(i,j)\times D(gcd(i,j))}\]
意思转化为求:\[\sum_{d=1}^n d\times D(d)\sum_{i=1}^{n/d}\sum_{j=1}^{n/d}[(i,j)==1]\]
参考小清新数论那题:\[\sum_{d=1}^n d\times D(d)\times (2\times P(\frac nd)-1)\]
又因为:\[\sum_{i=1}^n i\times D(i)=\sum_{i=1}^n i\times \sum_{d|i}1=\sum_{d=1}^n d\times \sum_{i=1}^\frac nd 1=\sum d\times \frac{\frac nd(\frac nd+1)}{2}\]

LL n, m;
int noprime[MXN], pp[MXN], pcnt;
int phi[MXN], mu[MXN], a1[MXN];
LL pre_phi[MXN], pre_g[MXN], g[MXN];
void init_prime() {//线性筛预处理
    noprime[0] = noprime[1] = 1;
    mu[1] = 1; phi[1] = 1; g[1] = 1;
    for(int i = 2; i < MXN; ++i) {
        if(!noprime[i]) pp[pcnt++] = i, phi[i] = i-1, mu[i] = -1, g[i] = 2, a1[i] = 1;
        for(int j = 0; j < pcnt && pp[j] * i < MXN; ++j) {
            noprime[pp[j]*i] = 1;
            phi[pp[j]*i] = (pp[j]-1)*phi[i];
            mu[pp[j]*i] = -mu[i];
            g[i*pp[j]] = g[i] * 2;
            a1[i*pp[j]] = 1;
            if(i % pp[j] == 0) {
                phi[pp[j]*i] = pp[j]*phi[i];
                mu[pp[j]*i] = 0;
                a1[i*pp[j]] = a1[i] + 1;
                g[i*pp[j]] = g[i]*(a1[i*pp[j]]+1)/a1[i*pp[j]];
                break;
            }
        }
    }
}
unordered_map<LL, LL> mp1, mp2;
LL solve_g(LL n) {
    if(n < MXN) return pre_g[n];
    if(mp1.find(n) != mp1.end()) return mp1[n];
    //if(mp1[n]) return mp1[n];
    //if(mp1.count(n)) return mp1[n];
    LL ans = 0;
    for(LL L = 1, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans + ((n/L)*(n/L+1)/2)%mod*((R-L+1)*(L+R)/2)%mod)%mod;
    }
    ans = (ans + mod) % mod;
    mp1[n] = ans;
    return ans;
}
LL solve_phi(LL n) {//杜教筛欧拉函数前缀和
    if(n < MXN) return pre_phi[n];
    if(mp2.find(n) != mp2.end()) return mp2[n];
    //if(mp2[n]) return mp2[n];
    //if(mp2.count(n)) return mp2[n];
    LL ans = n*(n+1)/2;
    for(LL L = 2, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans - (R-L+1)*solve_phi(n/L)%mod)%mod;
    }
    ans = (ans + mod) % mod;
    mp2[n] = ans;
    return ans;
}
int main() {
    init_prime();
    scanf("%lld%lld%lld", &n, &m, &mud);
    mod = mud - 1;
    for(int i = 1; i < MXN; ++i) {
        pre_phi[i] = (pre_phi[i-1] + phi[i])%mod;
        pre_g[i] = (pre_g[i-1] + i * g[i] % mod)%mod;
    }
    LL ans = 0;
    for(LL L = 1, R; L <= n; L = R + 1) {
        R = n/(n/L);
        ans = (ans + (solve_g(R)-solve_g(L-1))%mod*(solve_phi(n/L)*2LL%mod-1LL)%mod)%mod;
    }
    ans = (ans+mod)%mod;
    printf("%lld\n", ksm(m, ans, mud));
    return 0;
}
// 当x≥ϕ(p)时,有a^x ≡ a^(x mod ϕ(p)+ϕ(p)) mod p
// gcd(a,mod)=1  a^(x) = a^(x%ϕ(mod)) % mod

南昌邀请赛和湘潭邀请赛


插曲:二项式反演

\[F(n)=\sum_{s}^{n}C_n^iG(i)\]
\[G(n)=\sum_{i=s}^n(-1)^{n-i}C_n^iF(i)\]



插曲:集合反演

\(max(S)=\sum_{T\in S}(-1)^{|T|-1}\times min(T)\)
\(min(S)=\sum_{T\in S}(-1)^{|T|-1}\times max(T)\)


posted @ 2019-02-10 15:54 Cwolf9 阅读(...) 评论(...) 编辑 收藏

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