Transformation

Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 49 Accepted Submission(s): 16

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

Source

2013ACM-ICPC杭州赛区全国邀请赛

很裸的线段树的题目。

但是做起来比较麻烦。

我用sum1,sum2,sum3分别代表和、平方和、立方和。

懒惰标记使用三个变量：

lazy1:是加的数

lazy2:是乘的倍数

lazy3:是赋值为一个常数，为0表示没有。

更新操作需要注意很多细节。

 /* **********************************************

 Author      : kuangbin

 Created Time: 2013/8/10 13:24:03

 File Name   : F:\2013ACM练习\比赛练习\2013杭州邀请赛重现\1003.cpp

 *********************************************** */

 #include <stdio.h>

 #include <string.h>

 #include <iostream>

 #include <algorithm>

 #include <vector>

 #include <queue>

 #include <set>

 #include <map>

 #include <string>

 #include <math.h>

 #include <stdlib.h>

 using namespace std;

 const int MOD = ;

 const int MAXN = ;

 struct Node

 {

     int l,r;

     int sum1,sum2,sum3;

     int lazy1,lazy2,lazy3;

 }segTree[MAXN*];

 void build(int i,int l,int r)

 {

     segTree[i].l = l;

     segTree[i].r = r;

     segTree[i].sum1 = segTree[i].sum2 = segTree[i].sum3 = ;

     segTree[i].lazy1 = segTree[i].lazy3 = ;

     segTree[i].lazy2 = ;

     int mid = (l+r)/;

     if(l == r)return;

     build(i<<,l,mid);

     build((i<<)|,mid+,r);

 }

 void push_up(int i)

 {

     if(segTree[i].l == segTree[i].r)

         return;

     segTree[i].sum1 = (segTree[i<<].sum1 + segTree[(i<<)|].sum1)%MOD;

     segTree[i].sum2 = (segTree[i<<].sum2 + segTree[(i<<)|].sum2)%MOD;

     segTree[i].sum3 = (segTree[i<<].sum3 + segTree[(i<<)|].sum3)%MOD;

 }

 void push_down(int i)

 {

     if(segTree[i].l == segTree[i].r) return;

     if(segTree[i].lazy3 != )

     {

         segTree[i<<].lazy3 = segTree[(i<<)|].lazy3 = segTree[i].lazy3;

         segTree[i<<].lazy1 = segTree[(i<<)|].lazy1 = ;

         segTree[i<<].lazy2 = segTree[(i<<)|].lazy2 = ;

         segTree[i<<].sum1 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD;

         segTree[i<<].sum2 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD;

         segTree[i<<].sum3 = (segTree[i<<].r - segTree[i<<].l + )*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD*segTree[i<<].lazy3%MOD;

         segTree[(i<<)|].sum1 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD;

         segTree[(i<<)|].sum2 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD;

         segTree[(i<<)|].sum3 = (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD*segTree[(i<<)|].lazy3%MOD;

         segTree[i].lazy3 = ;

     }

     if(segTree[i].lazy1 !=  || segTree[i].lazy2 != )

     {

         segTree[i<<].lazy1 = ( segTree[i].lazy2*segTree[i<<].lazy1%MOD + segTree[i].lazy1 )%MOD;

         segTree[i<<].lazy2 = segTree[i<<].lazy2*segTree[i].lazy2%MOD;

         int sum1,sum2,sum3;

         sum1 = (segTree[i<<].sum1*segTree[i].lazy2%MOD + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD)%MOD;

         sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i<<].sum2 % MOD + *segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[i<<].sum1%MOD + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD;

         sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i<<].sum3 % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i<<].sum2) % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[i<<].sum1) % MOD;

         sum3 = (sum3 + (segTree[i<<].r - segTree[i<<].l + )*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD;

         segTree[i<<].sum1 = sum1;

         segTree[i<<].sum2 = sum2;

         segTree[i<<].sum3 = sum3;

         segTree[(i<<)|].lazy1 = ( segTree[i].lazy2*segTree[(i<<)|].lazy1%MOD + segTree[i].lazy1 )%MOD;

         segTree[(i<<)|].lazy2 = segTree[(i<<)|].lazy2 * segTree[i].lazy2 % MOD;

         sum1 = (segTree[(i<<)|].sum1*segTree[i].lazy2%MOD + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD)%MOD;

