51nod 1244 莫比乌斯函数之和

 

链接

 

题意

求a~b的莫比乌斯函数和，$a,b<=10^{10}$

 

分析

杜教筛+hash+记忆化

求$S(n) = \sum\limits_{i=1}^{n}μ(i)$

对于莫比乌斯函数有$\sum\limits_{d|i}μ(d)=[i=1]$

所以$S(n) = \sum\limits_{i=1}^{n}\sum\limits_{d|i}μ(d)-\sum\limits_{i=1}^{n}\sum\limits_{d|i,d≠i}μ(d)$

$=1-\sum\limits_{i=1}^{n}\sum\limits_{d|i,d≠1}μ(\lfloor\frac{i}{d}\rfloor)$

$=1-\sum\limits_{i=2}^{n}\sum\limits_{d=1}^{\lfloor\frac{n}{i}\rfloor}μ(d)$

$=1-\sum\limits_{i=2}^{n}S(\lfloor\frac{n}{i}\rfloor)$

 

code

#include<cstdio>
#include<cstring>
#include<iostream>

using namespace std;
const int N = 5000000;
const int mod = 1000007;
typedef long long LL;

int prime[N+10],mu[N+10],tot,cnt;
bool noprime[N+10];
int h[mod],nxt[mod];
LL v[mod],d[mod];

void getmu() {
    mu[1] = 1;
    for (int i=2; i<=N; ++i) {
        if (!noprime[i]) prime[++tot] = i,mu[i] = -1;
        for (int j=1; j<=tot&&i*prime[j]<=N; ++j) {
            noprime[i*prime[j]] = true;
            if (i % prime[j]==0) {mu[i*prime[j]] = 0;break;}
            mu[i * prime[j]] = -mu[i];
        }
    }
    for (int i=1; i<=N; ++i) mu[i] += mu[i-1];
}
void addhash(LL x,LL y) {
    int p = x % mod;
    d[++cnt] = x;v[cnt] = y;nxt[cnt] = h[p];h[p] = cnt;
}
int gethash(LL x) {
    int p = h[x % mod];
    while (p!=-1 && d[p]!=x) p=nxt[p];
    return p;
}
LL Calc(LL x) {
    if (x <= N) return (LL)mu[x];
    int p = gethash(x);
    if (p != -1) return v[p];
    LL l=2,r,ret = 1;
    while (l <= x) {
        r = x / (x / l);
        ret -= 1ll * (r - l + 1) * Calc(x / l);
        l = r + 1;
    }
    addhash(x,ret);
    return ret;
}
int main () {
    memset(h,-1,sizeof(h));
    memset(nxt,-1,sizeof(nxt));
    LL a,b;
    getmu();
    cin >> a >> b;
    cout << Calc(b) - Calc(a-1);
    return 0;
}