hdu6715(莫比乌斯函数)

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=6715

题解

结论题。。首先有 $\mu([i,j])=\mu(i)\mu(j)\mu((i,j))$

那么

直接预处理第一个卷积,后面的直接埃筛即可。。




代码

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/**
*         ┏┓    ┏┓
*         ┏┛┗━━━━━━━┛┗━━━┓
*         ┃       ┃  
*         ┃   ━    ┃
*         ┃ >   < ┃
*         ┃       ┃
*         ┃... ⌒ ...  ┃
*         ┃ ┃
*         ┗━┓ ┏━┛
*          ┃ ┃ Code is far away from bug with the animal protecting          
*          ┃ ┃ 神兽保佑,代码无bug
*          ┃ ┃           
*          ┃ ┃       
*          ┃ ┃
*          ┃ ┃           
*          ┃ ┗━━━┓
*          ┃ ┣┓
*          ┃ ┏┛
*          ┗┓┓┏━━━━━━━━┳┓┏┛
*           ┃┫┫ ┃┫┫
*           ┗┻┛ ┗┻┛
*/

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<queue>
#include<map>
#include<stack>
#include<set>
#include<bitset>
#include<cmath>
#include<assert.h>
#define inc(i,l,r) for(int i=l;i<=r;i++)
#define dec(i,l,r) for(int i=l;i>=r;i--)
#define link(x) for(edge *j=h[x];j;j=j->next)
#define mem(a) memset(a,0,sizeof(a))
#define ll unsigned long long
#define succ(x) (1ll<<x)
#define lowbit(x) (x&(-x))
#define sqr(x) ((x)*(x))
#define mid (x+y>>1)
#define NM 1000005
#define nm 50005
using namespace std;
const double pi=acos(-1);
const double eps=1e-8;
const ll inf=1e9+7;
ll read(){
ll x=0,f=1;char ch=getchar();
while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return f*x;
}




int n,m;
ll ans;
int tot,prime[NM],mu[NM],f[NM];
bool v[NM];
void init(){
n=1e6;f[1]=mu[1]=1;
inc(i,2,n){
if(!v[i])prime[++tot]=i,mu[i]=-1,f[i]=-2;
inc(j,1,tot){
if(i*prime[j]>n)break;
v[i*prime[j]]++;
if(i%prime[j]==0){
if(i/prime[j]%prime[j]){
f[i*prime[j]]=-f[i]/2;
}
break;
}
f[i*prime[j]]=-2*f[i];
mu[i*prime[j]]=-mu[i];
}
}
}

int main(){
init();
int _=read();while(_--){
n=read();m=read();
ans=0;
inc(i,1,n){
int s=0,t=0;
for(int j=i;j<=n;j+=i)s+=mu[j];
for(int j=i;j<=m;j+=i)t+=mu[j];
ans+=f[i]*s*t;
}
printf("%lld\n",ans);
}
return 0;
}