Description
求(sum_{i=1}^nsum_{i=1}^nvarphi(gcd(i,j)),Tleqslant 5 imes 10^3,nleqslant 10^7)
Solution
数论分块+莫比乌斯反演.
化式子
(sum_{i=1}^nsum_{i=1}^nvarphi(gcd(i,j)))
(=sum_dvarphi(d)sum_{i=1}^nsum_{j=1}^n[(i,j)=d])
(=sum_dvarphi(d)(sum_{i=1}^{lfloor frac{n}{d}
floor}sum_{j=1}^{lfloor frac{n}{d}
floor}[(i,j)=1]))
(=sum_dvarphi(d)(sum_{p}mu(p)sum_{i=1}^{lfloor frac{n}{pd}
floor}sum_{j=1}^{lfloor frac{n}{pd}
floor}))
( ext{Let T=pd})
(=sum_{T}lfloor frac{n}{T}
floorlfloor frac{n}{T}
floorsum_{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.
Code
/************************************************************** 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 ",ans); } return 0; }