hdu 5212(莫比乌斯反演)

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Code

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)

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

)

For each case:

The first line contains an integer

N(1≤N≤10000)

.

The next line contains

integers

a1,a2,...,aN(1≤ai≤10000)

Output

For each case:

Print an integer,denoting what the function returns.

Sample Input

51 3 4 2 4

Sample Output

64Hint
gcd(x,y) means the greatest common divisor of x and y.
 
解题思路：这道题关键是找到相同的gcd(x,y)的对数。可以用莫比乌斯反演。
参考博客：http://blog.csdn.net/Yukizzz/article/details/51541973
#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;const int maxn = 10005;const int mod = 10007;bool check[maxn];int mu[maxn],prime[maxn];int n,m,a[maxn],cnt[maxn];LL F[maxn];void moblus(){int tot = 0;memset(check,false,sizeof(check));mu[1] = 1;for(int i = 2; i <= maxn; i++){if(!check[i]){prime[tot++] = i;mu[i] = -1;}for(int j = 0; j < tot; j++){if(i * prime[j] > maxn) break;check[i * prime[j]] = true;if(i % prime[j] == 0){mu[i * prime[j]] = 0;break;}else mu[i * prime[j]] = -mu[i];}}}int main(){moblus();while(scanf("%d",&n)!=EOF){memset(cnt,0,sizeof(cnt));m = 0;for(int i = 1; i <= n; i++){scanf("%d",&a[i]);cnt[a[i]]++;m = max(m,a[i]);}for(int i = 1; i <= m; i++){F[i] = 0;for(int j = i; j <= m; j += i)F[i] += cnt[j];F[i] = F[i] * F[i];}LL tmp,ans = 0;for(int i = 1; i <= m; i++) //枚举最大公约数{tmp = 0;for(int j = i; j <= m; j += i)tmp = (tmp + mu[j / i] * F[j] % mod) % mod;ans = (ans + tmp * i % mod * (i - 1) % mod) % mod;}printf("%lld\n",ans);}return 0;}

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