HDU 4578-Transformation（线段树）

J - Transformation

Time Limit:8000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 4578

Appoint description:  System Crawler  (2015-11-11)

Description

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

 

这题的代码量太感人了，写到我想哭。这题其实难点在于要用公式ax+b来做，这样就不管是乘了先还是加了先。

AC代码：

#include
#include
#include
#include
using namespace std;
#define T 200005
#define mod 10007
#define lson (rt<<1)
#define rson (rt<<1|1)
typedef long long ll;
int n,m;
struct node
{
	int L,R,mid;
	int add,mul;
	int sum[3];
}tree[T<<2];
void PushUp(int rt)//向上更新节点值
{
	for(int i=0;i<3;++i)
	tree[rt].sum[i] = (tree[lson].sum[i]+tree[rson].sum[i])%mod;
}
void cal(int rt,int mul,int add)//向下更新
{
	int len = tree[rt].R - tree[rt].L + 1;//区间大小
	tree[rt].sum[0] = tree[rt].sum[0] * mul %mod;//一次方
	tree[rt].sum[1] = tree[rt].sum[1] * mul %mod*mul%mod;//二次方 
	tree[rt].sum[2] = tree[rt].sum[2] * mul %mod*mul%mod*mul%mod;//三次方

	/*
	这里用 val = a*x + c 来表示
	所以
	当是乘法时直接当a是mul
	当是加法时将直接c乘mul再加就行了
	*/
	tree[rt].mul = (tree[rt].mul * mul )%mod;//乘法累加
	tree[rt].add = ((tree[rt].add * mul )% mod + add)%mod;//加法累加
	
	/*
	(ax+c)^3 = a^3*x^3 + 3*a^2*c*x + 3*a*c^2*x + c^3
	其中
	tree.sum就是1-3次方
	*/

	tree[rt].sum[2] = (tree[rt].sum[2] + 3*add%mod*add%mod *tree[rt].sum[0]%mod)%mod;
	tree[rt].sum[2] = (tree[rt].sum[2] + 3*add%mod*tree[rt].sum[1]%mod)%mod ;
	tree[rt].sum[2] = (tree[rt].sum[2] + len * add%mod *add %mod *add%mod)%mod;
	
	tree[rt].sum[1] = (tree[rt].sum[1] + 2*add%mod *tree[rt].sum[0] %mod )%mod;
	tree[rt].sum[1] = (tree[rt].sum[1] + len*add%mod*add%mod)%mod;

	tree[rt].sum[0] = (tree[rt].sum[0] +len*add%mod)%mod;

}
void PushDown(int rt)
{
	if(tree[rt].L==tree[rt].R){return;}
	cal(lson,tree[rt].mul,tree[rt].add);
	cal(rson,tree[rt].mul,tree[rt].add);
	tree[rt].mul = 1;
	tree[rt].add = 0;
}
void Build(int rt,int L,int R)
{
	tree[rt].L = L,tree[rt].R = R;
	tree[rt].mid = (L+R)>>1;
	tree[rt].mul = 1,tree[rt].add = 0;
	tree[rt].sum[0] = tree[rt].sum[1] = tree[rt].sum[2] = 0;
	if(L==R){
		return;
	}
	Build(lson,L,tree[rt].mid);
	Build(rson,tree[rt].mid+1,R);
}
void Update(int rt,int L,int R,int mul,int add)
{
	if(L<=tree[rt].L&&R>=tree[rt].R){
		cal(rt,mul,add);//更新当前节点的值
		return;
	}
	PushDown(rt);
	if(R<=tree[rt].mid){
		Update(lson,L,R,mul,add);
	}
	else if(L>tree[rt].mid){
		Update(rson,L,R,mul,add);
	}
	else
	{
		Update(lson,L,tree[rt].mid,mul,add);
		Update(rson,tree[rt].mid+1,R,mul,add);
	}
	PushUp(rt);
}
int query(int rt,int L,int R,int num)
{
	if(L<=tree[rt].L&&R>=tree[rt].R){
			return tree[rt].sum[num-1];
	}
	PushDown(rt);
	if(R<=tree[rt].mid){
		return query(lson,L,R,num);
	}
	else if(L>tree[rt].mid){
		return query(rson,L,R,num);
	}
	else
	{
		return (query(lson,L,tree[rt].mid,num) + query(rson,tree[rt].mid+1,R,num))%mod;
	}
}
int main()
{
#ifdef zsc
	freopen("input.txt","r",stdin);
#endif
	int i,num,x,y,c;
	while(scanf("%d%d",&n,&m)&&(n&&m))
	{
		Build(1,1,n);
		for(i=0;i