多操作线段树

Transformation

 

Yuanfang is puzzled with the question below: 
There are n integers, a 1, a 2, …, 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

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <stack>
#include <cmath>
#include <vector>
using namespace std;
const int NUM=100005;
const int mod=10007;
struct point
{
    int l,r,w1,w2,w3,f1,f2,f3;
}tree[NUM*4];
int LL,RR,val,ans;
void build(int k,int LL,int RR)
{
    tree[k].l=LL,tree[k].r=RR;
    tree[k].f1=tree[k].f3=tree[k].w1=tree[k].w2=tree[k].w3=0;
    tree[k].f2=1;
    if(tree[k].l==tree[k].r){
        return;
    }
    int bet=(tree[k].l+tree[k].r)/2;
    build(k*2,LL,bet);
    build(k*2+1,bet+1,RR);
}
int aaa(int a,int b)
{
    int bbb=1;
    for(int i=0;i<b;i++){
        bbb=bbb*a%mod;
    }
    return bbb;
}
void down(int k)
{
    if(tree[k].l==tree[k].r)
        return;
    if(tree[k].f3){
        tree[k*2].f3=tree[k*2+1].f3=tree[k].f3;
        tree[k*2].w1=((tree[k*2].r-tree[k*2].l+1)*tree[k].f3)%mod;
        tree[k*2+1].w1=((tree[k*2+1].r-tree[k*2+1].l+1)*tree[k].f3)%mod;
        tree[k*2].w2=((tree[k*2].r-tree[k*2].l+1)*aaa(tree[k].f3,2))%mod;
        tree[k*2+1].w2=((tree[k*2+1].r-tree[k*2+1].l+1)*aaa(tree[k].f3,2))%mod;
        tree[k*2].w3=((tree[k*2].r-tree[k*2].l+1)*aaa(tree[k].f3,3))%mod;
        tree[k*2+1].w3=((tree[k*2+1].r-tree[k*2+1].l+1)*aaa(tree[k].f3,3))%mod;
        tree[k].f3=tree[k*2].f1=tree[k*2+1].f1=0;
        tree[k*2].f2=tree[k*2+1].f2=1;
    }
    if(tree[k].f2!=1){
        if(tree[k*2].f3) down(k*2);
        if(tree[k*2+1].f3) down(k*2+1);
        tree[k*2].f2=(tree[k*2].f2*tree[k].f2)%mod;
        tree[k*2+1].f2=(tree[k*2+1].f2*tree[k].f2)%mod;
        tree[k*2].w1=(tree[k*2].w1*tree[k].f2)%mod;
        tree[k*2+1].w1=(tree[k*2+1].w1*tree[k].f2)%mod;
        tree[k*2].w2=(tree[k*2].w2*aaa(tree[k].f2,2))%mod;
        tree[k*2+1].w2=(tree[k*2+1].w2*aaa(tree[k].f2,2))%mod;
        tree[k*2].w3=(tree[k*2].w3*aaa(tree[k].f2,3))%mod;
        tree[k*2+1].w3=(tree[k*2+1].w3*aaa(tree[k].f2,3))%mod;
        tree[k*2].f1=(tree[k*2].f1*tree[k].f2)%mod;
        tree[k*2+1].f1=(tree[k*2+1].f1*tree[k].f2)%mod;
        tree[k].f2=1;
    }
    if(tree[k].f1){
        if(tree[k*2].f3||tree[k*2].f2!=1) down(k*2);
        if(tree[k*2+1].f3||tree[k*2+1].f2!=1) down(k*2+1);
        tree[k*2].f1=(tree[k*2].f1+tree[k].f1)%mod;
        tree[k*2+1].f1=(tree[k*2+1].f1+tree[k].f1)%mod;
        tree[k*2].w3=(tree[k*2].w3+3*tree[k*2].w2*tree[k].f1+3*tree[k*2].w1*aaa(tree[k].f1,2)+(tree[k*2].r-tree[k*2].l+1)*aaa(tree[k].f1,3))%mod;
