BZOJ 4176 [莫比乌斯反演][杜教筛]

Description

求

∑i=1n∑j=1nd(ij)

Solution

通过陈老师r老师等式可以的得到该式子就等于

∑i=1n∑j=1n⌊ni⌋⌊nj⌋[(i,j)=1]

一波反演以后就得到

∑d=1nμ(d)(∑i=1⌊nd⌋⌊nid⌋)2

发现后面那个东西的取值只有 O(n√) 种，只需要枚举后面的值，前面的用杜教筛求就好了，时间复杂度为 O(n34) 。

#include 
#include 
#include 
#include 
using namespace std;
using namespace tr1;

const int MOD = 1000000007;
const int N = 1000100;
typedef long long ll;

int n, Pcnt, x, ans, res, tmp, pos;
int prime[N], mu[N], vis[N];
unordered_map<int, int> MU;

int Mu(int x) {
    if (x < N) return mu[x];
    if (MU.count(x)) return MU[x];
    int pos, res = 1;
    for (int i = 2; i <= x; i = pos + 1) {
        pos = x / (x / i);
        (res -= (ll)(pos - i + 1) * Mu(x / i) % MOD) % MOD;
    }
    return MU[x] = res;
}
int F(int x) {
    int res = 0, pos;
    for (int i = 1; i <= x; i = pos + 1) {
        pos = x / (x / i);
        (res += (ll)(x / i) * (pos - i + 1) % MOD) %= MOD;
    }
    return res;
}

int main(void) {
    mu[1] = 1; scanf("%d", &n);
    for (int i = 2; i < N; i++) {
        if (!vis[i]) {
            prime[++Pcnt] = i; mu[i] = -1;
        }
        for (int j = 1; j <= Pcnt && (x = prime[j] * i) < N; j++) {
            vis[x] = 1;
            if (i % prime[j]) {
                mu[x] = -mu[i];
            } else {
                mu[x] = 0; break;
            }
        }
    }
    for (int i = 2; i < N; i++) mu[i] += mu[i - 1];
    for (int i = 1; i <= n; i = pos + 1) {
        pos = n / (n / i); tmp = F(n / i);
        res = (Mu(pos) - Mu(i - 1) + MOD) % MOD;
        res = (ll)res * tmp % MOD * tmp % MOD;
        ans = (ans + res) % MOD;
    }
    printf("%d\n" ,ans);
    return 0;
}