SPOJ - PGCD Primes in GCD Table - 莫比乌斯反演

#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 10000000
#define MAXP 800000

int prime[MAXP+10],cntpr,mu[MAXN+10],a,b,sum[MAXN+10];
bool isprime[MAXN+10];

void CalMobius(int n)
{
mu[1]=1;
for(int i=2;i<=n;i++){
if(!isprime[i]){
prime[++cntpr]=i;
mu[i]=-1;
}
for(int j=1;prime[j]*i<=n&&j<=cntpr;j++){
isprime[prime[j]*i]=true;
if(i%prime[j]==0){
mu[prime[j]*i]=0;
break;
}
mu[prime[j]*i]=-mu[i];
}
}
for(int i=1;i<=n;i++)
sum[i]=mu[i]+sum[i-1];
}
long long Cal(int n,int m)
{
long long ret=0;
int side=min(n,m),last;
for(int i=1;i<=side;i=last+1){
last=min(n/(n/i),m/(m/i));
ret+=1LL*(sum[last]-sum[i-1])*(n/i)*(m/i);
}
return ret;
}
int main()
{
int T;
scanf("%d",&T);
CalMobius(MAXN);
while(T--){
scanf("%d%d",&a,&b);
int side=min(a,b);
long long ans=0;
for(int i=1;prime[i]<=side&&i<=cntpr;i++)
ans+=Cal(a/prime[i],b/prime[i]);
printf("%lld\n",ans);
}
}

#include<cstdio>
#include<algorithm>
using namespace std;
#define MAXN 10000000
int sum[MAXN+10],mu[MAXN+10],p[MAXN+10],pcnt,m,n,T;
long long ans;
bool f[MAXN+10];
void Read(int &x){
char c;
while(c=getchar(),c!=EOF)
if(c>='0'&&c<='9'){
x=c-'0';
while(c=getchar(),c>='0'&&c<='9')
x=x*10+c-'0';
ungetc(c,stdin);
return;
}
}
void prepare(){
int i,j;
for(i=2;i<=MAXN;i++){
if(!f[i])
p[++pcnt]=i,mu[i]=-1,sum[i]=1;
for(j=1;p[j]*i<=MAXN;j++){
f[p[j]*i]=1;
if(i%p[j]==0){
sum[i*p[j]]=mu[i];
mu[i*p[j]]=0;
break;
}
mu[i*p[j]]=-mu[i];
sum[i*p[j]]=mu[i]-sum[i];
}
sum[i]+=sum[i-1];
}
}
void solve(){
int i,last,t=min(m,n);
ans=0;
for(i=1;i<=t;i=last+1){
last=min(n/(n/i),m/(m/i));
ans+=1ll*(sum[last]-sum[i-1])*(n/i)*(m/i);
}
}
int main()
{
Read(T);
prepare();
while(T--){
Read(n),Read(m);
solve();
printf("%lld\n",ans);
}
}