hdu 5828 Rikka with Sequence 线段树

Rikka with Sequence

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=5828

Description

As we know, Rikka is poor at math. Yuta is worrying about this situation, so he gives Rikka some math tasks to practice. There is one of them:

Yuta has an array A with n numbers. Then he makes m operations on it.

There are three type of operations:

1 l r x : For each i in [l,r], change A[i] to A[i]+x

2 l r : For each i in [l,r], change A[i] to ⌊A−−√[i]⌋

3 l r : Yuta wants Rikka to sum up A[i] for all i in [l,r]

It is too difficult for Rikka. Can you help her?

Input

The first line contains a number t(1<=t<=100), the number of the testcases. And there are no more than 5 testcases with n>1000.

For each testcase, the first line contains two numbers n,m(1<=n,m<=100000). The second line contains n numbers A[1]~A[n]. Then m lines follow, each line describe an operation.

It is guaranteed that 1<=A[i],x<=100000.

Output

For each operation of type 3, print a lines contains one number -- the answer of the query.

Sample Input

1

5 5

1 2 3 4 5

1 3 5 2

2 1 4

3 2 4

2 3 5

3 1 5

Sample Output

5

6

Hint

题意

给你n个数，m个操作

一共有三类操作

区间开根号

区间加

区间求和

题解：

线段树，区间开根号最多十几次就会变成1，我们把相同值的数合并在一起去更新就好了

区间加，就正常做，和普通的线段树一样去做就好了。

区间求和也正常做……

其实就比普通的线段树多一个lazy，表示左右是否相同。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e5+7;
long long n;
int a[maxn];
int to[2000005];
typedef long long ll;
typedef long long SgTreeDataType;

const int BUF=40000000;
char Buf[BUF],*buf=Buf;
const int OUT=20000000;
char Out[OUT],*ou=Out;int Outn[30],Outcnt;
inline void write(int x){
  if(!x)*ou++=48;
  else{
    for(Outcnt=0;x;x/=10)Outn[++Outcnt]=x%10+48;
    while(Outcnt)*ou++=Outn[Outcnt--];
  }
}
inline void writell(ll x){
  if(!x)*ou++=48;
  else{
    for(Outcnt=0;x;x/=10)Outn[++Outcnt]=x%10+48;
    while(Outcnt)*ou++=Outn[Outcnt--];
  }
}
inline void writechar(char x){*ou++=x;}
inline void writeln(){*ou++='\n';}
inline void read(int&a){for(a=0;*buf<48;buf++);while(*buf>47)a=a*10+*buf++-48;}

inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}

inline int TransForm( long long x ){
    if( x <= 2000000 ) return to[x];
    return sqrt( x );
}

struct Sgtree{
    struct treenode{
        int l , r ;
        long long f , lzy , sum ;

        void Update( long long x ){
            lzy += x;
            sum += (r - l + 1) * x;
            if( ~f ) f += x;
        }

        void SetUp( long long x ){
            f = x;
            sum = x * ( r - l + 1 );
        }

        void SqrtUp(){
            f = TransForm( f );
            sum = f * ( r - l + 1 );
        }

    }tree[maxn << 2];

    void Push_Up( int o ){
        tree[o].sum = tree[o << 1].sum + tree[o << 1 | 1].sum;
        if( tree[o << 1].f == tree[o << 1 | 1].f ) tree[o].f = tree[o << 1].f;
        else tree[o].f = -1;
    }

    void ReleaseLabel( int o ){
        if( tree[o].lzy ){
            tree[o << 1].Update( tree[o].lzy );
            tree[o << 1 | 1].Update( tree[o].lzy );
            tree[o].lzy = 0;
        }
        if( ~tree[o].f ){
            tree[o << 1].SetUp( tree[o].f );
            tree[o << 1 | 1].SetUp( tree[o].f );
        }
    }

    void Build( int l , int r , int o ){
        tree[o].l = l , tree[o].r = r , tree[o].lzy = tree[o].sum = 0;
        if( r > l ){
            int mid = l + r >> 1;
            Build( l , mid , o << 1 );
            Build( mid + 1 , r , o << 1 | 1 );
            Push_Up( o );
        }else tree[o].f = tree[o].sum = a[l];
    }

    void Add( int ql , int qr , int v , int o ){
        int l = tree[o].l , r = tree[o].r;
        if( ql <= l && r <= qr ) tree[o].Update( v );
        else{
            int mid = l + r >> 1;
            ReleaseLabel( o );
            if( ql <= mid ) Add( ql , qr , v , o << 1 );
            if( qr > mid ) Add( ql , qr , v , o << 1 | 1 );
            Push_Up( o );
        }
    }

    void DFS( int o ){
        if( ~tree[o].f ){
            tree[o].SqrtUp( );
        }else{
            ReleaseLabel( o );
            DFS( o << 1 );
            DFS( o << 1 | 1 );
            Push_Up( o );
        }
    }

    void OpeSqrt( int ql , int qr , int o ){
        int l = tree[o].l , r = tree[o].r;
        if( ql <= l && r <= qr ) DFS( o );
        else{
            int mid = l + r >> 1;
            ReleaseLabel( o );
            if( ql <= mid ) OpeSqrt( ql , qr , o << 1 );
            if( qr > mid ) OpeSqrt( ql , qr , o << 1 | 1 );
            Push_Up( o );
        }
    }

    long long Ask( int ql , int qr , int o ){
        int l = tree[o].l , r = tree[o].r;
        if( ql <= l && r <= qr ) return tree[o].sum;
        else{
            int mid = l + r >> 1;
            ReleaseLabel( o );
            long long Result = 0;
            if( ql <= mid ) Result += Ask( ql , qr , o << 1 );
            if( qr > mid ) Result += Ask( ql , qr , o << 1 | 1 );
            Push_Up( o );
            return Result;
        }
    }

}Sgtree;

int main()
{
    fread(Buf,1,BUF,stdin);
    for(int i=0;i<=2000000;i++)
        to[i]=sqrt(i);
    int t;
    read(t);
    while(t--){
        int n,m;
        read(n);read(m);
        for(int i=1;i<=n;i++) read(a[i]);
        Sgtree.Build( 1 , n , 1 );
        while(m--)
        {
            int op,x,y,z;
            read(op),read(x),read(y);
            if(op==2) Sgtree.OpeSqrt( x , y , 1 );
            if(op==1){
                read(z);
                Sgtree.Add( x , y , z , 1 );
            }
            if(op==3)writell(Sgtree.Ask(x,y,1)),writeln();
        }
    }
    fwrite(Out,1,ou-Out,stdout);
    return 0;
}