hdu 4578 Transformation(线段树)

Transformation

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 3084    Accepted Submission(s): 749

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

 

 

写得比较恶心。。因为 p <= 3 ， 所以处理好3个sum值 。

push_down要写得优美才可以过

至于，过程中那些 + 与 * 的操作的话只是需要把 n 方公式分解好就可以了。

 

比如 说 （ c * a + b ）^ 3 = (c*a)^3 + 3*(c*a)^2*b + 3*(a*c)*b^2 + b^3 .

　　　　那么平方 ， 一次的操作也是这么进行

 

至于操作3的话就是直接把 操作1跟操作2的lazy清空掉就可以了

 

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstring>
#include <map>
#include <queue>
using namespace std;
typedef long long LL ;
typedef pair<int,int> pii ;
#define X first
#define Y second
#define root 1,n,1
#define lr rt<<1
#define rr rt<<1|1
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int N = 200010;
const int mod = 10007;

int sum1[N<<2] , sum2[N<<2] , sum3[N<<2] , lazy1[N<<2] , lazy2[N<<2] , lazy3[N<<2];
int n , m ;
void build( int l , int r , int rt ) {
    sum1[rt] = sum2[rt] = sum3[rt] = 0 ;
    lazy1[rt] = lazy3[rt] = 0 ;   // add  and clean
    lazy2[rt] = 1 ;                 // muti
    if( l == r ) return ;
    int mid = (l+r)>>1;
    build(lson) , build(rson);
}

void Up( int rt ) {
    sum1[rt] = ( sum1[lr] + sum1[rr] ) % mod ;
    sum2[rt] = ( sum2[lr] + sum2[rr] ) % mod ;
    sum3[rt] = ( sum3[lr] + sum3[rr] ) % mod ;
}

void Down( int l , int r , int rt ) {
    if( l == r ) return ;
    int mid = (l+r)>>1;
    if( lazy3[rt] != 0 ) {
        lazy3[lr] = lazy3[rr] = lazy3[rt] ;
        lazy1[lr] = lazy1[rr] = 0 ;
        lazy2[lr] = lazy2[rr] = 1 ;
        sum1[lr] = ( mid - l + 1 ) * lazy3[rt] % mod ;
        sum2[lr] = ( mid - l + 1 ) * lazy3[rt] % mod * lazy3[rt] % mod ;
        sum3[lr] = ( mid - l + 1 ) * lazy3[rt] % mod * lazy3[rt] % mod * lazy3[rt] % mod ;
        sum1[rr] = ( r - mid ) * lazy3[rt] % mod ;
        sum2[rr] = ( r - mid ) * lazy3[rt] % mod * lazy3[rt] % mod ;
        sum3[rr] = ( r - mid ) * lazy3[rt] % mod * lazy3[rt] % mod * lazy3[rt] % mod ;
        lazy3[rt] = 0 ;
    }
    if( lazy1[rt] != 0 || lazy2[rt] != 1 ) {
        lazy1[lr] = ( lazy1[lr] * lazy2[rt] % mod + lazy1[rt] ) % mod ;
        lazy2[lr] = lazy2[lr] * lazy2[rt] % mod ;
        sum3[lr] = ( lazy2[rt] * lazy2[rt] % mod * lazy2[rt] % mod * sum3[lr] % mod
                    + 3 * lazy2[rt] % mod * lazy2[rt] % mod * sum2[lr] % mod * lazy1[rt] % mod +
                    + 3 * lazy2[rt] % mod * sum1[lr] % mod * lazy1[rt] % mod * lazy1[rt] % mod
                    + ( mid - l + 1 ) * lazy1[rt] % mod * lazy1[rt] % mod * lazy1[rt] % mod
                    ) % mod ;
        sum2[lr] = ( lazy2[rt] * lazy2[rt] % mod * sum2[lr] % mod
                    + 2 * lazy1[rt] % mod * lazy2[rt] % mod * sum1[lr] % mod
                    + ( mid - l + 1 ) * lazy1[rt] % mod * lazy1[rt] % mod ) % mod ;
        sum1[lr] = ( sum1[lr] * lazy2[rt] % mod + lazy1[rt]*( mid - l + 1 ) % mod ) % mod;


