nyoj CO-PRIME 莫比乌斯反演

CO-PRIME

时间限制：1000 ms | 内存限制：65535 KB

难度：3

描述

This problem is so easy! Can you solve it?

You are given a sequence which contains n integers a1,a2……an, your task is to find how many pair(ai, aj)(i < j) that ai and aj is co-prime.

输入

There are multiple test cases.
Each test case conatains two line,the first line contains a single integer n,the second line contains n integers.
All the integer is not greater than 10^5.

输出

For each test case, you should output one line that contains the answer.

样例输入

3
1 2 3

样例输出

3

思路： http://blog.csdn.net/lyhvoyage/article/details/38455415应该是出题的人吧。

分析：莫比乌斯反演。

此题中，设F(d)表示n个数中gcd为d的倍数的数有多少对，f(d)表示n个数中gcd恰好为d的数有多少对，

则F(d)=∑f(n) (n % d == 0)

f(d)=∑mu[n / d] * F(n) (n %d == 0)

上面两个式子是莫比乌斯反演中的式子。

所以要求互素的数有多少对，就是求f(1)。

而根据上面的式子可以得出f(1)=∑mu[n] * F(n)。

所以把mu[]求出来，枚举n就行了，其中mu[i]为i的莫比乌斯函数。

 1 #include<iostream>
 2 #include<stdio.h>
 3 #include<cstring>
 4 #include<cstdlib>
 5 using namespace std;
 6 const int N = 1e5+1;
 7 
 8 int vis[N];
 9 int mu[N];
10 int prime[N],cnt;
11 int date[N];
12 long long ys[N];
13 int num[N];
14 void init()
15 {
16     memset(vis,0,sizeof(vis));
17     mu[1] = 1;
18     cnt = 0;
19     for(int i=2;i<N;i++)
20     {
21         if(!vis[i])
22         {
23             prime[cnt++] = i;
24             mu[i] = -1;
25         }
26         for(int j = 0;j<cnt&&i*prime[j]<N;j++)
27         {
28             vis[i*prime[j]] = 1;
29             if(i%prime[j]) mu[i*prime[j]] = -mu[i];
30             else
31             {
32                 mu [i *prime[j]] = 0;
33                 break;
34             }
35         }
36     }
37 }
38 int main()
39 {
40     int n,maxn;
41     init();
42     while(scanf("%d",&n)>0)
43     {
44         memset(num,0,sizeof(num));
45         memset(ys,0,sizeof(ys));
46         maxn = -1;
47         for(int i=1;i<=n;i++){
48             scanf("%d",&date[i]);
49             num[date[i]] ++;
50             if(date[i]>maxn) maxn = date[i];
51         }
52         /***计算F(N)*/
53         for(int i=1;i<=maxn;i++)
54         {
55             for(int j=i;j<=maxn;j=j+i)
56             {
57                 ys[i] = ys[i] + num[j];
58             }
59         }
60         long long sum = 0;
61         for(int i=1;i<=maxn;i++){
62             long long tmp = (long long)ys[i] *( ys[i]-1 )/2;
63              sum = sum + mu[i]*tmp;
64         }
65 
66         printf("%I64d
",sum);
67     }
68     return 0;
69 }