第十五届浙江省赛 F Now Loading!!!

Now Loading!!!

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;  
}