BZOJ 3930 [CQOI2015]选数分块+前缀和+玄_a n s = ( ∑ n l = 1 ∑ n r = l gcd ( a l , , a r -优快云博客

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BZOJ 3930 [CQOI2015]选数

Solution

题目要求:

\sum a 1 = L R \sum a 2 = L R \dots \sum a N = L R [g c d (a 1, a 2, \dots, a N) = K]

$\sum_{a_1=L}^{R}\sum_{a_2=L}^{R}\cdots\sum_{a_N=L}^{R}[gcd(a_1,a_2,\cdots,a_N)=K]$

设要求的答案为 $f(K)$ ,表示

\sum a 1 = L R \sum a 2 = L R \dots \sum a N = L R [g c d (a 1, a 2, \dots, a N) = K]

$\sum_{a_1=L}^{R}\sum_{a_2=L}^{R}\cdots\sum_{a_N=L}^{R}[gcd(a_1,a_2,\cdots,a_N)=K]$
考虑构造一个函数

F(K)F(K) $F(K)$ ,表示

\sum a 1 = L R \sum a 2 = L R \dots \sum a N = L R [K | g c d (a 1, a 2, \dots, a N)]

$\sum_{a_1=L}^{R}\sum_{a_2=L}^{R}\cdots\sum_{a_N=L}^{R}[K|gcd(a_1,a_2,\cdots,a_N)]$

那么 $f,F$ 之间有如下关系
$F(n)=\sum_{n|d}^{R}f(d)$

进行莫比乌斯反演
得到 $f(n)=\sum_{n|d}^{R}\mu(\frac{d}{n})F(d)$

$F(n)=(\left \lfloor \frac{R}{n} \right \rfloor-\left \lfloor \frac{L-1}{n} \right \rfloor)^N$

所以原式化为:

f (K) = \sum K | d R μ (d K) (⌊ R d ⌋ - ⌊ L - 1 d ⌋) N

$f(K)=\sum_{K|d}^{R}\mu(\frac{d}{K})(\left \lfloor \frac{R}{d} \right \rfloor-\left \lfloor \frac{L-1}{d} \right \rfloor)^N$

设 $T=\frac{d}{K}$ ,枚举 $T$ ,得:

f (K) = \sum_{T = 1}^{⌊ \frac{R}{K} ⌋} μ (T) (⌊ \frac{R}{K T} ⌋ - ⌊ \frac{L - 1}{K T} ⌋)^{N}

$f(K)=\sum_{T=1}^{\left \lfloor \frac{R}{K} \right \rfloor}\mu(T) (\left \lfloor \frac{R}{KT} \right \rfloor-\left \lfloor \frac{L-1}{KT} \right \rfloor)^N$

用一个分块+前缀和搞定
看上去好像做完了的样子
但是若 $K=1,R=1e9$ 就会神奇的发现
前缀和爆了
显然只能开到 $1e7$ (某些奇怪的OJ只有128MB内存)
但是要求你求到 $1e9$

所以说需要一些神奇的方法(玄学到自己都不清不白)
因为 $\sum_{d|i}\mu(d)=[i==1]$
可得

\sum i = 1 n \sum d | i μ (d) = 1

$\sum_{i=1}^{n}\sum_{d|i}\mu(d)=1$
枚举

dd $d$ (约数)可得

\sum_{d = 1}^{n} \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} μ (d) = 1

$\sum_{d=1}^{n}\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor}\mu(d)=1$
等价于

\sum i = 1, d = 1 i * d < = n d * μ (i)

$\sum_{i=1,d=1}^{i*d<=n}d*\mu(i)$
因为

i,ji,j $i,j$ 等价,所以

\sum i = 1, d = 1 i * d < = n i * μ (d)

$\sum_{i=1,d=1}^{i*d<=n}i*\mu(d)$
等价于

\sum d = 1 n \sum i = 1 ⌊ n d ⌋ μ (i) = 1

$\sum_{d=1}^{n}\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor}\mu(i)=1$
等价于枚举

dd $d$ (次数)

\sum_{d = 1}^{n} \sum_{i = 1}^{⌊ \frac{n}{d} ⌋} μ (i) = 1

$\sum_{d=1}^{n}\sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor}\mu(i)=1$

设

s u m (n) = \sum i = 1 n μ (i)

$sum(n)=\sum_{i=1}^{n}\mu(i)$

那么

\sum d = 1 n s u m (⌊ n d ⌋) = 1

$\sum_{d=1}^{n}sum(\left \lfloor \frac{n}{d} \right \rfloor)=1$

s u m (n) = 1 - \sum d = 2 n s u m (⌊ n d ⌋)

$sum(n)=1-\sum_{d=2}^{n}sum(\left \lfloor \frac{n}{d} \right \rfloor)$
这样我们就可以递归的将

sum(n)sum(n) $sum(n)$ 求出
而且将前

1e61e6 $1e6$ 的

sumsum $sum$ 打表出来,直接代入,可极大加速

详细代码如下:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int INF = 1000000010;
const int N = 1000005;
const int mod = 1000000007;
unordered_map <int,long long> mu_sum;
int n,k,l,r;
ll pri[N],tot,mu[N];
ll s[N];
bool mark[N];
void get() {
    mu[1]=1;
    for(int i=2;i<=1000000;++i) {
        if(!mark[i]) {
            pri[++tot]=i;
            mu[i]=-1;
        }
        for(int j=1;j<=tot && i*pri[j]<=1000000;++j) {
            mark[i*pri[j]]=1;
            if(i%pri[j]==0) {break;}
            mu[i*pri[j]]=-mu[i];
        }
    }
    for(int i=1;i<=1000000;++i) {
        mu[i]+=mu[i-1];
        s[i]=s[i-1]+mu[i];
    }
}
ll qpow(ll a,ll b) {
    ll ans=1;
    while(b) {
        if(b&1) {
            ans*=a;
            ans%=mod;
        }
        b>>=1;
        a*=a;
        a%=mod;
    }
    return ans;
}
ll get_mu(int x) {
    if(x<=1000000) return mu[x];
    if(mu_sum.find(x)!=mu_sum.end())
        return mu_sum[x];
    int pos;
    ll ans=1;
    for(int i=1;i<=x;i=pos+1) {
        pos=x/(x/i);
        if(x/i-1) ans-=(get_mu(pos)-get_mu(i-1))*(x/i-1);
    }
    return mu_sum[x]=ans;
}
int main() {
    scanf("%d%d%d%d",&n,&k,&l,&r);
    r/=k;l=(l-1)/k;
    get();
    ll ans=0;
    int pos=0;
    for(int i=1;i<=r;i=pos+1) {
        pos=min(r/(r/i),l/i?l/(l/i):INF);
        ans+=(get_mu(pos)-get_mu(i-1))*qpow((r/i)-(l/i),n);
        ans%=mod;
    }
    printf("%lld\n",(ans+mod)%mod);
    return 0;
}