莫比乌斯反演
根据约数和个数公式
$ans = \sum_{i=1}^{n}\sum_{j=1}^{n}\sum_{x|i}\sum_{y|j}{[gcd(i, j)==1]}$
交换枚举顺序
$ans = \sum_{x=1}^{n}\sum_{y=1}^{n}{[\frac{n}{x}][\frac{n}{y}]*[gcd(x, y)==1]}$
$=\sum_{x=1}^{n}\sum_{y=1}^{n}{[\frac{n}{x}][\frac{n}{y}]\sum_{d|x,y}{\mu(d)}}$
再交换枚举顺序
$=\sum_{d=1}^{n}{\mu(d)\sum_{i=1}^{\frac{n}{d}}{\frac{n}{di}}\sum_{j=1}^{\frac{n}{d}}{\frac{n}{dj}}}$
设$f(n)=\sum_{i=1}^{n}{\frac{n}{i}}$
那么$=\sum_{d=1}^{n}{\mu(d)f(\frac{n}{d})^{2}}$
这就可以分块求了，$\mu$用杜教筛求，$f$用二次分块
反演就是不断交换求和顺序或者改变枚举变量
#include <cstdio> #include <cstring> #include <algorithm> #include <map> using namespace std; typedef long long ll; const int N = 1e7 + 5, P = 1e9 + 7; int n; ll ans; int p[N], mark[N], mu[N]; map<int, ll> sum_1; void ini() { mu[1] = 1; for(int i = 2; i < N; ++i) { if(!mark[i]) { p[++p[0]] = i; mu[i] = -1; } for(int j = 1; j <= p[0] && i * p[j] < N; ++j) { mark[i * p[j]] = 1; if(i % p[j] == 0) { mu[i * p[j]] = 0; break; } mu[i * p[j]] = -mu[i]; } } for(int i = 1; i < N; ++i) { mu[i] += mu[i - 1]; } } ll dj_m(int n) { if(n < N) { return mu[n]; } if(sum_1.find(n) != sum_1.end()) { return sum_1[n]; } ll ret = 1; for(int i = 2, j = 0; i <= n; i = j + 1) { j = n / (n / i); ret = (ret - (ll)(j - i + 1) * dj_m(n / i) % P + P) % P; } return sum_1[n] = ret; } ll calc(int n) { ll ret = 0; for(int i = 1, j = 0; i <= n; i = j + 1) { j = n / (n / i); ret = (ret + (ll)(n / i) * (j - i + 1) % P) % P; } return ret; } int main() { ini(); scanf("%d", &n); for(int i = 1, j = 0; i <= n; i = j + 1) { j = n / (n / i); ll a = ((dj_m(j) - dj_m(i - 1)) % P + P) % P, b = calc(n / i); ans = (ans + a * b % P * b % P) % P; } printf("%lld\n", ans); return 0; }
