题目来源于糖教主浅谈一类积性函数的前缀和...
51Nod 1244 莫比乌斯函数之和
考虑$mu(x)$的性质：$[n==1]=sum _{dmid n} mu(d)$
可以用上面哪个公式来推导：
$f(n)=sum _{i=1}^{n}$
$1=sum _{i=1}^{n} [i==1]$
$=sum _{i=1}^{n} sum _{dmid i} mu (d)$
$=sum _{frac{i}{d}=1}^{n} sum _{d=1}^{frac{n}{frac{i}{d}}} mu (d)$
$=sum _{i=1}^{n}sum _{d=1}^{frac{n}{i}} mu(d)$
$=sum _{i=1}^{n}f(frac{n}{i})$
$f(n)=1-sum _{i=2}^{n} f(frac{n}{i})$
然后，我们预处理出前$n^{frac{2}{3}}$个的$f(x)$，然后对于大于$n^{frac{2}{3}}$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...
代码：
#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
",calc(m)-calc(n-1));
return 0;
}
51Nod 1239 欧拉函数之和
和上面的题目差不多...
这次利用的是$phi(x)$的这个性质：$sum _{dmid n} phi(d)=n$
$phi(n)=n-sum _{dmid n d<n}phi(d)$
$f(n)=sum _{i=1}^{n} (i-sum _{dmid i d<i} phi(d))$
$=frac{n(n+1)}{2}-sum _{i=2}^{n} sum _{dmid i d<i} phi(d)$
$=frac{n(n+1)}{2}-sum _{i=2}^{n}sum _{d=1}^{frac{n}{i}} phi(d)$
$=frac{n(n+1)}{2}-sum _{i=2}^{n} f(frac{n}{i})$
代码：
#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
",calc(n));
return 0;
}
51Nod 1190 最小公倍数之和 V2
推了好久的式子发现本来还挺对的，后来越来越麻烦...一定是化简方向错了...所以看了题解...发现果然如此...
懒得写一遍式子了...一起%题解吧...
需要注意的是，对于枚举判断一个数的质因子的时候，记得加上一个小的剪枝，否则就是把9000+的素数全部枚举一边，绝逼GG啊...
if(1LL*pri[i]*pri[i]>tmp) break;
代码：
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
//by NeighThorn
using namespace std;
const int maxn=100000+5,mod=1e9+7;
int n,m,ans,cas,cnt,inv,tot,tot_pri,pri[maxn],exp[maxn],vis[maxn],fact[maxn],fac_pri[maxn];
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){
for(int i=2;i<=100000;i++){
if(!vis[i])
vis[i]=1,pri[++cnt]=i;
for(int j=1;j<=cnt&&1LL*i*pri[j]<=100000;j++){
vis[i*pri[j]]=1;
if(i%pri[j]==0) break;
}
}
}
inline void dfs(int id,int fac){
if(id>tot_pri){
fact[++tot]=fac;return;
}
dfs(id+1,fac);
for(int i=1;i<=exp[id];i++)
fac*=fac_pri[id],dfs(id+1,fac);
}
inline void getfact(void){
int tmp=m;
for(int i=1;i<=cnt;i++){
if(1LL*pri[i]*pri[i]>tmp) break;
if(tmp%pri[i]==0){
fac_pri[++tot_pri]=pri[i],exp[tot_pri]=0;
while(tmp%pri[i]==0)
exp[tot_pri]++,tmp/=pri[i];
}
}
if(tmp>1) fac_pri[++tot_pri]=tmp,exp[tot_pri]=1;
dfs(1,1);
}
signed main(void){
// freopen("51nod_Problem_1190_Test_16_In.txt","r",stdin);
// freopen("out.txt","w",stdout);
scanf("%d",&cas);prework();inv=power(2,mod-2);
while(cas--){
scanf("%d%d",&n,&m);
tot_pri=tot=ans=0;getfact();
for(int i=1,tmp,ttmp,lala;i<=tot;i++){
tmp=1;ttmp=n+fact[i]-1;lala=fact[i];
for(int j=1;j<=tot_pri;j++)
if(fact[i]%fac_pri[j]==0)
tmp=1LL*tmp*(1-fac_pri[j]+mod)%mod;
tmp=1LL*tmp*inv%mod;
tmp=1LL*tmp*((ttmp/lala+m/lala)%mod)%mod;
tmp=1LL*tmp*(m/lala-ttmp/lala+1)%mod;
ans=(ans+tmp)%mod;
}
ans=1LL*ans*m%mod;
printf("%d
",ans);
}
return 0;
}
BZOJ 3944: Sum
就是第一道题和第二道题的结合...
