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
思路：
对于a[i]的范围为[2,1e9]，我们可以知道log p (a[i])的值一定小于30，因此，我们可以把输入的a[ ]先预处理一下，把floor(a[i]/k)（k属于[1,30]）的值的前缀和先存下来（注意：这里要用vector，ll sum[30][5e5+10]的数组开不下，会导致段错误）。
接下来，我们要确定对于当前p，log p (a[i])的值具体为多少：
易得，当a[i]<=p的时候，log p (a[i])向上取整的值为1；
当p<a[i]<=p^2的时候，log p (a[i])向上取整的值为2；
当p^2<a[i]<=p^3的时候，log p (a[i])向上取整的值为3；
......
因此，我们可以把p，p^2，p^3，...，p^n存下来，只要存到p^n>1e9即可。
得到这两个vector之后，我们就可以通过遍历p[i]来计算答案了：
对于一个p，我们先在a[i]中找到<=p的下标位置，ans+=这个位置之前的a[i]除以1后的前缀和；再在a[i]中找到<=p^2的下标位置，ans+=这个位置之前的a[i]除以2后的前缀和；...；知道p^n的值大于a[i]中的所有元素就结束循环。
至于如何找，当然是二分了，注意要先把a[i]排个序。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
typedef long long ll;
const int MAX = 5e5 + 10;
const ll MOD = 1e9;
ll a[MAX], p[MAX];
vector<ll>sum[40];
vector<ll>mu[MAX];
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
int n, m;
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i++)
scanf("%lld", &a[i]);
sort(a + 1, a + n + 1);
for (int i = 0; i < m; i++)
scanf("%lld", &p[i]);
for (int i = 0; i <= 35; i++)
sum[i].clear();
for (int i = 1; i <= 32; i++)
{
sum[i].push_back(0);
}
for (int i = 1; i <= 32; i++)
{
sum[i].push_back(a[1] / i);
}
for (int i = 2; i <= n; i++)
{
for (int j = 1; j <= 32; j++)
{
ll temp = sum[j][i - 1] + a[i] / j;
sum[j].push_back(temp);
}
}
for (int i = 0; i <= m; i++)
mu[i].clear();
for (int i = 0; i < m; i++)
{
ll res = p[i];
mu[i].push_back(p[i]);
while (true)
{
res = res*p[i];
mu[i].push_back(res);
if (res > (1e9 + 100))
break;
}
}
ll ans = 0;
for (int i = 0; i < m; i++)
{
ll tmp = 0;
int pre = 0;
for (int j = 0; j < mu[i].size(); j++)
{
int pos = upper_bound(a + 1, a + n + 1, mu[i][j]) - a;
pos = pos - 1;
if (pos == 0) continue;
tmp = (tmp + sum[j + 1][pos] - sum[j + 1][pre]) % MOD;
if (pos == n) break;
pre = pos;
}
ans = (ans + tmp*(i + 1) % MOD) % MOD;
}
printf("%lld\n", ans);
}
return 0;
}