         sum2 = (segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[(i<<)|].sum2 % MOD + *segTree[i].lazy1*segTree[i].lazy2%MOD * segTree[(i<<)|].sum1%MOD + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD*segTree[i].lazy1%MOD)%MOD;

         sum3 = segTree[i].lazy2 * segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[(i<<)|].sum3 % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[(i<<)|].sum2) % MOD;

         sum3 = (sum3 + *segTree[i].lazy2 % MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD * segTree[(i<<)|].sum1) % MOD;

         sum3 = (sum3 + (segTree[(i<<)|].r - segTree[(i<<)|].l + )*segTree[i].lazy1%MOD * segTree[i].lazy1 % MOD * segTree[i].lazy1 % MOD) % MOD;

         segTree[(i<<)|].sum1 = sum1;

         segTree[(i<<)|].sum2 = sum2;

         segTree[(i<<)|].sum3 = sum3;

         segTree[i].lazy1 = ;

         segTree[i].lazy2 = ;

     }

 }

 void update(int i,int l,int r,int type,int c)

 {

     if(segTree[i].l == l && segTree[i].r == r)

     {

         c %= MOD;

         if(type == )

         {

             segTree[i].lazy1 += c;

             segTree[i].lazy1 %= MOD;

             segTree[i].sum3 = (segTree[i].sum3 + *segTree[i].sum2%MOD*c%MOD + *segTree[i].sum1%MOD*c%MOD*c%MOD + (segTree[i].r - segTree[i].l + )*c%MOD*c%MOD*c%MOD)%MOD;

             segTree[i].sum2 = (segTree[i].sum2 + *segTree[i].sum1%MOD*c%MOD + (segTree[i].r - segTree[i].l + )*c%MOD*c%MOD)%MOD;

             segTree[i].sum1 = (segTree[i].sum1 + (segTree[i].r - segTree[i].l + )*c%MOD)%MOD;

         }

         else if(type == )

         {

             segTree[i].lazy1 = segTree[i].lazy1*c%MOD;

             segTree[i].lazy2 = segTree[i].lazy2*c%MOD;

             segTree[i].sum1 = segTree[i].sum1*c%MOD;

             segTree[i].sum2 = segTree[i].sum2*c%MOD*c%MOD;

             segTree[i].sum3 = segTree[i].sum3*c%MOD*c%MOD*c%MOD;

         }

         else

         {

             segTree[i].lazy1 = ;

             segTree[i].lazy2 = ;

             segTree[i].lazy3 = c%MOD;

             segTree[i].sum1 = c*(segTree[i].r - segTree[i].l + )%MOD;

             segTree[i].sum2 = c*(segTree[i].r - segTree[i].l + )%MOD*c%MOD;

             segTree[i].sum3 = c*(segTree[i].r - segTree[i].l + )%MOD*c%MOD*c%MOD;

         }

         return;

     }

     push_down(i);

     int mid = (segTree[i].l + segTree[i].r)/;

     if(r <= mid)update(i<<,l,r,type,c);

     else if(l > mid)update((i<<)|,l,r,type,c);

     else

     {

         update(i<<,l,mid,type,c);

         update((i<<)|,mid+,r,type,c);

     }

     push_up(i);

 }

 int query(int i,int l,int r,int p)

 {

     if(segTree[i].l == l && segTree[i].r == r)

     {

         if(p == )return segTree[i].sum1;

         else if(p== )return segTree[i].sum2;

         else return segTree[i].sum3;

     }

     push_down(i);

     int mid = (segTree[i].l + segTree[i].r )/;

     if(r <= mid)return query(i<<,l,r,p);

     else if(l > mid)return query((i<<)|,l,r,p);

     else return (query(i<<,l,mid,p)+query((i<<)|,mid+,r,p))%MOD;

 }

 int main()

 {

     //freopen("in.txt","r",stdin);

     //freopen("out.txt","w",stdout);

     int n,m;

     while(scanf("%d%d",&n,&m) == )

     {

         if(n ==  && m == )break;

         build(,,n);

         int type,x,y,c;

         while(m--)

         {

             scanf("%d%d%d%d",&type,&x,&y,&c);

             if(type == )printf("%d\n",query(,x,y,c));

             else update(,x,y,type,c);

         }

     }

     return ;

 }

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HDU 4578 Transformation （线段树）

Transformation

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