zoukankan      html  css  js  c++  java
  • HDU4893:Wow! Such Sequence!(段树lazy)

    Problem Description
    Recently, Doge got a funny birthday present from his new friend, Protein Tiger from St. Beeze College. No, not cactuses. It's a mysterious blackbox.

    After some research, Doge found that the box is maintaining a sequence an of n numbers internally, initially all numbers are zero, and there are THREE "operations":

    1.Add d to the k-th number of the sequence.
    2.Query the sum of ai where l ≤ i ≤ r.
    3.Change ai to the nearest Fibonacci number, where l ≤ i ≤ r.
    4.Play sound "Chee-rio!", a bit useless.

    Let F0 = 1,F1 = 1,Fibonacci number Fn is defined as Fn = Fn - 1 + Fn - 2 for n ≥ 2.

    Nearest Fibonacci number of number x means the smallest Fn where |Fn - x| is also smallest.

    Doge doesn't believe the machine could respond each request in less than 10ms. Help Doge figure out the reason.
     

    Input
    Input contains several test cases, please process till EOF.
    For each test case, there will be one line containing two integers n, m.
    Next m lines, each line indicates a query:

    1 k d - "add"
    2 l r - "query sum"
    3 l r - "change to nearest Fibonacci"

    1 ≤ n ≤ 100000, 1 ≤ m ≤ 100000, |d| < 231, all queries will be valid.
     

    Output
    For each Type 2 ("query sum") operation, output one line containing an integer represent the answer of this query.
     

    Sample Input
    1 1 2 1 1 5 4 1 1 7 1 3 17 3 2 4 2 1 5
     

    Sample Output
    0 22
    #include <stdio.h>
    #include <string.h>
    #include <algorithm>
    using namespace std;
    #define ll __int64
    #define maxn 100005
    #define ls l,mid,2*i
    #define rs mid+1,r,2*i+1
    #define lson 2*i
    #define rson 2*i+1
    struct node
    {
        int l,r;
        ll e,f;//e为该区间的和,f为其近期的斐波那契数
        int flag,len;//flag,标记这个区间内是否斐波那契数,len为长度
    } a[maxn<<2];
    int n,m;
    ll f[90] = {1,1};
    
    ll pabs(ll a)
    {
        return a<0?-a:a;
    }
    
    void PushDown(int i,int m)
    {
        if(a[i].flag)
        {
            a[lson].flag = a[rson].flag = a[i].flag;
            a[lson].len = a[i].flag*(m-m>>1);
            a[rson].len = a[i].flag*(m>>1);
            a[lson].e = a[lson].f;
            a[rson].e = a[rson].f;
            a[i].flag = 0;
        }
    }
    
    void PushUp(int i)
    {
        a[i].e = a[lson].e+a[rson].e;
        a[i].f = a[lson].f+a[rson].f;
    }
    
    void init(int l,int r,int i)
    {
        a[i].flag = a[i].len = 0;
        a[i].l = l;
        a[i].r = r;
        a[i].e = 0;
        if(l == r)
        {
            a[i].f = 1;
            return;
        }
        int mid = (l+r)>>1;
        init(ls);
        init(rs);
        PushUp(i);
    }
    
    void add(int pos,int m,int l,int r,int i)
    {
        if(pos<l || pos>r) return ;
        if(l == r)
        {
            if(a[i].flag)
            {
                a[i].e = m+a[i].f;
                a[i].flag = 0;
                a[i].len = 0;
            }
            else a[i].e+=m;
            int p = lower_bound(f,f+80,a[i].e)-f;
            if(!p)
                a[i].f = 1;
            else if(pabs(a[i].e-f[p])<pabs(a[i].e-f[p-1]))
                a[i].f = f[p];
            else
                a[i].f = f[p-1];
            return ;
        }
        PushDown(i,r-l+1);
        int mid = (l+r)>>1;
        if(pos<=mid) add(pos,m,ls);
        else add(pos,m,rs);
        PushUp(i);
    }
    
    ll query(int L,int R,int l,int r,int i)
    {
        if(R<l || L>r) return 0;
        else if(L<=l && R>=r) return a[i].e;
        PushDown(i,r-l+1);
        ll ans = 0;
        int mid = (l+r)>>1;
        if(L<=mid)
            ans += query(L,R,ls);
        if(R>mid)
            ans += query(L,R,rs);
        return ans;
    }
    
    void change(int L,int R,int l,int r,int i)
    {
        if(R<l || L>r) return ;
        if(L<=l && R>=r)
        {
            a[i].e = a[i].f;
            a[i].flag = 1;
            a[i].len = r-l+1;
            return ;
        }
        PushDown(i,r-l+1);
        int mid = (l+r)>>1;
        if(L<=mid)
            change(L,R,ls);
        if(R>mid)
            change(L,R,rs);
        PushUp(i);
    }
    
    int main()
    {
        int i,j,x,k,d,l,r;
        for(i = 2; i<80; i++)
            f[i] = f[i-1]+f[i-2];
        while(~scanf("%d%d",&n,&m))
        {
            init(1,n,1);
            while(m--)
            {
                scanf("%d",&x);
                if(x == 1)
                {
                    scanf("%d%d",&k,&d);
                    add(k,d,1,n,1);
                }
                else
                {
                    scanf("%d%d",&l,&r);
                    if(x == 2)
                        printf("%I64d
    ",query(l,r,1,n,1));
                    else
                        change(l,r,1,n,1);
                }
            }
        }
    
        return 0;
    }
    



  • 相关阅读:
    keras系列︱迁移学习:利用InceptionV3进行fine-tuning及预测、完美案例(五)
    keras系列︱人脸表情分类与识别:opencv人脸检测+Keras情绪分类(四)
    keras系列︱图像多分类训练与利用bottleneck features进行微调(三)
    keras系列︱Application中五款已训练模型、VGG16框架(Sequential式、Model式)解读(二)
    将C++资源文件读取出来
    windows驱动程序中的预处理含义
    win10网上邻居看不到别的共享电脑怎么样办
    #pragma alloc_text 与 ALLOC_PRAGMA
    IoAllocateMdl,MmProbeAndLockPages的用法
    Composer三步曲:安装、使用、发布
  • 原文地址:https://www.cnblogs.com/mengfanrong/p/5040296.html
Copyright © 2011-2022 走看看