【题解】51nod1575 LCM and GCD (min25筛)
http://www.51nod.com/Challenge/Problem.html#problemId=1575
设(f(n)=sum_limits{i,j} [(i,n),(n,j)])，打出以下的表，发现是个积性函数
1 :: 1
2 :: 7
3 :: 19
4 :: 42
5 :: 61
6 :: 133
7 :: 127
8 :: 228
9 :: 273
10 :: 427
11 :: 331
12 :: 798
13 :: 469
14 :: 889
15 :: 1159
16 :: 1160
17 :: 817
18 :: 1911
19 :: 1027
20 :: 2562
什么，你要问证明?
[f(n)={sum_{d_1|n}{sum_{d_2|n}{ ext{lcm}(d_1,d_2)varphi(frac{n}{d_1})varphi(frac{n}{d_2})}}}\={sum_{d_1|n}{sum_{d_2|n}{sum_{d|d_1,d|d_2}{sum_{d'|frac{d_1}{d},d'|frac{d_2}{d}}{mu(d')d'^2frac{d_1}{dd'}frac{d_2}{dd'}d}}varphi(frac{n}{frac{d_1}{dd'}dd'})varphi(frac{n}{frac{d_2}{dd'}dd'})}}}\={sum_{d|n}{sum_{d'|frac{n}{d}}{dmu(d')d'^2sum_{frac{d_1}{dd'}|frac{n}{dd'}}{frac{d_1}{dd'}varphi(frac{frac{n}{dd'}}{frac{d_1}{dd'}})}sum_{frac{d_2}{dd'}|frac{n}{dd'}}{frac{d_2}{dd'}varphi(frac{frac{n}{dd'}}{frac{d_2}{dd'}})}}}}\={sum_{d|n}{sum_{d'|frac{n}{d}}{dmu(d')d'^2(sum_{frac{d''}{dd'}|frac{n}{dd'}}{frac{d''}{dd'}varphi(frac{frac{n}{dd'}}{frac{d''}{dd'}})})^2}}}\={sum_{xyz=i}{xmu(y)y^2(sum_{d|z}{dvarphi(frac{z}{d})})^2}}\={(idast id_2muast(idast varphi)^2)(n)} ]抄自这
考虑下min25筛，考虑求质数次幂时的值
[f(p^k)=sum_i sum_j [(p^k,i),(p^k,j)]
\
=sum_{p|i} sum_{p|j} [(p^k,i),(p^k,j)] +2sum_{p
ot mid i}sum_{p|j} [(p^k,i),(p^k,j)]+sum_{p
ot mid i}sum_{p
ot mid j}[(p^k,i),(p^k,j)]
\
=2 sum_i^k ip^i -sum_i^kp^i+2varphi(p^k)sum_i^k p^i +varphi(p^k)^2
\
=(2k+1)p^{2k}-(2k+1)p^{2k-1}+p^{k-1}
]
而
[f(p)=3p^2-3p+1
]
Min25 核tao心lu式子
[g(n,j)=g(n,j-1)-p^kig(g(lfloor {nover p} floor,j-1)-{ m sump} (j-1)ig) ]
//@winlere
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std; typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int uint;
inline int qr(){
int ret=0,f=0,c=getchar();
while(!isdigit(c)) f|=c==45,c=getchar();
while( isdigit(c)) ret=ret*10+c-48,c=getchar();
return f?-ret:ret;
}
const int maxn=1e5+5;
uint k2[maxn],k1[maxn],k0[maxn],tot[maxn],sump2[maxn],sump1[maxn],sump0[maxn],sump[maxn];
int id1[maxn],id2[maxn],val[maxn],pr[maxn],usd[maxn],cnt,n,N,id;
uint ksm(uint base,uint p){
uint ret=1;
while(p){
if(p&1) ret*=base;
base*=base;
p>>=1;
}
return ret;
}
int&getId(int val){return val>N?id2[n/val]:id1[val];}
uint sum1(uint x){return 1ull*x*(x+1)>>1;}
uint sum2(ull x){
int k=x%6;
if(k==0) return x*(x+1)/6*(2*x+1);
if(k==2) return x*(x+1)/6*(2*x+1);
if(k==3) return x*(x+1)/6*(2*x+1);
if(k==5) return (x+1)*(2*x+1)/6*x;
if(k==1) return (x+1)*(2*x+1)/6*x;
return (2*x+1)*x/6*(x+1);
}
uint fPrime(uint v,int k){
return (2u*k+1)*(ksm(v,2*k)-ksm(v,2*k-1))+ksm(v,k-1);
}
void seive(int n){
for(int t=2;t<=n;++t){
if(!usd[t]) pr[++cnt]=t,sump[cnt]=sump[cnt-1]+3*t*t-3*t+1,sump2[cnt]=sump2[cnt-1]+1*t*t,sump1[cnt]=sump1[cnt-1]+1*t,sump0[cnt]=sump0[cnt-1]+1;
for(int i=1,v;i<=cnt;++i){
v=pr[i]*t;
if(v>n) break;
usd[v]=1;
if(t%pr[i]==0) break;
}
}
}
uint S(int n,int j){
if(n==0||pr[j]>n) return 0;
int k=getId(n);
uint ret=tot[k]-sump[j-1];
for(int t=j;t<=cnt&&pr[t]*pr[t]<=n;++t)
for(ll p=pr[t],e=1;p*pr[t]<=n;++e,p*=pr[t])
ret+=fPrime(pr[t],e+1)+fPrime(pr[t],e)*S(n/p,t+1);
return ret;
}
void init(){
id=0;
for(int l=1,r=1;r<=n;l=++r){
r=n/(n/l);
getId(n/l)=++id;
k2[id]=sum2(n/l)-1,k1[id]=sum1(n/l)-1,k0[id]=n/l-1;
val[id]=n/l;
}
for(int t=1;t<=cnt&&pr[t]*pr[t]<=val[1];++t)
for(int i=1;i<=id&&pr[t]*pr[t]<=val[i];++i){
int k=getId(val[i]/pr[t]);
k2[i]-=pr[t]*pr[t]*(k2[k]-sump2[t-1]);
k1[i]-= pr[t]*(k1[k]-sump1[t-1]);
k0[i]-= (k0[k]-sump0[t-1]);
}
for(int t=1;t<=id;++t) tot[t]=3u*k2[t]-3u*k1[t]+k0[t];
printf("%u
",S(n,1)+1u);
}
int main(){
int T=qr();
seive(maxn-2);
while(T--) n=qr(),N=sqrt(n),init();
return 0;
}