然后BZOJ就tm有人卡OJ了...
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
//by NeighThorn
using namespace std;
const int maxn=5000000+5;
bool vis[maxn];
int n,cas,cnt,pri[maxn/5];
long long f1[maxn],f2[maxn];
map<int,long long> mp1,mp2;
inline void prework(void){
f2[1]=f1[1]=1;
for(int i=2;i<=5000000;i++){
if(!vis[i])
vis[i]=1,pri[++cnt]=i,f2[i]=-1,f1[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){
f2[i*pri[j]]=0;
f1[i*pri[j]]=f1[i]*pri[j];
break;
}
f1[i*pri[j]]=f1[i]*(pri[j]-1),f2[i*pri[j]]=-f2[i];
}
}
for(int i=1;i<=5000000;i++) f1[i]+=f1[i-1],f2[i]+=f2[i-1];
}
inline long long calc1(long long n){
if(n<=5000000) return f1[n];
if(mp1.find(n)!=mp1.end()) return mp1[n];
long long ans=1LL*n*(n+1)/2;
for(long long i=2,r;i<=n;i=r+1){
r=n/(n/i);
ans-=(r-i+1)*calc1(n/i);
}
return mp1[n]=ans;
}
inline long long calc2(long long n){
if(n<=5000000) return f2[n];
if(mp2.find(n)!=mp2.end()) return mp2[n];
long long ans=1;
for(long long i=2,r;i<=n;i=r+1){
r=n/(n/i);
ans-=(r-i+1)*calc2(n/i);
}
return mp2[n]=ans;
}
signed main(void){
scanf("%d",&cas);prework();
while(cas--){
scanf("%d",&n);
printf("%lld %lld
",calc1(n),calc2(n));
}
return 0;
}
看到题目的第一眼觉得是莫比乌斯反演...
然后看了看数据范围，貌似是杜教筛...
考虑$g(n)=sum _{i=1}^{n} f(i)$
$sum _{i=1}^{n} sum _{dmid i} f(d)$
$=sum _{i=1}^{n} f(i)left lfloor frac{n}{i} ight floor $
$=sum _{i=1}^{n} g(left lfloor frac{n}{i} ight floor)$
又因为$sum _{i=1}^{n} sum _{dmid i} f(d)=sum _{i=1}^{n} (i-1)*(i-2)$
所以说$g(n)=sum _{i=1}^{n} (i-1)(i-2)-sum _{i=2}^{n} g(left lfloor frac{n}{i} ight floor)$
至于预处理我是用莫比乌斯反演写的...
