HDU 4578 Transformation 线段树

Transformation

Time Limit: 15000/8000 MS (Java/Others)    Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 2374    Accepted Submission(s): 558

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 53 3 5 71 2 4 44 1 5 22 2 5 84 3 5 30 0

 

Sample Output

3077489

 

Source

2013ACM-ICPC杭州赛区全国邀请赛

  

传送门：HDU 4578 Transformation
题目大意：
给你一个初始全0的数组。有四种操作：
1.[L,R]范围内的数全部+c
2.[L,R]范围内的数全部*c
3.[L,R]范围内的数全部变为c
4.输出[L,R]范围内的数的k次方和（1 <= k <= 3 )
题目分析：
就是在裸的线段树上稍作变化，注意lazy标记中的乘法标记和加法标记有结合律，然后拆项转换即可。

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std ;#define lson l , m , o << 1#define rson m + 1 , r , o << 1 | 1const int mod = 10007 ;const int maxN = 500000 ;int sum1[maxN] , sum2[maxN] , sum3[maxN] ;int add[maxN] , mul[maxN] , set[maxN] ;void PushUp ( int o ) {sum1[o] = ( sum1[o << 1] + sum1[o << 1 | 1] ) % mod ;sum2[o] = ( sum2[o << 1] + sum2[o << 1 | 1] ) % mod ;sum3[o] = ( sum3[o << 1] + sum3[o << 1 | 1] ) % mod ;}void PushDown ( int o , int l , int r ) {int m = ( l + r ) >> 1 ;if ( set[o] ) {set[o << 1] = set[o << 1 | 1] = set[o] ;add[o << 1] = add[o << 1 | 1] = 0 ;mul[o << 1] = mul[o << 1 | 1] = 1 ;sum1[o << 1] = ( m - l + 1 ) * set[o] % mod ;sum2[o << 1] = sum1[o << 1] * set[o] % mod ;sum3[o << 1] = sum2[o << 1] * set[o] % mod ;sum1[o << 1 | 1] = ( r - m ) * set[o] % mod ;sum2[o << 1 | 1] = sum1[o << 1 | 1] * set[o] % mod ;sum3[o << 1 | 1] = sum2[o << 1 | 1] * set[o] % mod ;set[o] = 0 ;}if ( add[o] || mul[o] != 1 ) {add[o << 1] = ( add[o << 1] * mul[o] + add[o] ) % mod ;mul[o << 1] = mul[o << 1] * mul[o] % mod ;add[o << 1 | 1] = ( add[o << 1 | 1] * mul[o] + add[o] ) % mod ;mul[o << 1 | 1] = mul[o << 1 | 1] * mul[o] % mod ;int _sum1 , _sum2 , _sum3 ;_sum1 = ( sum1[o << 1] * mul[o] % mod + add[o] * ( m - l + 1 ) % mod ) % mod ;_sum2 = ( mul[o] * mul[o] % mod * sum2[o << 1] % mod+ 2 * add[o] % mod * mul[o] % mod * sum1[o << 1] % mod+ add[o] * add[o] % mod * ( m - l + 1 ) % mod ) % mod ;_sum3 = ( mul[o] * mul[o] % mod * mul[o] % mod * sum3[o << 1] % mod+ 3 * mul[o] % mod * mul[o] % mod * add[o] % mod * sum2[o << 1] % mod+ 3 * mul[o] % mod * add[o] % mod * add[o] % mod * sum1[o << 1] % mod+ add[o] * add[o] % mod * add[o] % mod * ( m - l + 1 ) % mod ) % mod ;sum1[o << 1] = _sum1 ;sum2[o << 1] = _sum2 ;sum3[o << 1] = _sum3 ;_sum1 = ( sum1[o << 1 | 1] * mul[o] % mod + add[o] * ( r - m ) % mod ) % mod ;_sum2 = ( mul[o] * mul[o] % mod * sum2[o << 1 | 1] % mod+ 2 * add[o] % mod * mul[o] % mod * sum1[o << 1 | 1] % mod+ add[o] * add[o] % mod * ( r - m ) % mod ) % mod ;_sum3 = ( mul[o] * mul[o] % mod * mul[o] % mod * sum3[o << 1 | 1] % mod+ 3 * mul[o] % mod * mul[o] % mod * add[o] % mod * sum2[o << 1 | 1] % mod+ 3 * mul[o] % mod * add[o] % mod * add[o] % mod * sum1[o << 1 | 1] % mod+ add[o] * add[o] % mod * add[o] % mod * ( r - m ) % mod ) % mod ;sum1[o << 1 | 1] = _sum1 ;sum2[o << 1 | 1] = _sum2 ;sum3[o << 1 | 1] = _sum3 ;add[o] = 0 ;mul[o] = 1 ;}}void Build ( int l , int r , int o ) {sum1[o] = sum2[o] = sum3[o] = 0 ;add[o] = set[o] = 0 ;mul[o] = 1 ;if ( l == r ) return ;int m = ( l + r ) >> 1 ;Build ( lson ) ;Build ( rson ) ;}void Update ( int ch , int v , int L , int R , int l , int r , int o ) {if ( L <= l && r <= R ) {int len = r - l + 1 ;if ( 1 == ch ) {add[o] = ( add[o] + v ) % mod ;int _sum1 , _sum2 , _sum3 ;_sum1 = ( sum1[o] + v * len ) % mod ;_sum2 = ( sum2[o] + 2 * sum1[o] % mod * v % mod + v * v % mod * len % mod ) % mod ;_sum3 = ( sum3[o] + 3 * v % mod * v % mod * sum1[o] % mod+ 3 * v % mod * sum2[o] % mod + v * v % mod * v % mod * len % mod ) % mod ;sum1[o] = _sum1 ;sum2[o] = _sum2 ;sum3[o] = _sum3 ;}if ( 2 == ch ) {mul[o] = mul[o] * v % mod ;add[o] = add[o] * v % mod ;sum1[o] = sum1[o] * v % mod ;sum2[o] = sum2[o] * v % mod * v % mod ;sum3[o] = sum3[o] * v % mod * v % mod * v % mod ;}if ( 3 == ch ) {set[o] = v ;add[o] = 0 ;mul[o] = 1 ;sum1[o] = v * len % mod ;sum2[o] = v * sum1[o] % mod ;sum3[o] = v * sum2[o] % mod ;}return ;}PushDown ( o , l , r ) ;int m = ( l + r ) >> 1 ;if ( L <= m ) Update ( ch , v , L , R , lson ) ;if ( m <  R ) Update ( ch , v , L , R , rson ) ;PushUp ( o ) ;}int Query ( int v , int L , int R , int l , int r , int o ) {if ( L <= l && r <= R ) {if ( 1 == v ) return sum1[o] ;if ( 2 == v ) return sum2[o] ;if ( 3 == v ) return sum3[o] ;}PushDown ( o , l , r ) ;int ans = 0 , m = ( l + r ) >> 1 ;if ( L <= m ) ans = ( ans + Query ( v , L , R , lson ) ) % mod ;if ( m <  R ) ans = ( ans + Query ( v , L , R , rson ) ) % mod ;return ans ;}void work () {int n , m , ch , v , l , r ;while ( ~scanf ( "%d%d" , &n , &m ) && ( n || m ) ) {Build ( 1 , n , 1 ) ;while ( m -- ) {scanf ( "%d%d%d%d" , &ch , &l , &r , &v ) ;if ( ch == 4 ) {int ans = Query ( v , l , r , 1 , n , 1 ) ;printf ( "%d\n" , ans ) ;}else Update ( ch , v , l , r , 1 , n , 1 ) ;}}}int main () {work () ;return 0 ;}