The 15th ZPCPC ||ZOJ Problem Set - 4029 Now Loading!!!【数论】

Now Loading!!!

Time Limit: 2000/1000 MS(Java/Others)    Memory Limit: 262144/131072 K (Java/Others)

Problem Description

DreamGrid has n integers a1,a2,a3...an. DreamGrid also has m queries, and each time he would like to know the value of

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

【题目链接】 link

【题意】

RT

【思路】

由于n，m比较大,说明时间复杂度最多再乘以log。

这里由于log(a[i])最大只有30，那么我们考虑去预处理出a[i]/k的前缀和(k为分母)

然后每次对一个读入的p，令分母确定为tmp,那么a[i]的范围应该是[pow(p,tmp-1)+1,pow(p,tmp)],利用二分找出两个位置，然后用一下前缀和即可。

时间复杂度O(n*k+mloglog)

#include <cstdio>
#include <bits/stdc++.h>
#include <cmath>
#include <map>
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)

typedef long long ll;
const int maxn = 500005;
const ll mod = 1e9;
const int INF = 1e9;
const double eps = 1e-6;

int n,m;
ll a[maxn];
ll sum[maxn][32];

int main()
{
    rush()
    {
        scanf("%d%d",&n,&m);
        for(int i=1;i<=n;i++) scanf("%lld",&a[i]);
        sort(a+1,a+1+n);
        for(int k=1;k<=30;k++)
        for(int i=1;i<=n;i++)
        {
            sum[i][k]=sum[i-1][k]+a[i]/k;
        }
        ll ans=0;
        for(int i=1;i<=m;i++)
        {
            ll x;
            scanf("%lld",&x);
            int k=1;
            ll cnt=0;
            int pos;
            ll up=x;
            for(int j=1;j<=n;j=pos+1)
            {
                pos=j-1;
                int l=j,r=n;
                while(l<=r)
                {
                    int mid=(l+r)/2;
                    if(a[mid]<=up)
                    {
                        pos=mid;
                        l=mid+1;
                    }
                    else r=mid-1;
                }
                cnt+=sum[pos][k]-sum[j-1][k];
                k++;
                up*=x;
            }
            ans+=cnt%mod*(ll)i%mod;
            ans%=mod;
        }
        printf("%lld\n",ans);
    }
}