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\n",sum);

67     }

68     return 0;

69 }

 

你可能感兴趣的:(Prim)