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

Sample Output

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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HOME Back

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

 1 #pragma GCC optimize(2)
 2 #pragma G++ optimize(2)
 3 #include<cstdio>
 4 #include<iostream>
 5 #include<cstring>
 6 #include<cmath>
 7 #define ll long long
 8 #define mod 1000000007
 9 #define N 50005
10 using namespace std;
11 inline int read()
12 {
13     int x=0,f=1;char ch=getchar();
14     while (ch<'0'||ch>'9'){if (ch=='-') f=-1;ch=getchar();}
15     while (ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
16     return x*f;
17 }
18 
19 bool flag[N];
20 int tot,p[N],miu[N],n,m,pos;
21 ll ans;
22 
23 void pre()
24 {
25     miu[1]=1;
26     for (int i=2;i<N;i++)
27     {
28         if (!flag[i]) p[++tot]=i,miu[i]=-1;
29         for (int j=1;j<=tot&&p[j]*i<N;j++)
30         {
31             flag[i*p[j]]=1;
32             if (i%p[j]==0) break;
33             miu[i*p[j]]=-miu[i];
34         }
35     }
36 }
37 int main()
38 {
39     pre();
40     scanf("%d",&n);m=sqrt(n);
41     for (int d=1;d<=m;d++)
42     {
43         for (int i=2;i<=m/d;i++)
44         {
45             int last=n/(d*d*i);
46             for (int x=i+1,p=0;x<=2*i-1&&x<=last;x=pos+1)
47             {
48                 pos=last/(last/x);
49                 ans+=1LL*miu[d]*(min(pos,2*i-1)-x+1)*(last/x);
50             }
51         }
52     } 
53     printf("%lld",ans);
54 }