BZOJ3309: DZY Loves Math

推G(x)的时候好神啊！！！

题解：http://blog.csdn.net/popoqqq/article/details/42122413

#include<cstdio>
#include<iostream>
using namespace std;
#define ll long long
char c;
inline void read(int&a)
{
    a=0;do c=getchar();while(c<'0'||c>'9');
    while(c<='9'&&c>='0')a=(a<<3)+(a<<1)+c-'0',c=getchar();
}
const int Maxn=10000001; 
int Prime[Maxn],tot,G[Maxn],Pow[Maxn],Maxp[Maxn];
bool Check[Maxn];
ll ans; 
inline int min(int a,int b){return a<b?a:b;} 
int main()
{ 
  ll k;
  int T,n,m,i,j;
  for(i=2;i<Maxn;i++)
   { 
    if(!Check[i])Prime[++tot]=i,G[i]=Pow[i]=1,Maxp[i]=i; 
    for(j=1;j<=tot;j++)
    { 
      k=i*Prime[j];
      if(k>=Maxn)break; 
      Check[k]=1; 
      if(i%Prime[j])
      { 
        Pow[k]=1,Maxp[k]=Prime[j]; 
        if(Pow[i]==1)G[k]=-G[i]; 
      }
       else
       { 
         Pow[k]=Pow[i]+1,Maxp[k]=Maxp[i]*Prime[j],n=i/Maxp[i]; 
          if(n==1)G[k]=1;
         else G[k]=Pow[n]==Pow[k]?-G[n]:0; 
        break; 
       } 
    } 
  } 
  for(i=2;i<Maxn;i++)
    G[i]+=G[i-1]; 
  read(T);
  while(T--)
  {  
    read(n),read(m);
    for(ans=0,i=1;i<=n&&i<=m;i=j+1)
     j=min(n/(n/i),m/(m/i)),ans+=(ll)(G[j]-G[i-1])*(n/i)*(m/i); 
    printf("%lld\n",ans); 
  }

  return 0; 
}