hdu 4578 线段树(标记处理)

Transformation

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 4095    Accepted Submission(s): 1008

Problem Description

Yuanfang is puzzled with the question below: 
There are n integers, a₁, a₂, …, a_n. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between a_x and a_y inclusive. In other words, do transformation a_k<---a_k+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between a_x and a_y inclusive. In other words, do transformation a_k<---a_k×c, k = x,x+1,…,y.
Operation 3: Change the numbers between a_x and a_y to c, inclusive. In other words, do transformation a_k<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between a_x and a_y inclusive. In other words, get the result of a_x^p+a_x+1^p+…+a_y ^p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him. 

 

Input

There are no more than 10 test cases.
For each case, the first line contains two numbers n and m, meaning that there are n integers and m operations. 1 <= n, m <= 100,000.
Each the following m lines contains an operation. Operation 1 to 3 is in this format: "1 x y c" or "2 x y c" or "3 x y c". Operation 4 is in this format: "4 x y p". (1 <= x <= y <= n, 1 <= c <= 10,000, 1 <= p <= 3)
The input ends with 0 0.

 

Output

For each operation 4, output a single integer in one line representing the result. The answer may be quite large. You just need to calculate the remainder of the answer when divided by 10007.

 

Sample Input

5 5
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0

Sample Output

307
7489

/*
hdu 4578 线段树(标记处理)

给你n个初始化为0的数进行以下操作：
1 x y c 给[x,y]上的数全加上c     add
2 x y c 给[x,y]上的数全乘上c     mult
3 x y c 将[x,y]上面的数全置为c   same
4 x y c 查询[x,y]上所有数的c次方的和,然后对10007取模

首先我们可以发现 加法和乘法都无法直接维护我们想要的到的立方和,但对于same而言
sum = (r-l+1)*(tree[i].same^p).
如果每次查询我们都查找到单点,有极大的可能TLE。所以考虑查询的时候直接查找same标记,而且p也很小。
然后就是如何处理add,mult,same这三个标记的冲突.
就是same而言,更新到一个区间，那么先前这个区间上的所有标记都会作废
对于add和mult很明显会冲突,到后面你并不能知道是先处理add还是mult.所以
add和mult不能同时共处一个区间,而且先前到达的标记要先更新下去.

于是对于add和mult分3种情况:  //就add而言
1.如果当前区间有same,那愉快地更新same就好了
2.如果当前区间有mult,那先对当前区间进行update_down,把mult标记先更新下去
3.如果只有add这个标记,那么更新一下即可

感觉就标记下放这方面,主要是注意标记相互之间的影响。
hhh-2016-04-04 09:41:07
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <functional>
using namespace std;
#define lson  (i<<1)
#define rson  ((i<<1)|1)
typedef long long ll;
const int mod = 10007;
const int maxn = 100050;
struct node
{
    int l,r;
    ll mult,add,same;
    int mid()
    {
        return (l+r)>>1;
    }
    int len()
    {
        return (r-l+1) ;
    }
} tree[maxn<<2];

void update_up(int i)
{
}

void build(int i,int l,int r)
{
    tree[i].l = l,tree[i].r = r;
    tree[i].mult=1;
    tree[i].add=0;
    tree[i].same = -1;
    if(l == r)
    {
        tree[i].same = 0;
        return ;
    }
    build(lson,l,tree[i].mid());
    build(rson,tree[i].mid()+1,r);
    update_up(i);
}

void update_down(int i)
{
    if(tree[i].same != -1)
    {
        tree[lson].add = tree[rson].add  = 0;
        tree[lson].mult= tree[rson].mult = 1;
        tree[lson].same = tree[rson].same = tree[i].same;
        tree[i].same = -1;
    }
    if(tree[i].add)
    {
        if(tree[lson].same != -1)
            tree[lson].same = (tree[lson].same+tree[i].add)%mod;
        else if(tree[lson].mult > 1)
        {
            update_down(lson);
            tree[lson].add = tree[i].add;
        }
        else
            tree[lson].add = (tree[lson].add+tree[i].add)%mod;
        if(tree[rson].same != -1)
            tree[rson].same = (tree[rson].same+tree[i].add)%mod;
        else if(tree[rson].mult > 1)
        {
            update_down(rson);
            tree[rson].add = tree[i].add;
        }
        else
            tree[rson].add = (tree[rson].add+tree[i].add)%mod;
        tree[i].add = 0;
    }
    if(tree[i].mult > 1)
    {
        if(tree[lson].same != -1)
            tree[lson].same = (tree[lson].same*tree[i].mult)%mod;
        else if(tree[lson].add)
        {
            update_down(lson);
            tree[lson].mult = tree[i].mult;
        }
        else
            tree[lson].mult = (tree[lson].mult*tree[i].mult)%mod;
        if(tree[rson].same != -1)
            tree[rson].same = (tree[rson].same*tree[i].mult)%mod;
        else if(tree[rson].add)
        {
            update_down(rson);
            tree[rson].mult = tree[i].mult;
        }
        else
            tree[rson].mult = (tree[rson].mult*tree[i].mult)%mod;
        tree[i].mult = 1;
    }
}

void update(int i,int l,int r,int flag,ll val)
{
    if(tree[i].l >= l && tree[i].r <= r)
    {
        if(flag == 1)
        {
            if(tree[i].same != -1)
                tree[i].same = (tree[i].same+val)%mod;
            else if(tree[i].mult > 1)
            {
                update_down(i);
                tree[i].add = val;
            }
            else
                tree[i].add =(tree[i].add+val)%mod;
        }
        else if(flag == 2)
        {
            if(tree[i].same != -1)
                tree[i].same = (tree[i].same*val)%mod;
            else if(tree[i].add)
            {
                update_down(i);
                tree[i].mult = val;
            }
            else
                tree[i].mult = (tree[i].mult * val) %mod;
        }
        else if(flag == 3)
        {
            tree[i].same = val;
            tree[i].same %= mod;
            tree[i].add = 0;
            tree[i].mult = 1;
        }
        return ;
    }
    int mid = tree[i].mid();
    update_down(i);
    if(l <= mid)
        update(lson,l,r,flag,val);
    if(r > mid)
        update(rson,l,r,flag,val);
    update_up(i);
}

ll query(int i,int l,int r,int p)
{
    if(tree[i].l == tree[i].r)
    {
        ll ans = 1;
        for(int j =1; j <= p; j++)
            ans =(ll)(ans*tree[i].same)%mod;
        return ans%mod;
    }
    if(tree[i].l >= l && tree[i].r <= r && tree[i].same != -1)
    {
        ll ans = 1;
        for(int j =1; j <= p; j++)
            ans =(ll)(ans*tree[i].same)%mod;
        ans = (ll)ans*(tree[i].len()%mod)%mod;
        return ans%mod;
    }
    ll all = 0;
    update_down(i);
    int mid = tree[i].mid();
    if(l <= mid)
        all =(all+query(lson,l,r,p))%mod;
    if(r > mid)
        all = (all+query(rson,l,r,p))%mod;
    return all;
}

int main()
{
    int t,n,m;
    while(scanf("%d%d",&n,&m) && n && m)
    {
        build(1,1,n);
        for(int i = 1; i <= m; i++)
        {
            int op,x,y;
            ll c;
            scanf("%d%d%d%I64d",&op,&x,&y,&c);
            if(op <= 3)
                update(1,x,y,op,c);
            else
                printf("%I64d\n",query(1,x,y,c));
        }
    }
    return 0;
}