HDU 4578 Transformation （线段树）
Transformation
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 49 Accepted Submission(s): 16
There are n integers, a1, a2, …, an. The initial values of them are 0. There are four kinds of operations.
Operation 1: Add c to each number between ax and ay inclusive. In other words, do transformation ak<---ak+c, k = x,x+1,…,y.
Operation 2: Multiply c to each number between ax and ay inclusive. In other words, do transformation ak<---ak×c, k = x,x+1,…,y.
Operation 3: Change the numbers between ax and ay to c, inclusive. In other words, do transformation ak<---c, k = x,x+1,…,y.
Operation 4: Get the sum of p power among the numbers between ax and ay inclusive. In other words, get the result of axp+ax+1p+…+ay p.
Yuanfang has no idea of how to do it. So he wants to ask you to help him.
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.
3 3 5 7
1 2 4 4
4 1 5 2
2 2 5 8
4 3 5 3
0 0
7489
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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