2018ACM浙江省赛 ZOJ 4029 Now Loading!!!（二分）

最新推荐文章于 2019-05-04 17:32:00 发布

转载最新推荐文章于 2019-05-04 17:32:00 发布 · 772 阅读

本文介绍了一个算法竞赛题目，重点在于如何高效地处理一系列查询请求，通过排序和预处理前缀和的方法来加速查询过程，特别关注了在特定范围内对整数集合的操作。

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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

1136645619

Author: LIN, Xi

translator Jiang Haonan

Source: The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple

Submit Status

【题意】

对于每一次询问，求上式的值

【分析】

把a排序

由于分母的范围很小 [2,30]，可以枚举；

对于每一次询问p，枚举分母 i 时，可以找出a中分母等于 i 的那一段，预处理前缀和，用于此时直接加。

复杂度n * log * log

【代码】


  #include<bits/stdc++.h>
using namespace std;
typedef long long ll;
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;//2^31=2147483648 约等于2e9
    }
    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++) // a/log 的前缀和
        {
            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);
    }
}