luoguP4466 [国际集训队]和与积 莫比乌斯反演

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自然想到枚举\(gcd(a, b)\)，不妨设其为\(d\)，并且\(a = di, b = dj(a > b)\)

那么\(\frac{ab}{a + b} = \frac{dij}{i + j}\)

由于此时有\((i,j) = 1\)，因此\((i, i + j) = (j, i + j) = 1\)

那么，当且仅当\(i + j | d\)时，\((i, j)\)数对对答案有贡献

对答案有多少的贡献呢？\(\frac{n}{i(i + j)}\) 没有想到这一步

理由是\(d = k(i + j)\)，那么只需满足\(ki(i + j) \leq n\)

当\(i > \sqrt n\)时，\((i,j)\)对答案绝对没有贡献

所以答案为\(\sum \limits_{d = 1}^{\sqrt n} \sum \limits_{i = 1}^{\sqrt n / d} \sum \limits_{j} [(i, j) = 1]\frac{n}{i(i + j)}\)

莫比乌斯反演，得到

\(\sum \limits_{d = 1}^{\sqrt n} \mu(d) \sum \limits_{i = 1}^{\sqrt n / d} \sum \limits_{j} \frac{n}{d^2i(i + j)}\)

对内层数论分块统计答案即可一开始把不分块的复杂度算错了，以为能过

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分析一下复杂度上界

首先考虑对于确定的\(d\)，枚举\(i, j\)的复杂度

\(\sum \limits_{i = 1}^{\sqrt n / d} \sqrt \frac{n}{d^2 i} = \frac{\sqrt n}{d} * \sum \limits_{i = 1}^{\sqrt n / d} \frac{1}{\sqrt i}\)

用归纳法可以证明，右边那个东西\(\leq 2 \sqrt {\sqrt n / d}\)

所以对于一个\(d\)而言，需要\(\frac{n^{\frac{3}{4}}}{d^{\frac{3}{2}}}\)的复杂度

由于\(\frac{1}{1} + \frac{1}{2^{1.5}} + ... + \frac{1}{n^{1.5}} \leq 3\)

所以复杂度就是\(O(n^{\frac{3}{4}})\)

然后绝对跑不到这个上界....

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#include <cmath>
#include <ctime>
#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;

#define ri register int
#define ll long long
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int sid = 1e5 + 5;
    
ll ans;
int n, sq, tot;
int mu[sid], nop[sid], pr[sid];
inline void Sieve(int m) {
    mu[1] = 1;
    rep(i, 2, m) {
        if(!nop[i]) {
            pr[++ tot] = i;
            mu[i] = -1;
        }
        rep(j, 1, tot) {
            int p = i * pr[j];
            if(p > m) break; nop[p] = 1;
            if(i % pr[j] == 0) break;
            mu[p] = -mu[i];
        }
    }
}


int main() {
    cin >> n;
    sq = sqrt(n) + 1;
    Sieve(sq);
    rep(d, 1, sq) {
        if(!mu[d]) continue;
        int p = d * d;
        rep(i, 1, sq / d) {
            int fs = n / d / d / i;
            for(ri ii = i + 1, jj; ii <= (i << 1) - 1 && ii <= fs; ii = jj + 1) {
                jj = min(fs / (fs / ii), (i << 1) - 1);
                ans += 1ll * mu[d] * (jj - ii + 1) * (fs / ii);
            }
        }
    }
    printf("%lld\n", ans);
    return 0;
}