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

Now Loading!!!

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Time Limit: 1 Second       Memory Limit: 131072 KB

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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

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Author:  LIN, Xi
Source:  The 15th Zhejiang Provincial Collegiate Programming Contest Sponsored by TuSimple
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【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
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;
    }
    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>1;
                        if(a[mid]>down)r=mid;
                        else l=mid+1;
                    }
                int L=r;

                l=1,r=n+1;
                if(up!=-1)
                    while(l>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);
    }
}