第十五届浙江省赛 F Now Loading!!!

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Time Limit: 1 Second      Memory Limit: 131072 KB

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

 

for a given number , where , .

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

题解：  
对于a^n+1与a^(n+1),以a为底取log向上取整后是一样的。  
因此可以按区间计算，然后二分查找区间。  
注意：先除在求和与先求和在除是不一样的，赛场上没想到  
这一点导致这题没过。  
代码：  
#include<bits/stdc++.h>  
using namespace std;  
#define ll long long  
const int maxn=500005;  
const int mod=1e9;  
ll a[maxn];  
int b[maxn][33];  
int main()  
{  
    int T;scanf("%d",&T);  
    while(T--)  
    {  
        int n,m;scanf("%d%d",&n,&m);  
        ll ma=0,ans=0;  
        for(int i=1;i<=n;i++)  
        {  
            scanf("%lld",&a[i]);  
            ma=max(ma,a[i]);  
        }  
        sort(a+1,a+n+1);  
        for(ll i=1;i<33;i++)  
        {  
            for(int j=1;j<=n;j++)  
                b[j][i]=(b[j-1][i]+a[j]/i)%mod;  
        }  
        for(int i=1;i<=m;i++)  
        {  
            ll x;scanf("%lld",&x);  
            bool bb=0;ll sum=0;  
            int a1=1;  
            for(ll j=1,k=1;;k++)  
            {  
                if(j<=ma)j=j*x;  
                else break;  
                int id=upper_bound(a+a1,a+n+1,j)-a-1;  
                sum=(sum+b[id][k]-b[a1-1][k])%mod;  
                if(sum<0)sum+=mod;  
                a1=id+1;  
            }  
            ans=(ans+i*sum)%mod;  
        }  
        printf("%lld
",ans);  
    }  
    return 0;  
}