bzoj 2671 莫比乌斯反演

Calc

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 451  Solved: 234
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Description

　　给出N，统计满足下面条件的数对(a,b)的个数：
　　1.1<=a<b<=N
　　2.a+b整除a*b

　

Input

　一行一个数N

　

Output

　一行一个数表示答案

Sample Input

15

Sample Output

4

HINT

数据规模和约定

Test N Test N

1 <=10 11 <=5*10^7

2 <=50 12 <=10^8

3 <=10^3 13 <=2*10^8

4 <=5*10^3 14 <=3*10^8

5 <=2*10^4 15 <=5*10^8

6 <=2*10^5 16 <=10^9

7 <=2*10^6 17 <=10^9

8 <=10^7 18 <=2^31-1

9 <=2*10^7 19 <=2^31-1

10 <=3*10^7 20 <=2^31-1

Source

 

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http://blog.csdn.net/popoqqq/article/details/45095601

 #pragma GCC optimize(2)
 #pragma G++ optimize(2)
 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<cmath>
 #define ll long long
 #define mod 1000000007
 #define N 50005
 using namespace std;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while (ch<''||ch>''){if (ch=='-') f=-;ch=getchar();}
     while (ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }

 bool flag[N];
 int tot,p[N],miu[N],n,m,pos;
 ll ans;

 void pre()
 {
     miu[]=;
     for (int i=;i<N;i++)
     {
         if (!flag[i]) p[++tot]=i,miu[i]=-;
         for (int j=;j<=tot&&p[j]*i<N;j++)
         {
             flag[i*p[j]]=;
             if (i%p[j]==) break;
             miu[i*p[j]]=-miu[i];
         }
     }
 }
 int main()
 {
     pre();
     scanf("%d",&n);m=sqrt(n);
     for (int d=;d<=m;d++)
     {
         for (int i=;i<=m/d;i++)
         {
             int last=n/(d*d*i);
             for (int x=i+,p=;x<=*i-&&x<=last;x=pos+)
             {
                 pos=last/(last/x);
                 ans+=1LL*miu[d]*(min(pos,*i-)-x+)*(last/x);
             }
         }
     }
     printf("%lld",ans);
 }