求\(\sum_{i=1}^n\sum_{i=1}^n\varphi(gcd(i,j)),T\leqslant 5\times 10^3,n\leqslant 10^7\)
数论分块+莫比乌斯反演.
化式子
\(\sum_{i=1}^n\sum_{i=1}^n\varphi(gcd(i,j))\)
\(=\sum_d\varphi(d)\sum_{i=1}^n\sum_{j=1}^n[(i,j)=d]\)
\(=\sum_d\varphi(d)(\sum_{i=1}^{\lfloor \frac{n}{d}\rfloor}\sum_{j=1}^{\lfloor \frac{n}{d}\rfloor}[(i,j)=1])\)
\(=\sum_d\varphi(d)(\sum_{p}\mu(p)\sum_{i=1}^{\lfloor \frac{n}{pd}\rfloor}\sum_{j=1}^{\lfloor \frac{n}{pd}\rfloor})\)
\(\text{Let T=pd}\)
\(=\sum_{T}\lfloor \frac{n}{T}\rfloor\lfloor \frac{n}{T}\rfloor\sum_{p}\mu(p)\varphi(\frac{T}{p})\)
因为积性函数的狄利克雷前缀和也是积性函数，并且因为\(\mu\)的存在这个式子还是很好筛的.
/**************************************************************
Problem: 4804
User: BeiYu
Language: C++
Result: Accepted
Time:4272 ms
Memory:128240 kb
****************************************************************/
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 10000050;
int pr[N],cp;
bool b[N];
LL f[N];
void pre() {
f[1]=1;
for(int i=2;i<N;i++) {
if(!b[i]) pr[++cp]=i,f[i]=i-2;
for(int j=1;j<=cp && (LL)i*pr[j]<N;j++) {
b[i*pr[j]]=1;
if(i%pr[j]) f[i*pr[j]]=f[i]*f[pr[j]];
else {
if(i/pr[j]%pr[j]) f[i*pr[j]]=f[i/pr[j]]*(pr[j]-1)*(pr[j]-1);
else f[i*pr[j]]=f[i]*pr[j];
break;
}
}
}for(int i=2;i<N;i++) f[i]+=f[i-1];
}
int T,n;
int main() {
pre();
for(scanf("%d",&T);T--;) {
scanf("%d",&n);
LL ans=0;
for(int i=1,j;i<=n;i=j+1) {
j=n/(n/i);
ans+=1LL*(n/i)*(n/i)*(f[j]-f[i-1]);
}printf("%lld\n",ans);
}
return 0;
}
原文：http://www.cnblogs.com/beiyuoi/p/6721189.html