Code

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

Total Submission(s): 209    Accepted Submission(s): 85

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.

 

Source

BestCoder Round #39 ($)

题意: 简单易懂,就给你一段代码,叫你优化;

题解: 莫比乌斯反演, 首先我们设f(d)表示在给出的所有数中有f(d)对最大公约数是d. cnt(n) 表示在给出的所有数中有cnt(n)个是n的倍数(包含n). 假设我们已经知道了这两个函数,然后就可以用莫比乌斯反演做了. 首先F(n) = cnt(n) * cnt(n); 表示大于等于n的所有数组成的对数,那么有F(n) = f(n) + f(n * 2) + f(n * 3) + .....f(max);

做完以上步骤,就可以套用莫比乌斯反演做了.

莫比无私反演的公式在bin神的博客上有,去搬吧! 啊啊啊啊啊啊啊啊啊啊啊!!!!!!!!!!!!

AC代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int MAXN = 1e4;
typedef long long ll;
const int mod = 10007;
bool check[MAXN + 10];
ll prime[MAXN + 10],cnt[MAXN + 10];
ll mu[MAXN + 10],F[MAXN + 10];

void Moblus()
{
    memset(check,false,sizeof(check));
    mu[1] = 1;
    int tot = 0;
    for(int i = 2; i <= MAXN; i++)
    {
        if(!check[i])
        {
            prime[tot++] = i;
            mu[i] = -1;
        }
        for(int j = 0; i * prime[j] <= MAXN; j++)
        {
            check[i * prime[j]] = true;
            if(i % prime[j] == 0)
            {
                mu[i * prime[j]] = 0;
                break;
            }
            mu[i * prime[j]] = -mu[i];
        }
    }
}

int main()
{
    //freopen("in","r",stdin);
    Moblus();
    int n,x;
    while(~scanf("%d",&n))
    {
        memset(cnt,0,sizeof(cnt));
        for(int i = 0; i < n; i++)
        {
            scanf("%d",&x);
            cnt[x]++;
        }
        for(int i = 1; i <= MAXN; i++)
            for(int j = i * 2; j <= MAXN; j += i)
                cnt[i] += cnt[j];
        for(int i = 1; i <= MAXN; i++) F[i] = cnt[i] * cnt[i];
        ll res = 0;
        for(int i = 1; i <= MAXN; i++)
        {
            ll tp = 0;
            for(int j = i; j <= MAXN; j += i)
            {
                tp += mu[j / i] * F[j];
                if(tp >= mod) tp %= mod;
            }
            res += tp * i % mod * (i - 1);
            if(res >= mod) res %= mod;
        }
        printf("%I64d\n",res);
    }
    return 0;
}