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
分母范围很小，暴力分母存一个前缀和
每次加上就行了
转自点击打开链接
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <math.h>
#include <cstring>
#include <string>
#include <queue>
#include <deque>
#include <stack>
#include <stdlib.h>
#include <list>
#include <map>
#include <utility>
#include <set>
#include <bitset>
#include <vector>
#define pi acos(-1.0)
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fLL
#define ms(a,b) memset(a,b,sizeof(a))
using namespace std;
typedef long long ll;
int min3(int a,int b,int c){return min(min(a,b),c);}
int max3(int a,int b,int c){return max(max(a,b),c);}
int gcd(int x, int y){if(y==0)return x;return gcd(y, x%y);}
inline int read(){int x=0,f=1;char ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();return x*f;}
const int N=1e6+5;
const int INF=0x3f3f3f3f;
const int mod=1e9;
int n,m;
int a[500010],p;
int sum[32][500010];
ll qpow(ll n,ll m)
{
ll ans=1;
while(m){
if(m&1)ans*=n;
n*=n;
m>>=1;
if(ans>20001001000)return -1;
}
return ans;
}
int main()
{
int T;cin>>T;
while(T--)
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)scanf("%d",&a[i]);
sort(a+1,a+1+n);
for(int i=1;i<=31;i++) //pre sum of a/log
{
sum[i][0]=0;
for(int j=1;j<=n;j++)
sum[i][j]=(sum[i][j-1]+a[j]/i)%mod;
}
ll ans=0;
for(int k=1;k<=m;k++)
{
scanf("%d",&p);
ll res=0;
for(int i=1;i<=31;i++) //log p ai
{
ll down=qpow(p,i-1); //> it
ll up=qpow(p,i); // <=it
if(down==-1)break;
int l=1,r=n+1,mid;
if(down!=-1)
while(l<r){
mid=(l+r)>>1;
if(a[mid]>down)r=mid;
else l=mid+1;
}
int L=r;
l=1,r=n+1;
if(up!=-1)
while(l<r){
mid=(l+r)>>1;
if(a[mid]>up)r=mid;
else l=mid+1;
}
int R=r-1;
if(L>n)break;
if(L>R)continue;
res=(res+sum[i][R]-sum[i][L-1])%mod;
res=(res+mod)%mod;
}
ans=(ans+res*k%mod)%mod;
}
printf("%lld\n",ans);
}
}