        tree[k*2+1].w3=(tree[k*2+1].w3+3*tree[k*2+1].w2*tree[k].f1+3*tree[k*2+1].w1*aaa(tree[k].f1,2)+(tree[k*2+1].r-tree[k*2+1].l+1)*aaa(tree[k].f1,3))%mod;
        tree[k*2].w2=(tree[k*2].w2+2*tree[k*2].w1*tree[k].f1+(tree[k*2].r-tree[k*2].l+1)*aaa(tree[k].f1,2))%mod;
        tree[k*2+1].w2=(tree[k*2+1].w2+2*tree[k*2+1].w1*tree[k].f1+(tree[k*2+1].r-tree[k*2+1].l+1)*aaa(tree[k].f1,2))%mod;
        tree[k*2].w1=((tree[k*2].r-tree[k*2].l+1)*tree[k].f1+tree[k*2].w1)%mod;
        tree[k*2+1].w1=((tree[k*2+1].r-tree[k*2+1].l+1)*tree[k].f1+tree[k*2+1].w1)%mod;
        tree[k].f1=0;
    }
}
void change(int k,int ch)
{
    if(LL<=tree[k].l&&RR>=tree[k].r){
        down(k);
        if(ch==1){
            tree[k].f1=(val+tree[k].f1)%mod;
            tree[k].w3=(tree[k].w3+3*tree[k].w2*val+3*tree[k].w1*aaa(val,2)+(tree[k].r-tree[k].l+1)*aaa(val,3))%mod;
            tree[k].w2=(tree[k].w2+2*tree[k].w1*val+(tree[k].r-tree[k].l+1)*aaa(val,2))%mod;
            tree[k].w1=((tree[k].r-tree[k].l+1)*val+tree[k].w1)%mod;
            return;
        }
        if(ch==2){
            tree[k].f2=(val*tree[k].f2)%mod;
            tree[k].w1=(tree[k].w1*val)%mod;
            tree[k].w2=(tree[k].w2*aaa(val,2))%mod;
            tree[k].w3=(tree[k].w3*aaa(val,3))%mod;
            return;
        }
        if(ch==3){
            tree[k].f3=val;
            tree[k].w1=(tree[k].r-tree[k].l+1)*val%mod;
            tree[k].w2=(tree[k].r-tree[k].l+1)*aaa(val,2)%mod;
            tree[k].w3=(tree[k].r-tree[k].l+1)*aaa(val,3)%mod;
            return;
        }
    }
    if(tree[k].f1||tree[k].f2!=1||tree[k].f3) down(k);
    int bet=(tree[k].l+tree[k].r)/2;
    if(LL<=bet) change(k*2,ch);
    if(RR>bet) change(k*2+1,ch);
    tree[k].w1=(tree[k*2].w1+tree[k*2+1].w1)%mod;
    tree[k].w2=(tree[k*2].w2+tree[k*2+1].w2)%mod;
    tree[k].w3=(tree[k*2].w3+tree[k*2+1].w3)%mod;
}
void ask(int k,int ch)
{
    if(LL<=tree[k].l&&RR>=tree[k].r){
        if(ch==1)
            ans=(tree[k].w1+ans)%mod;
        if(ch==2)
            ans=(tree[k].w2+ans)%mod;
        if(ch==3)
            ans=(tree[k].w3+ans)%mod;
        return;
    }
    if(tree[k].f1||tree[k].f2!=1||tree[k].f3) down(k);
    int bet=(tree[k].l+tree[k].r)/2;
    if(LL<=bet) ask(k*2,ch);
    if(RR>bet) ask(k*2+1,ch);
}
int main()
{
    int n,m;
    while(scanf("%d %d",&n,&m)!=-1){
        if(n==0&&m==0)
            break;
        build(1,1,n);
        int ch;
        for(int i=0;i<m;i++){
            scanf("%d %d %d %d",&ch,&LL,&RR,&val);
            if(ch==4){
                ans=0;
                ask(1,val);
                printf("%d\n",ans);
            }
            else{
                change(1,ch);
            }
        }
    }
    return 0;
}