zoukankan      html  css  js  c++  java
  • hdu4893Wow! Such Sequence! (线段树)

    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

    Source

    Recommend
    hujie | We have carefully selected several similar problems for you: 4896 4895 4894 4892 4891
    解题:线段树结点含两个和,一个真正和,还有一个是3号操作的备用和。

    #include<iostream>
    #include<stdio.h>
    using namespace std;
    struct tree
    {
        int l,r,t;
        __int64 s,ts;
    }ss[100010*3];
    __int64 f[100]={1,1};
    void setit(int l,int r,int d)
    {
        ss[d].l=l;
        ss[d].r=r;
        ss[d].t=0;//推断子结点是否须要进行3号操作
        ss[d].s=0;//当前区间内的总和
        ss[d].ts=1;//当前区间进行3号操作的备用和
        if(l==r)
            return;
        setit(l,(l+r)/2,d*2);
        setit((l+r)/2+1,r,d*2+1);
        ss[d].ts=ss[d*2].ts+ss[d*2+1].ts;
    }
    __int64 doublekill(__int64 k)//查找与k最相近的Fibonacci
    {
        int l=0,r=77,mid;
        __int64 x1,x2;
        while(l<=r)
        {
            mid=(l+r)/2;
            if(f[mid]==k)return f[mid];
            if(f[mid]<k)l=mid+1;
            else r=mid-1;
        }
        x1=f[l]-k;
        if(x1<0)x1=-x1;
        if(l>0)
        {
            x2=f[l-1]-k;
            if(x2<0)x2=-x2;
            if(x2<=x1)return f[l-1];
        }
        return f[l];
    }
    void insert1(__int64 e,int k,int d)//点更新
    {
        int l,r,mid;
        l=ss[d].l;
        r=ss[d].r;
        mid=(l+r)/2;
        if(l==r)
        {
            if(ss[d].t)
            {
                ss[d].s=ss[d].ts; ss[d].t=0;
            }
            ss[d].s+=e; ss[d].ts=doublekill(ss[d].s);//同一时候更新备用
            return;
        }
        if(ss[d].t)//假设当前段须要进行3号操作
        {
            ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;
            ss[d*2].s=ss[d*2].ts;
            ss[d*2+1].s=ss[d*2+1].ts;
        }
        if(k>mid)insert1(e,k,d*2+1);
        else insert1(e,k,d*2);
        ss[d].s=ss[d*2].s+ss[d*2+1].s;
        ss[d].ts=ss[d*2].ts+ss[d*2+1].ts;
    }
    void insert2(int l,int r,int d)//区间进行3号操作
    {
        int ll,rr,mid;
        ll=ss[d].l;
        rr=ss[d].r;
        mid=(ll+rr)/2;
        if(ll>=l&&rr<=r)
        {
            ss[d].s=ss[d].ts; ss[d].t=1;//子结点须要3号操作更新
            return;
        }
        if(ss[d].t)
        {
            ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;
            ss[d*2].s=ss[d*2].ts;
            ss[d*2+1].s=ss[d*2+1].ts;
        }
        if(l<=mid)insert2(l,r,d*2);
        if(r>mid)insert2(l,r,d*2+1);
         ss[d].s=ss[d*2].s+ss[d*2+1].s;
    }
    __int64 find(int l,int r,int d)//求和
    {
        __int64 ll,rr,mid,s=0;
        ll=ss[d].l;
        rr=ss[d].r;
        mid=(ll+rr)/2;
        if(ll>=l&&rr<=r)
        {
            return ss[d].s;
        }
        if(ss[d].t)
        {
            ss[d*2].t=1; ss[d*2+1].t=1; ss[d].t=0;
            ss[d*2].s=ss[d*2].ts;
            ss[d*2+1].s=ss[d*2+1].ts;
        }
        if(l<=mid)s+=find(l,r,d*2);
        if(r>mid)s+=find(l,r,d*2+1);
        ss[d].s=ss[d*2].s+ss[d*2+1].s;
        return s;
    }
    int main (void)
    {
        int n,m,i,j,k,l;
        __int64 d;
        for(i=2;i<78;i++)
        f[i]=f[i-1]+f[i-2];
        while(scanf("%d%d",&n,&m)>0)
        {
            setit(1,n,1);
            while(m--)
            {
                scanf("%d%d",&j,&k);
                if(j==1)
                {
                    scanf("%I64d",&d);insert1(d,k,1);
                }
                else
                {
                    scanf("%d",&l);
                    if(j==2)printf("%I64d
    ",find(k,l,1));
                    else insert2(k,l,1);
                }
            }
        }
        return 0;
    }
    


  • 相关阅读:
    4、提取看似无用的委托变量,减少构造开销
    Cloud Foundry buildpack
    mysql中engine=innodb和engine=myisam的区别
    Maven中dependencyManagement的作用
    bean
    servlet
    web服务器 应用服务器区别 web框架
    Java和Python的Web开发
    spring mvc controller 高并发问题
    es 批量添加数据
  • 原文地址:https://www.cnblogs.com/lcchuguo/p/5264752.html
Copyright © 2011-2022 走看看