POI2007_zap 莫比乌斯反演

题意：http://hzwer.com/4205.html

同hdu1695

 #include <iostream>
 #include <cstring>
 #include <cmath>
 #include <cstdio>
 using namespace std;
 #define LL long long
 #define MMX 50010
 int mu[MMX],msum[MMX];
 LL n;
 bool check[MMX];
 int prime[MMX];

 void Moblus()                //莫比乌斯反演，mu[i]为莫比乌斯函数
 {
     memset(check,false,sizeof(check));
     mu[] = ;
     int tot = ;
     for(int i = ; i <= MMX; i++)
     {
         if( !check[i] )
         {
             prime[tot++] = i;
             mu[i] = -;
         }
         for(int j = ; j < tot; j++)
         {
             if(i * prime[j] > MMX) break;
             check[i * prime[j]] = true;
             if( i % prime[j] == )
             {
                 mu[i * prime[j]] = ;
                 break;
             }
             else
             {
                 mu[i * prime[j]] = -mu[i];
             }
         }
     }
     msum[]=mu[];                //求和，求G的时候用
     for (int i=;i<=MMX;i++)
         msum[i]=msum[i-]+mu[i];
 }

 LL G(int n,int m)           //G(x,y)表示有多少组x<=n，y<=m，且x,y互质  （(x,y)和(y,x)算两组）
 {
     LL ans = ;
     if(n > m)   swap(n,m);
     for(int i = , la = ; i <= n; i = la+)
     {
         la = min(n/(n/i),m/(m/i));
         ans += (LL)(msum[la] - msum[i-])*(n/i)*(m/i);      //事先预处理：msum[n]=SUM(mu[1..n])
     }
     return ans;
 }

 int main()
 {
     int T;
     cin>>T;
     Moblus();
     for (int zy=; zy<=T; zy++)
     {
         int a,b,c,d,k;
         //cin>>b>>d>>k;
         scanf("%d%d%d\n",&b,&d,&k);
         if (b>d) swap(b,d);     //assume b<d
         b=b/k;
         d=d/k;

         LL ans1 = G(b,d);

         printf("%lld\n",ans1);
     }
 }

1020

root

1003

Accepted

564K

3819MS

G++

1.62K

2014-11-16 18:08:01