DreamGrid has  integers . DreamGrid also has  queries, and each time he would like to know the value of

Input

There are multiple test cases. The first line of input is an integer  indicating the number of test cases. For each test case:

The first line contains two integers  and  () -- the number of integers and the number of queries.

The second line contains  integers  ().

The third line contains  integers  ().

It is guaranteed that neither the sum of all  nor the sum of all  exceeds .

Output

For each test case, output an integer , where  is the answer for the -th query.

Sample Input

2
3 2
100 1000 10000
100 10
4 5
2323 223 12312 3
1232 324 2 3 5

Sample Output

11366
45619

【題意】

對於每一次詢問，求上式的值

【分析】

把a排序

由於分母的範圍很小 [2,30]，可以列舉；

對於每一次詢問p，列舉分母 i 時，可以找出a中分母等於 i 的那一段，預處理字首和，用於此時直接加。

複雜度n * log * log

【程式碼】

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=1e6+5;
const int INF=0x3f3f3f3f;
const int mod=1e9;

int n,m;
int a[500010],p;
int sum[32][500010];
ll qpow(ll n,ll m)
{
    ll ans=1;
    while(m){
        if(m&1)ans*=n;
        n*=n;
        m>>=1;
        if(ans>20001001000)return -1;
    }
    return ans;
}
int main()
{
    int T;cin>>T;
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++)scanf("%d",&a[i]);
        sort(a+1,a+1+n);
        for(int i=1;i<=31;i++) //pre sum of a/log
        {
            sum[i][0]=0;
            for(int j=1;j<=n;j++)
                sum[i][j]=(sum[i][j-1]+a[j]/i)%mod;
        }
        ll ans=0;
        for(int k=1;k<=m;k++)
        {
            scanf("%d",&p);
            ll res=0;
            for(int i=1;i<=31;i++) //log p ai
            {
                ll down=qpow(p,i-1); //> it
                ll up=qpow(p,i); // <=it
                if(down==-1)break;

                int l=1,r=n+1,mid;
                if(down!=-1)
                    while(l<r){
                        mid=(l+r)>>1;
                        if(a[mid]>down)r=mid;
                        else l=mid+1;
                    }
                int L=r;

                l=1,r=n+1;
                if(up!=-1)
                    while(l<r){
                        mid=(l+r)>>1;
                        if(a[mid]>up)r=mid;
                        else l=mid+1;
                    }
                int R=r-1;
                if(L>n)break;
                if(L>R)continue;

                res=(res+sum[i][R]-sum[i][L-1])%mod;
                res=(res+mod)%mod;
            }
            ans=(ans+res*k%mod)%mod;
        }
        printf("%lld\n",ans);
    }
}