HDU 4578(线段树区间更新)

Transformation

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)

Total Submission(s): 3726    Accepted Submission(s): 900

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

 

//只有一个字 艹

//每个节点存是纯色的val 

#include <iostream>
#include <stdio.h>
#include <algorithm>
using namespace std;
const int maxn=100000+10;
struct Node
{
    __int64 l,r,col;  //纯色 0 不纯为1
    __int64 val;
    int mid()
    {
        return (l+r)/2;
    }
}tree[maxn*6];
__int64 Mod=10007;
__int64 n,m,ans=0;
void Buildtree(__int64 rt,__int64 l,__int64 r)
{
    tree[rt].l=l;
    tree[rt].r=r;
    tree[rt].val=tree[rt].col=0;
    if(l!=r)
    {
        Buildtree(2*rt,l,tree[rt].mid());
        Buildtree(2*rt+1,tree[rt].mid()+1,r);
    }
}
void Pushdown(__int64 rt)
{
        tree[2*rt].val=tree[2*rt+1].val=tree[rt].val;
        tree[2*rt].col=tree[2*rt+1].col=0;
}
void Update(__int64 rt,__int64 l,__int64 r,__int64 type,__int64 val)
{
    if(tree[rt].l==l&&tree[rt].r==r&&tree[rt].col==0)
    {
        if(type==1)
            tree[rt].val=(tree[rt].val+val)%10007;  //这里让我错了N多发 fuck...
        else if(type==2)
            tree[rt].val=(tree[rt].val*val)%10007;
        else
            tree[rt].val=val%10007;
        tree[rt].col=0;
        return;
    }
    if(tree[rt].col==0)
        Pushdown(rt);
    if(r<=tree[rt].mid())
        Update(2*rt,l,r,type,val);
    else if(l>tree[rt].mid())
        Update(2*rt+1,l,r,type,val);
    else
    {
        Update(2*rt,l,tree[rt].mid(),type,val);
        Update(2*rt+1,tree[rt].mid()+1,r,type,val);
    }
    tree[rt].col=(tree[rt*2].col||tree[2*rt+1].col||tree[2*rt].val!=tree[2*rt+1].val);  //回溯
    if(tree[2*rt].col==0&&tree[rt*2+1].col==0&&tree[2*rt].val==tree[rt*2+1].val)
    {
        tree[rt].val=tree[2*rt].val;
        tree[rt].col=0;
    }
}
void Query(__int64 rt,__int64 l,__int64 r,__int64 val)
{
    __int64 t=tree[rt].val;
    if(tree[rt].l==l&&tree[rt].r==r&&tree[rt].col==0)
    {
        if(val==1)
            ans=(ans+((tree[rt].r-tree[rt].l+1)*t))%10007;
        else if(val==2)
            ans=(ans+((tree[rt].r-tree[rt].l+1)*t%10007*t))%10007;
        else if(val==3)
            ans=(ans+((tree[rt].r-tree[rt].l+1)*t%10007*t%10007*t%10007))%10007;
        return;
    }
    if(tree[rt].col==0)
        Pushdown(rt);
    if(r<=tree[rt].mid())
        Query(2*rt,l,r,val);
    else if(l>tree[rt].mid())
        Query(2*rt+1,l,r,val);
    else
    {
        Query(2*rt,l,tree[rt].mid(),val);
        Query(2*rt+1,tree[rt].mid()+1,r,val);
    }
}
int main()
{
    while(~scanf("%I64d%I64d",&n,&m)&&(n+m))
    {
        Buildtree(1,1,n);
        while(m--)
        {
            __int64 type,l,r,val;
            scanf("%I64d%I64d%I64d%I64d",&type,&l,&r,&val);
            if(type<=3)
                Update(1,l,r,type,val);
            else
            {
                ans=0;
                Query(1,l,r,val);
                printf("%I64d\n",ans%10007);
            }
        }
    }
    return 0;
}