题目链接：

CA Loves GCD

Time Limit: 6000/3000 MS (Java/Others)   

 Memory Limit: 262144/262144 K (Java/Others)

Problem Description

CA is a fine comrade who loves the party and people; inevitably she loves GCD (greatest common divisor) too. 
Now, there are N different numbers. Each time, CA will select several numbers (at least one), and find the GCD of these numbers. In order to have fun, CA will try every selection. After that, she wants to know the sum of all GCDs. 
If and only if there is a number exists in a selection, but does not exist in another one, we think these two selections are different from each other.

Input

First line contains T denoting the number of testcases.
T testcases follow. Each testcase contains a integer in the first time, denoting N, the number of the numbers CA have. The second line is N numbers. 
We guarantee that all numbers in the test are in the range [1,1000].
1≤T≤50

Output

T lines, each line prints the sum of GCDs mod 100000007.

Sample Input

2

2

2 4

3

1 2 3

Sample Output

8

10

题意：

给n个数,问n个数的全组合的gcd的和是多少;

思路：

注意到数都在1000以内,所以把所以得组合都找出来,但是数量巨大所以相同的要压缩;

AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
ll dp[1003],num[1003],p[1003][1003];
int n,a[1003];
const ll mod=1e8+7;//注意是8,比赛的时候这个地方直接wa到cry
int gcd(int x,int y)
{
    if(y==0)return x;
    return gcd(y,x%y);
}
int main()
{
    for(int i=1;i<=1000;i++)
    {
        for(int j=i;j<=1000;j++)
        {
            p[i][j]=gcd(i,j);
        }
    }
    int t;
    scanf("%d",&t);
    while(t--)
    {
        for(int i=1;i<=1000;i++)
        {
            num[i]=0;
        }
        ll ans=0;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
        }
        sort(a+1,a+n+1);
        for(int i=1;i<=n;i++)
        {
            dp[i]=(ll)a[i];
            for(int j=1;j<=a[i];j++)
            {
                if(num[j])
                {
                    num[j]%=mod;
                    dp[i]+=num[j]*p[j][a[i]];//压缩的地方
                    dp[i]%=mod;
                    num[p[j][a[i]]]+=num[j];
                    num[p[j][a[i]]]%=mod;
                }
            }
            num[a[i]]++;
            ans+=dp[i];
            ans%=mod;
        }
        cout<<ans<<"\n";
    }

    return 0;
}