        lazy1[rr] = ( lazy1[rr] * lazy2[rt] % mod + lazy1[rt] ) % mod ;
        lazy2[rr] = lazy2[rr] * lazy2[rt] % mod ;
        sum3[rr] = ( lazy2[rt] * lazy2[rt] % mod * lazy2[rt] % mod * sum3[rr] % mod
                    + 3 * lazy2[rt] % mod * lazy2[rt] % mod * sum2[rr] % mod * lazy1[rt] % mod +
                    + 3 * lazy2[rt] % mod * sum1[rr] % mod * lazy1[rt] % mod * lazy1[rt] % mod
                    + ( r - mid  ) * lazy1[rt] % mod * lazy1[rt] % mod * lazy1[rt] % mod
                    ) % mod ;
        sum2[rr] = ( lazy2[rt] * lazy2[rt] % mod * sum2[rr] % mod
                    + 2 * lazy1[rt] % mod * lazy2[rt] % mod * sum1[rr] % mod
                    + ( r - mid ) * lazy1[rt] % mod * lazy1[rt] % mod ) % mod ;
        sum1[rr] = ( sum1[rr] * lazy2[rt] % mod + lazy1[rt]*( r - mid ) % mod ) % mod;
        lazy1[rt] = 0;  lazy2[rt] = 1;
        }
}

void update( int l , int r , int rt , int L , int R , int c , int op ) {
    if( l == L && r == R ) {                // suppose lazy1 = lazy2 = 0 ; op1
        c %= mod ;
        if( op == 1 ) {
            lazy1[rt] = ( c + lazy1[rt] ) % mod;
            sum3[rt] = ( sum3[rt]
                       + 3 * sum2[rt] % mod * c % mod
                       + 3 * sum1[rt] % mod * c % mod * c % mod
                       + c * c % mod * c % mod * ( r - l + 1 ) % mod
                       ) % mod;
            sum2[rt] = ( sum2[rt]
                        + 2 * sum1[rt] % mod * c % mod
                        + c * c % mod * ( r - l + 1 ) % mod
                        ) % mod ;
            sum1[rt] = ( sum1[rt] + ( r - l + 1 ) * c % mod ) % mod;
        }
        else if( op == 2 ) {
            lazy1[rt] = lazy1[rt] * c % mod ;
            lazy2[rt] = lazy2[rt] * c % mod ;
            sum1[rt] = sum1[rt] * c % mod ;
            sum2[rt] = sum2[rt] * c % mod * c % mod ;
            sum3[rt] = sum3[rt] * c % mod * c % mod * c % mod ;
        }
        else {
            lazy1[rt] = 0  ; lazy2[rt] = 1 ; lazy3[rt] = c % mod ;
            sum1[rt] = (r-l+1) * c % mod ;
            sum2[rt] = (r-l+1) * c % mod * c % mod;
            sum3[rt] = (r-l+1) * c % mod * c % mod *c % mod;
        }
        return ;
    }
    Down(l,r,rt);
    int mid = (l+r)>>1;
    if( R <= mid ) update(lson,L,R,c,op);
    else if( L > mid ) update(rson,L,R,c,op);
    else update(lson,L,mid,c,op),update(rson,mid+1,R,c,op);
    Up(rt);
}

int query( int l , int r , int rt , int L , int R , int c ) {
    if( l == L && r == R ) {
        if( c == 1 ) return sum1[rt] ;
        else if( c == 2 ) return sum2[rt];
        else return sum3[rt];
    }
    Down(l,r,rt);
    int mid = (l+r)>>1;
    if( R <= mid ) return query(lson,L,R,c);
    else if( L > mid ) return query(rson,L,R,c);
    else return (query(lson,L,mid,c)+query(rson,mid+1,R,c))%mod;
}

int main()
{
    #ifdef LOCAL
       freopen("in.txt","r",stdin);
       //freopen("out.txt","w",stdout);
    #endif // LOCAL
    while( ~scanf("%d%d",&n,&m ) ) {
        if( n == 0 && m == 0 ) break ;
        build( root ) ;
        int op , x , y , c ;
        while( m-- ) {
            scanf("%d%d%d%d",&op,&x,&y,&c);
            if( op != 4 ) update(root,x,y,c,op);
            else printf("%d
",query(root,x,y,c));
        }
    }
}