zoukankan      html  css  js  c++  java
  • ural Ambitious Experiment 树状数组

    During several decades, scientists from planet Nibiru are working to create an engine that would allow spacecrafts to fall into hyperspace and move there with superluminal velocity. To check whether their understanding of properties of hyperspace is right, scientists have developed the following experiment.
    A chain of n particles is placed in hyperspace. Positions of particles in the chain are numbered from 1 to n. Initially, ith particle has charge a i.
    According to the current theory, if particle number i got special radiation with power d, oscillations would spread by hyperspace and increase by d charge of particles with numbers i, 2 i, 3 i and so on (i.e. with numbers divisible by i).
    Using a special device, scientists can direct the radiation of the same power at a segment of adjacent particles. For example, suppose that initially there were 6 particles with zero charges, and scientists have sent radiation with power five to particles with numbers 2 and 3. Then charge of 2nd, 3rd, and 4th particles will increase to five, and charge of 6th particle will increase to ten (the oscillations will reach it twice). Charge of other particles won’t change.
    Charge of particles can’t change without impact of the device.
    During the experiment, the scientists plan to perform actions of the following types:
    1. Measure current charge of the particle number i.
    2. Direct radiation with power d at particles with numbers from l to r inclusive.
    Your program will be given a list of performed actions. For every action of the first type the program should output value of expected charge of the particle calculated in accordance with the current theory described above.
    If the expected charges of the particles coincide with charges measured during the experiment, it will turn out that scientists’ understanding of hyperspace is right, and they will be able to start building of the hyperdrives. Then inhabitants of Nibiru will finally meet their brothers from Earth in just a few years!

    Input

    The first line contains a single integer n — number of particles (1 ≤ n ≤ 3 · 10 5).
    The second line contains n integers a i separated by spaces — initial charges of the particles (0 ≤ a i ≤ 10 6).
    The third line contains a single integer q — number of actions in the experiment (1 ≤ q ≤ 3 · 10 5).
    Each of the following q lines contain two or four integers — a description of the next action in one of the following formats:
    • i — measure current charge of the particle number i (1 ≤ i ≤ n).
    • l r d — direct radiation with power d at particles with numbers from l to r inclusive (1 ≤ l ≤ r ≤ n, 0 ≤ d ≤ 106).

    Output

    For each query output the expected charge of the ith particle.

    Example

    inputoutput
    3
    1 2 3
    2
    2 1 3 5
    1 2
    
    12
    
    6
    1 2 1 4 5 6
    5
    2 2 4 2
    1 3
    1 4
    2 3 5 1
    1 5
    
    3
    8
    6
    这题花了三个小时没做出来,可能是最近有点脑残。。刚开始用树状数组做了tle,又改成线段树,发现线段树还是tle,最后又改成树状数组。。有一点技巧
    #include<map>
    #include<set>
    #include<cmath>
    #include<queue>
    #include<stack>
    #include<vector>
    #include<cstdio>
    #include<iomanip>
    #include<cstdlib>
    #include<cstring>
    #include<iostream>
    #include<algorithm>
    #define pi acos(-1)
    #define ll long long
    #define mod 1000000007
    
    using namespace std;
    
    const int N=300000+5,maxn=1000000+5,inf=1e9+5;
    
    ll a[N],m[N];
    void add(int i,int v)
    {
        while(i<=N){
            m[i]+=v;
            i+=i&(-i);
        }
    }
    ll sum(int i)
    {
        ll ans=0;
        while(i>0){
            ans+=m[i];
            i-=i&(-i);
        }
        return ans;
    }
    int main()
    {
        ios::sync_with_stdio(false);
        cin.tie(0);
        int n;
        cin>>n;
        memset(m,0,sizeof m);
        for(int i=1;i<=n;i++)cin>>a[i];
        int k;
        cin>>k;
        while(k--){
            int p,l,r,d;
            cin>>p;
            if(p==1)
            {
                cin>>d;
                ll ans=0;
                for(int i=1;i*i<=d;i++)
                {
                    if(d%i==0)
                    {
                        ans+=sum(i);
                        if(i*i!=d)ans+=sum(d/i);
                    }
                }
                cout<<ans+a[d]<<endl;
            }
            else
            {
                cin>>l>>r>>d;
                add(l,d);
                add(r+1,-d);
            }
        }
        return 0;
    }
    View Code
  • 相关阅读:
    hdu 5648 DZY Loves Math 组合数+深搜(子集法)
    hdu 5647 DZY Loves Connecting 树形DP
    hdu 4550 卡片游戏 贪心
    hdu 5646 DZY Loves Partition 二分+数学分析+递推
    hdu 2196 Computer 树形DP
    poj 2342 Anniversary party 树形DP入门
    Vijos P1003 等价表达式 随机数+单调栈
    【BZOJ】1044: [HAOI2008]木棍分割 二分+区间DP
    【BZOJ】1925: [Sdoi2010]地精部落 DP+滚动数组
    【BZOJ】1012: [JSOI2008]最大数maxnumber 树状数组求区间最值
  • 原文地址:https://www.cnblogs.com/acjiumeng/p/6857144.html
Copyright © 2011-2022 走看看