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本文介绍了一种处理区间查询问题的有效算法。通过预先计算特定区间的累积和并利用二分查找技术，该方法能够快速响应查询请求，即计算给定数值a的n次幂加1与a的n+1次幂的向上取整后的值是否相等。文章还强调了除法与求和顺序的不同可能会影响最终结果，并提供了实现这一算法的C++代码。

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Time Limit: 1 Second Memory Limit: 131072 KB

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\n",ans);
    }
    return 0;
}