HDU 5212 莫比乌斯反演

Code

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1306    Accepted Submission(s): 540

Problem Description

WLD likes playing with codes.One day he is writing a function.Howerver,his computer breaks down because the function is too powerful.He is very sad.Can you help him?

The function:

int calc
{
  
  int res=0;
  
  for(int i=1;i<=n;i++)
    
    for(int j=1;j<=n;j++)
    
    {
      
      res+=gcd(a[i],a[j])*(gcd(a[i],a[j])-1);
      
      res%=10007;
    
    }
  
  return res;

}

 

Input

There are Multiple Cases.(At MOST 10)

For each case:

The first line contains an integer N(1≤N≤10000).

The next line contains N integers a1,a2,...,aN(1≤ai≤10000).

 

Output

For each case:

Print an integer,denoting what the function returns.

 

Sample Input

5
1 3 4 2 4

 

Sample Output

64

Hint

gcd(x,y) means the greatest common divisor of x and y.

 

 

题意：给定序列1≤i,j≤n，求gcd(a[i],a[j])∗(gcd(a[i],a[j])−1)之和。

思路：倍数莫比乌斯反演。

代码：

 #include<bits/stdc++.h>
 #define  ll long long
 using namespace std;
 const int N = 1e5 + ;
 const int mod=;
 int t;
 //线性筛法求莫比乌斯函数
 bool vis[N + ];
 int pri[N + ];
 int mu[N + ];
 int sum[N];

 void mus() {
     memset(vis, , sizeof(vis));
     mu[] = ;
     int tot = ;
     for (int i = ; i < N; i++) {
         if (!vis[i]) {
             pri[tot++] = i;
             mu[i] = -;
         }
         for (int j = ; j < tot && i * pri[j] < N; j++) {
             vis[i * pri[j]] = ;
             if (i % pri[j] == ) {
                 mu[i * pri[j]] = ;
                 break;
             }
             else  mu[i * pri[j]] = -mu[i];
         }
     }
     sum[]=;
     for(int i=;i<N;i++) sum[i]=sum[i-]+mu[i];
 }
 int n,m,k;

 int a[N];
 int b[N];
 int F[N];
 int main() {
     mus();
     while(scanf("%d",&n)==){
         int ma=;
         memset(F,, sizeof(F));
         memset(b,, sizeof(b));
         for(int i=;i<n;i++){
             scanf("%d",&a[i]);
             b[a[i]]++;//b[a[i]]的个数
             ma=max(ma,a[i]);
         }
         for(int i=;i<=ma;i++)
             for(int j=i;j<=ma;j+=i) F[i]+=b[j];//在范围内i的倍数的个数
         ll ans=;
         for(int i=;i<=ma;i++){
             ll res=;
             for(int j=i;j<=ma;j+=i){
                 if(!F[j]) continue;
                 res+=1ll*mu[j/i]*F[j]*F[j]%mod;//公约数均为i
             }
             ans=(ans+res*i%mod*(i-))%mod;//同时乘上i*i-1
         }
         ans%=mod;
         ans+=mod;
         ans%=mod;
         printf("%lld\n",ans);
     }
     return ;
 }