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BZOJ2301 莫比乌斯反演

题意:a<=x<=b,c<=y<=d,求满足gcd(x,y)=k的数对(x,y)的数量         ((x,y)和(y,x)不算同一个)

 

比hdu1695多加了个下界,还有顺序不一样的也算上了。

因为G(x,y)本来就是顺序不一样的算不同方案,所以这题的公式就是:

Ans=G(b/k,d/k)-G((a-1)/k,d/k)-G(b/k,(c-1)/k)+G((a-1)/k,(c-1)/k)

 

但是本题数据很大,直接计算会TLE,

有一个优化:http://www.cnblogs.com/zhsl/p/3269288.html

 

//calc G(a,b)

LL _G(int a,int b)          //朴素算法
{
    int tx=min(a,b),ty=max(a,b);
    LL ans = 0;
    for(int i = 1; i <= tx; i++)
        ans += (LL)mu[i]*(tx/i)*(ty/i);
    return ans;
}

LL G(int n,int m)           //加分块优化
{
    LL ans = 0;
    if(n > m)   swap(n,m);
    for(int i = 1, la = 0; i <= n; i = la+1)
    {
        la = min(n/(n/i),m/(m/i));
        ans += (LL)(msum[la] - msum[i-1])*(n/i)*(m/i);      //事先预处理:msum[n]=SUM(mu[1..n])
    }
    return ans;
}

 

 

不知为什么自己测AC了结果bzoj上一直RuntimeError......要完蛋了orz,改成scanf和printf就过了

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cmath>
 4 #include <cstdio>
 5 using namespace std;
 6 #define LL long long
 7 #define MMX 50100
 8 LL mu[MMX],msum[MMX];
 9 bool check[MMX];
10 int prime[MMX];
11 int a,b,c,d,k,T;
12 
13 void Moblus()
14 {
15     memset(check,false,sizeof(check));
16     mu[1] = 1;
17     int tot = 0;
18     for(int i = 2; i <= MMX; i++)
19     {
20         if( !check[i] )
21         {
22             prime[tot++] = i;
23             mu[i] = -1;
24         }
25         for(int j = 0; j < tot; j++)
26         {
27             if(i * prime[j] > MMX) break;
28             check[i * prime[j]] = true;
29             if( i % prime[j] == 0)
30             {
31                 mu[i * prime[j]] = 0;
32                 break;
33             }
34             else
35             {
36                 mu[i * prime[j]] = -mu[i];
37             }
38         }
39     }
40     msum[1]=mu[1];
41     for (int i=2;i<=MMX;i++)
42         msum[i]=msum[i-1]+mu[i];
43 }
44 
45 //calc G(a,b)
46 LL _G(int a,int b)          //朴素算法
47 {
48     int tx=min(a,b),ty=max(a,b);
49     LL ans = 0;
50     for(int i = 1; i <= tx; i++)
51         ans += (LL)mu[i]*(tx/i)*(ty/i);
52     return ans;
53 }
54 
55 LL G(int n,int m)           //加分块优化
56 {
57     LL ans = 0;
58     if(n > m)   swap(n,m);
59     for(int i = 1, la = 0; i <= n; i = la+1)
60     {
61         la = min(n/(n/i),m/(m/i));
62         ans += (LL)(msum[la] - msum[i-1])*(n/i)*(m/i);      //事先预处理:msum[n]=SUM(mu[1..n])
63     }
64     return ans;
65 }
66 
67 int main()
68 {
69     //freopen("in.txt","r",stdin);
70     //freopen("out.txt","w",stdout);
71 
72     cin>>T;
73     Moblus();
74     while (T--)
75     {
76         scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
77         //cout<<G(b/k,d/k)<<"  "<<G((a-1)/k,d/k)<<"  "<<G(b/k,(c-1)/k)<<"  "<<G((a-1)/k,(c-1)/k)<<endl;
78         LL res=G(b/k,d/k)-G((a-1)/k,d/k)-G(b/k,(c-1)/k)+G((a-1)/k,(c-1)/k);
79         printf("%lld\n",res);
80     }
81     return 0;
82 }
View Code

 

实测:

朴素算法:   

1005 root 1002 Accepted 380K 47400MS G++ 1.6K 2014-11-15 15:37:03

优化算法:

1014 root 1002 Accepted 952K 6011MS G++ 1.81K 2014-11-15 17:17:59

 

posted on 2014-11-15 15:57  Pentium.Labs  阅读(...)  评论(...编辑  收藏



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