标签：net namespace 前缀 isp ali 一个 最小 begin sdn
求 $\sum_{i=a}^b \sum_{j=1}^i \frac{lcm(i,j)}{i}$.
只需要求出前缀和，
$$\begin{aligned}
\sum_{i=1}^n \sum_{j=1}^i \frac{lcm(i,j)}{i} &= \sum_{i=1}^n \sum_{j=1}^i \frac{j}{gcd(i,j)} \\
&= \sum_{d=1}^n \sum _{i=1}^n \sum_{j=1}^i \frac{j}{d} \cdot [gcd(i,j)=1] \\
&= \sum_{d=1}^n \sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor} \sum_{j=1}^i j \cdot [gcd(i,j)=1]
\end{aligned}$$
其后面部分提出来，即求 $\sum_{i=1}^n i\cdot [gcd(i,n)=1]$，对于这种一个值固定的gcd求和有一个套路，即倒序两两配对：
若 $n=1$，和为1；
若 $n>1$，因为 $gcd(i, n) = gcd(n-i, n)$ 且 $\displaystyle \sum_{i=1}^n i\cdot [gcd(i,n)=1] = \sum_{i=1}^{n-1} i\cdot [gcd(i,n)=1]$，
$\displaystyle \sum_{i=1}^{n-1} i \cdot [gcd(i,n)=1] + \sum_{i=n-1}^1i\cdot [gcd(i,n)=1] = n\varphi (n)$，所以和为 $ n\varphi (n) /2$.
综合得 $\displaystyle \sum_{i=1}^n i\cdot [gcd(i,n)=1] = \frac{n\varphi (n)+[n=1]}{2}$.
具体实现上，$[i=1]$ 只成立 $n$，除2可以提出来，即 原式 = $\displaystyle \frac{1}{2}(\sum_{d=1}^n \sum_{i=1}^{\left \lfloor \frac{n}{d} \right \rfloor} i\varphi (i) + n)$.
现在唯一得问题是如何求 $\displaystyle S(n) = \sum_{i=1}^n i\varphi (i)$.
根据杜教筛，
设 $\displaystyle S(n) = \sum_{i=1}^n f(i)$，$f(n) = n\varphi (n), \ g(n) = n$.
$\displaystyle f*g = \sum_{d|n} d \varphi (d) \cdot \frac{n}{d} = n\sum_{d|n}\varphi (d) = n^2$.
因此 $\displaystyle S(n) = \sum_{i=1}^n i^2 - \sum_{d=2}^n d\cdot S(\left \lfloor \frac{n}{d} \right \rfloor)$.
再最外层套个数论分块即可。
#include <algorithm> #include <cstdio> #include <cstring> #include <map> #include<unordered_map> using namespace std; const int maxn = 2000010; typedef long long ll; const ll mod = 1000000007; const ll inv2 = (mod+1)>>1; const ll inv6 = 166666668; //166666668 ll T, a, b, pri[maxn], tot, phi[maxn], sum_phi_d[maxn]; bool vis[maxn]; unordered_map<ll, ll> mp_phi_d; //可换成unordered_map,约快3倍 ll S_phi_d(ll x) { if (x < maxn) return sum_phi_d[x]; if (mp_phi_d[x]) return mp_phi_d[x]; ll ret = x * (x+1) % mod * (2*x%mod+1) % mod * inv6 % mod; //%敲成*，浪费一个小时 for (ll i = 2, j; i <= x; i = j + 1) { j = x / (x / i); ret =(ret - S_phi_d(x / i) * (i+j) % mod * (j-i+1) % mod * inv2 % mod + mod) % mod; } return mp_phi_d[x] = ret; } void initPhi_d() { phi[1] = 1; for (int i = 2; i < maxn; i++) { if (!vis[i]) pri[++tot] = i, phi[i] = i-1; for (int j = 1; j <= tot && i * pri[j] < maxn; j++) { vis[i * pri[j]] = true; if (i % pri[j] == 0) { phi[i * pri[j]] = phi[i] * pri[j] % mod; break; } else { phi[i * pri[j]] = phi[i] * phi[pri[j]] % mod; } } } for (int i = 1; i < maxn; i++) sum_phi_d[i] = (sum_phi_d[i - 1] + phi[i]*i) % mod; } //ll G(ll n) //{ // //printf("G:%lld\n", n); // return (S_phi_d(n)+1) % mod; //} ll solve(ll n) { ll res = 0; for(ll i = 1, j;i <= n;i = j+1) { j = n / (n / i); res = (res + S_phi_d(n/i) * (j-i+1) % mod) % mod; } return (res+n)*inv2%mod; } int main() { initPhi_d(); scanf("%lld%lld", &a, &b); printf("%lld\n", (solve(b)-solve(a-1)+mod) % mod); return 0; }
参考链接：
1. https://blog.csdn.net/FromATP/article/details/74999989
2. https://www.cnblogs.com/owenyu/p/7397687.html
[51nod1227]平均最小公倍数（莫比乌斯反演+杜教筛）
标签：net namespace 前缀 isp ali 一个 最小 begin sdn
原文地址：https://www.cnblogs.com/lfri/p/11701388.html
