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
题意：给出数组A，和m次询问，计算以上公式
题解：初始的想法是对于每次询问，直接去算公式，这样的时间复杂度是O(n*m)对于题给数据显然是超时的，
现在我们考虑怎么优化，可以想到的是每次查询时，都会重复除以同一个数。为什么这么说呢，因为[logp(ai)]最大为31
(当p = 2,ai = 在[2^30+1,2^31-1]之间时)
考虑到公式中的分母如此之小，题目应该从这里入手
我们可以预先处理A数组的前缀和，注意应该是处理A[]除以分母的前缀和，我们不能在查询的时候计算出区间的值然后再除，这样不符合公式的定义。
枚举分母然后就是二分查找上下界即可。
temp = (temp + sum[i][r-1] - sum[i][l-1]) % mod;temp有可能减到负数
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 1e9;
const ll INF = (ll)1<<33;
const int maxn = 5e5+10;
ll arr[maxn],sum[32][maxn];
ll n,m;
namespace IO
{
template <typename T>
inline bool scan_d (T &ret)
{
char c;
int sgn;
if (c = getchar(), c == EOF)return false; //EOF
while (c != '-' && (c < '0' || c > '9') )
if((c = getchar()) == EOF) return false;
sgn = (c == '-') ? -1 : 1;
ret = (c == '-') ? 0 : (c - '0');
while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
ret *= sgn;
return true;
}
template<typename T>
void print(T x)
{
static char s[33], *s1;
s1 = s;
if (!x) *s1++ = '0';
if (x < 0) putchar('-'), x = -x;
while(x) *s1++ = (x % 10 + '0'), x /= 10;
while(s1-- != s) putchar(*s1);
}
template<typename T>
void println(T x)
{
print(x);
putchar('\n');
}
};
int main()
{
int caset;IO::scan_d(caset);
while(caset--) {
IO::scan_d(n);IO::scan_d(m);
for(int i=1;i<=n;i++) IO::scan_d(arr[i]);
sort(arr+1,arr+1+n);
for(int i=1;i<=31;i++) {
sum[i][0] = 0;
for(int j=1;j<=n;j++) {
sum[i][j] = (sum[i][j-1] + arr[j] / i) % mod;
}
}
ll ans = 0,p;
for(ll j=1;j<=m;j++) {
ll temp = 0,l,r;
IO::scan_d(p);
ll pos = 1;
for(ll i=1;i<=31;i++) {
l = upper_bound(arr+1,arr+n+1,pos) - arr;
if(l > n) break;
r = upper_bound(arr+1,arr+n+1,pos*p) - arr;
///if(l == n) continue;
if(l == n+1) continue;
///temp = (temp + ((sum[r-1] - sum[l-1])/i)) % mod;
temp = (temp + sum[i][r-1] - sum[i][l-1]) % mod;
temp = (temp + mod) % mod;
pos *= p;
}
ans = (ans + temp * j % mod) % mod;
}
IO::println(ans);
}
return 0;
}