51nod1222最小公倍数计数

51nod1222

http://210.33.19.103/contest/1113/problem/2

同学的神仙做法：

首先考虑先去掉X<=Y的限制，也就是先计算满足要求的任意有序pair(X,Y)的数量，再用一些简单操作（略去）得到目标答案

化简式子可以得到$\sum_{d}\mu(d)\sum_a\sum_b\sum_c[abc<={\lfloor}\frac{n}{d^2}{\rfloor}]$

可以强行给a,b,c规定一个顺序。设a是最小的，则a只需要枚举到$(\frac{n}{d^2})^{1/3}$就行

剩下有很多做法了，略去了

 %:pragma GCC optimize()
 #include<cstdio>
 #include<algorithm>
 #include<cstring>
 #include<vector>
 #include<cmath>
 using namespace std;
 #define fi first
 #define se second
 #define mp make_pair
 #define pb push_back
 typedef long long ll;
 typedef unsigned long long ull;
 int mu[];
 int prime[],len;
 bool nprime[];
 /*
 ll calc0(ll n)
 {
     ll d,i,j,k,tan1,tan2,tan3,t1;
     for(d=1;;++d)
     {
         if(d*d>n)    break;
         t1=n/(d*d);
         tan1=tan2=tan3=0;
         for(i=1;i<=100;++i)
             for(j=1;j<=100;++j)
                 for(k=1;k<=100;++k)
                     if(i*j*k<=t1)
                         if(i==j&&j==k)
                             ++tan3;
                         else if(i==j||j==k||i==k)
                             ++tan2;
                         else ++tan1;
         printf("at%lld %lld %lld\n",tan1,tan2,tan3);
     }
     return 233;
 }
 */
 ll calc(ll n)
 {
     ll d,t1,t2,endj,i,j,j1,tan1,tan2,tan3,ans=,endi;
     for(d=;;++d)
         if(mu[d])
         {
             if(d*d>n)    break;
             t1=n/(d*d);
             tan1=tan2=tan3=;
             for(i=;;++i)
             {
                 if(i*i*i>t1)    break;
                 t2=t1/i;
                 endj=ll(sqrt(t2+0.5));
                 for(j=i+;j<=endj;j=j1+)
                 {
                     j1=min(endj,t2/(t2/j));
                     tan1+=(t2/j)*(j1-j+);
                 }
                 if(i+<=endj)
                     tan1-=(endj+i+)*(endj-i)/;
             }
             tan1*=;
             endi=ll(sqrt(t1+0.5));
             for(i=;i<=endi;++i)
                 tan2+=t1/(i*i);
             for(i=;;++i)
             {
                 if(i*i*i>t1)    break;
                 ++tan3;
             }
             tan2=(tan2-tan3)*;
             ans+=mu[d]*(tan1+tan2+tan3);
             //printf("1t%lld %lld %lld %lld\n",d,tan1,tan2,tan3);
         }
     return ans;
 }
 int main()
 {
     ll i,j,t;
     mu[]=;
     for(i=;i<=;i++)
     {
         if(!nprime[i])    prime[++len]=i,mu[i]=-;
         for(j=;j<=len&&(t=i*prime[j])<=;j++)
         {
             nprime[t]=;
             if(i%prime[j]==)    {mu[t]=;break;}
             else    mu[t]=-mu[i];
         }
     }
     ll a,b;
     scanf("%lld%lld",&a,&b);
     printf("%lld\n",(calc(b)-calc(a-)+b-a+)/);
     //calc0(b);calc0(a-1);
     return ;
 }