zoukankan      html  css  js  c++  java
  • 浙江省赛 ZOJ4029

    Now Loading!!!
    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

    这题前缀和+二分

    我们发现分母只有1-30;

    我们构造一个sum/i(i-30)

    二分查找 a^(i -1)<p<a^(i)  分母 就是 i 

    你就把a[i]分成30段。

    用flag 去保存位子。

    每一段就是 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
    ",(sum+mod)%mod);
        }
        return 0;
    }
  • 相关阅读:
    《人月神话》阅读笔记02
    《人月神话》阅读笔记01
    第十四周学习进度条
    我们做的作品 请大家多多支持我们
    Beta阶段项目总结
    Alpha阶段项目总结
    Alpha版总结会议
    站立会议10(第二次冲刺)
    站立会议09(第二次冲刺)
    站立会议08(第二次冲刺)
  • 原文地址:https://www.cnblogs.com/2014slx/p/9007556.html
Copyright © 2011-2022 走看看