标签：ring 递归 clu class work ons htm logs amp
题目来源于糖教主浅谈一类积性函数的前缀和...
考虑$\mu(x)$的性质：$[n==1]=\sum _{d\mid n} \mu(d)$
可以用上面哪个公式来推导：
$f(n)=\sum _{i=1}^{n}$
$1=\sum _{i=1}^{n} [i==1]$
$=\sum _{i=1}^{n} \sum _{d\mid i} \mu (d)$
$=\sum _{\frac{i}{d}=1}^{n} \sum _{d=1}^{\frac{n}{\frac{i}{d}}} \mu (d)$
$=\sum _{i=1}^{n}\sum _{d=1}^{\frac{n}{i}} \mu(d)$
$=\sum _{i=1}^{n}f(\frac{n}{i})$
$f(n)=1-\sum _{i=2}^{n} f(\frac{n}{i})$
然后，我们预处理出前$\sqrt{n}$个的$f(x)$，然后对于大于$\sqrt{n}$的数的答案，分块递归计算...
复杂度的证明请见糖教主的文章...
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
//by NeighThorn
using namespace std;
const int maxn=5000000+5;
int cnt,mu[maxn],pri[maxn],vis[maxn];
long long n,m,f[maxn];
map<long long,long long> mp;
inline void prework(void){
mu[1]=1;
for(int i=2;i<=5000000;i++){
if(!vis[i])
vis[i]=1,pri[++cnt]=i,mu[i]=-1;
for(int j=1;j<=cnt&&1LL*i*pri[j]<=5000000;j++){
vis[i*pri[j]]=1;
if(i%pri[j]==0){
mu[i*pri[j]]=0;
break;
}
mu[i*pri[j]]=-mu[i];
}
}
for(int i=1;i<=5000000;i++) f[i]=f[i-1]+mu[i];
}
inline long long calc(long long x){
if(x<=5000000) return f[x];
if(mp.find(x)!=mp.end()) return mp[x];
long long ans=1;
for(long long i=2,r;i<=x;i=r+1){
r=x/(x/i);
ans-=calc(x/i)*(r-i+1);
}
return mp[x]=ans;
}
signed main(void){
prework();scanf("%lld%lld",&n,&m);
printf("%lld\n",calc(m)-calc(n-1));
return 0;
}
和上面的题目差不多...
这次利用的是$\phi(x)$的这个性质：$\sum _{d\mid n} \phi(d)=n$
$\phi(n)=n-\sum _{d\mid n d<n}\phi(d)$
$f(n)=\sum _{i=1}^{n} (i-\sum _{d\mid i d<i} \phi(d))$
$=\frac{n(n+1)}{2}-\sum _{i=2}^{n} \sum _{d\mid i d<i} \phi(d)$
$=\frac{n(n+1)}{2}-\sum _{i=2}^{n}\sum _{d=1}^{\frac{n}{i}} \phi(d)$
$=\frac{n(n+1)}{2}-\sum _{i=2}^{n} f(\frac{n}{i})$
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
//by NeighThorn
using namespace std;
const int maxn=5000000+5,mod=1e9+7;
int cnt,f[maxn],pri[maxn],phi[maxn],vis[maxn];
long long n;
map<long long,int> mp;
inline int mul(long long x,long long y){
int res=0;x%=mod;
while(y){
if(y&1) res=(res+x)%mod;
x=(x+x)%mod,y>>=1;
}
return res;
}
inline int power(int x,int y){
int res=1;
while(y){
if(y&1) res=1LL*res*x%mod;
x=1LL*x*x%mod,y>>=1;
}
return res;
}
inline void prework(void){
phi[1]=1;
for(int i=2;i<=5000000;i++){
if(!vis[i])
pri[++cnt]=i,vis[i]=1,phi[i]=i-1;
for(int j=1;j<=cnt&&1LL*i*pri[j]<=5000000;j++){
vis[i*pri[j]]=1;
if(i%pri[j]==0){
phi[i*pri[j]]=phi[i]*pri[j];
break;
}
phi[i*pri[j]]=phi[i]*(pri[j]-1);
}
}
for(int i=1;i<=5000000;i++) f[i]=(f[i-1]+phi[i])%mod;
}
inline int calc(long long n){
if(n<=5000000) return f[n];
if(mp.find(n)!=mp.end()) return mp[n];
int ans=mul(mul(n,n+1),power(2,mod-2));
for(long long i=2,r;i<=n;i=r+1){
r=n/(n/i);
ans=(ans-1LL*calc(n/i)*((r-i+1)%mod)%mod+mod)%mod;
}
return mp[n]=ans;
}
signed main(void){
prework();scanf("%lld",&n);
printf("%lld\n",calc(n));
return 0;
}
标签：ring 递归 clu class work ons htm logs amp
原文地址：http://www.cnblogs.com/neighthorn/p/6683741.html
