Zhejiang Province Competition ZOJ4029

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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
Author: LIN, Xi
Source: The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple
Submit    Status

This question prefix sum + two points

We find that the denominator is only 1-30;

We construct a sum/i(i-30)

Binary search a^(i -1)<p<a^(i) The denominator is i 

You divide a[i] into 30 segments.

Use flag to save the seat.

Each segment is k[j][flag[j]]-k[j][flag[j-1]]

#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn=500005;
const int mod=1e9;
int  a[maxn],p[maxn],k[35][maxn];
int n,m;
int main()
{
    int t,flag[maxn],cnt;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&n,&m);
        for(int i=1; i<=n; i++)
            scanf("%d",&a[i]);
        sort(a+1,a+n+1);
        for(int i=1; i<=m; i++)
            scanf("%d",&p[i]);
        for(int i=1; i<=30; i++)
        {
            k[i][0]=0;
            for(int j=1; j<=n; j++)
            {
                k[i][j]=(k[i][j-1]+(a[j]/i))%mod;
            }
        }
        ll sum=0,ans;
        ll temp;
        for(int i=1; i<=m; i++)
        {
            temp=1;
            cnt=0;
            ans=0;
            while(temp*p[i]<=a[n])
            {
                temp*=p[i];
                int l=1,r=n;
                while(l<=r)
                {
                    int mid=(l+r)/2;
                    if(a[mid]<=temp)
                        l=mid+1;
                    else
                        r=mid-1;
                }
                flag[++cnt]=r;
            }
            if(flag[cnt]<n)
                flag[++cnt]=n;
            for(int j=1; j<=cnt; j++)
            {
                ans = (ans+(k[j][flag[j]]-k[j][flag[j- 1 ]]))% mod;
            }
            sum=(sum+ans*i)%mod;
        }
        printf("%lld\n",(sum+mod)%mod);
    }
    return 0;
}