$sum _{dmid n}f(d)=F(n)$
$f(n)=sum _{dmid n} mu(d) F(frac{n}{d})$
代码：
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
//by NeighThorn
using namespace std;
const int mod=1e9+7,maxn=1000000+5;
int n,cas,cnt,inv,f[maxn],mu[maxn],pri[maxn],vis[maxn];
map<int,int> mp;
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 int F(int x){
return 1LL*(x-1)*(x-2)%mod;
}
inline void prework(void){
mu[1]=1;
for(int i=2;i<=1000000;i++){
if(!vis[i])
vis[i]=1,pri[++cnt]=i,mu[i]=-1;
for(int j=1;j<=cnt&&1LL*i*pri[j]<=1000000;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<=1000000;i++)
for(int j=i;j<=1000000;j+=i){
f[j]+=mu[i]*F(j/i);
if(f[j]<0) f[j]+=mod;
f[j]%=mod;
}
for(int i=1;i<=1000000;i++) f[i]=(f[i]+f[i-1])%mod;
}
inline int calc(int n){
if(n<=1000000) return f[n];
if(mp.find(n)!=mp.end()) return mp[n];
int ans=1LL*n*(n-1)%mod*(n-2)%mod*inv%mod;
for(int i=2,r;i<=n;i=r+1){
r=n/(n/i);
ans=ans-1LL*(r-i+1)*calc(n/i)%mod;
if(ans<0) ans+=mod;
ans%=mod;
}
return mp[n]=ans;
}
signed main(void){
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
scanf("%d",&cas);inv=power(3,mod-2);prework();
while(cas--){
scanf("%d",&n);
printf("%d
",calc(n));
}
return 0;
}
51Nod 1238 最小公倍数之和 V3
$sum _{i=1}^{n} sum _{j=1}^{m} lcm(i,j)$
$=sum _{i=1}^{n} sum _{j=1}^{m} frac {i*j}{gcd(i,j)}$
$=sum _{d=1}^{n} d sum _{i=1}^{frac{n}{d}} sum _{j=1}^{frac{m}{d}} i*j*[gcd(i,j)==1]$
$=sum _{d=1}^{n} d sum _{i=1}^{frac{n}{d}} i sum _{j=1}^{frac{m}{d}} j*[gcd(i,j)==1]$
因为有以下的公式：
$sum _{i=1}^{n} i*[gcd(i,n)==1]=frac{nphi(n)}{2}$
证明请见HDU 3501
于是可以接着化简得到如下式子：
$sum _{d=1}^{n} sum _{i=1}^{frac{n}{d}} i*i*phi(i)$
那么我们的问题就剩下了如何快速求$Sum(n)=sum _{i=1}^{n}f(i)=sum _{i=1}^{n} i*i*phi(i)$
考虑$sum _{i=1}^{n} sum _{dmid i} phi(d) =frac{n(n+1)}{2}$
$sum _{i=1}^{n} sum _{dmid i} phi(d) i^2=frac{n^2(n+1)^2}{4}$
$=sum _{i=1}^{n} sum _{dmid i} f(d)(frac{i}{d})^2$
$=sum _{i=1}^{n} sum _{d=1}^{frac{n}{i}} f(d)i^2$
$Sum(n)=frac{n^2(n+1)^2}{4}-sum _{i=2}^{n} i^2Sum(left lfloor frac{n}{i} ight floor$
没有推出来的原因：不知道$sum _{i=1}^{n} i*[gcd(i,n)==1]=frac{nphi(n)}{2}$这个公式，推了很久发现自己的方向是错误的...
对于杜教筛的化简一定要考虑一个积性函数前缀和的形式...
代码：
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
//by NeighThorn
using namespace std;
const int mod=1e9+7,maxn=5000000+5;
int cnt,inv,inv2,f[maxn],phi[maxn],pri[maxn],vis[maxn];
long long n;
map<long long,int> mp;
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])
vis[i]=1,pri[++cnt]=i,phi[i]=i-1;
for(int j=1;j<=cnt&&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]=1LL*i*i%mod*phi[i]%mod;
for(int i=1;i<=5000000;i++) f[i]=(f[i]+f[i-1])%mod;
}
inline int calc(long long n){
if(n<=5000000) return f[n];
if(mp.find(n)!=mp.end()) return mp[n];
int ans,tmp;tmp=n%mod;ans=1LL*tmp*(tmp+1)%mod*inv2%mod;ans=1LL*ans*ans%mod;
for(long long i=2,r,x,y;i<=n;i=r+1){
r=n/(n/i);x=(i-1)%mod,y=r%mod;
tmp=1LL*y*(y+1)%mod*(y<<1|1)%mod*inv%mod;
tmp-=1LL*x*(x+1)%mod*(x<<1|1)%mod*inv%mod;
if(tmp<0) tmp+=mod;
ans=(ans-1LL*tmp*calc(n/i)%mod);
if(ans<0) ans+=mod;
}
return mp[n]=ans;
}
signed main(void){
scanf("%lld",&n);prework();
int ans=0;inv=power(6,mod-2),inv2=power(2,mod-2);
for(long long i=1,r,x;i<=n;i=r+1){
r=n/(n/i);x=(n/i)%mod;
ans=(ans+1LL*x*(x+1)%mod*inv2%mod*(calc(r)-calc(i-1)+mod)%mod)%mod;
}
printf("%d
",ans);
return 0;
}
By NeighThorn