[hdoj4578][多延迟标记的线段树]

时间：2019-05-03 14:46:11 阅读：132 评论：0 收藏：0 [点我收藏+]

Transformation

Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)
Total Submission(s): 9392 Accepted Submission(s): 2408

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

题意：有4种操作，区间加，区间乘，区间修改以及区间幂之和。

题解：区间加==(mul==1,add==x) 区间乘==(mul==x,add==0)区间修改==(mul==0,add==x)区间幂通过展开式转化为区间乘和加，lazy标记用来标记当前点已经被更新过，pushdown(rt)用于更新子节点

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define debug(x) cout<<"["<<#x<<"]"<<" is "<<x<<endl
  4 #define forp(x) for(int i=1;i<=x;i++)
  5 #define scai(x) scanf("%d",&x)
  6 #define scal(x) scanf("%lld",&x)
  7 #define pri(x) printf("%d\n",x)
  8 #define prl(x) printf("%lld\n",x)
  9 typedef long long ll;
 10 const int maxn=1e5+5;
 11 const ll mod=1e4+7;
 12 struct node{
 13     int l;
 14     int r;
 15     ll ad;
 16     ll mu;
 17     ll a[4];
 18 }N[maxn<<2];
 19 void pushup(int rt){
 20     N[rt].a[1]=N[rt<<1].a[1]+N[(rt<<1)|1].a[1];
 21     N[rt].a[2]=N[rt<<1].a[2]+N[(rt<<1)|1].a[2];
 22     N[rt].a[3]=N[rt<<1].a[3]+N[(rt<<1)|1].a[3];
 23     N[rt].mu%=mod;
 24     N[rt].ad%=mod;
 25     N[rt].a[1]%=mod;
 26     N[rt].a[2]%=mod;
 27     N[rt].a[3]%=mod;
 28 }
 29 void pushdown(int rt){
 30     if(N[rt].mu!=1){
 31         ll m=N[rt].mu;
 32         N[rt].mu=1;
 33         N[rt<<1].ad=N[rt<<1].ad*m%mod;
 34         N[(rt<<1)|1].ad=N[(rt<<1)|1].ad*m%mod;
 35         N[rt<<1].mu=N[rt<<1].mu*m%mod;
 36         N[(rt<<1)|1].mu=N[(rt<<1)|1].mu*m%mod;
 37         N[rt<<1].a[1]=N[rt<<1].a[1]*m%mod;
 38         N[rt<<1].a[2]=N[rt<<1].a[2]*m%mod*m%mod;
 39         N[rt<<1].a[3]=N[rt<<1].a[3]*m%mod*m%mod*m%mod;
 40         N[(rt<<1)|1].a[1]=N[(rt<<1)|1].a[1]*m%mod;
 41         N[(rt<<1)|1].a[2]=N[(rt<<1)|1].a[2]*m%mod*m%mod;
 42         N[(rt<<1)|1].a[3]=N[(rt<<1)|1].a[3]*m%mod*m%mod*m%mod;
 43     }
 44     if(N[rt].ad!=0){
 45         ll m=N[rt].ad;
 46         N[rt].ad=0;
 47         N[rt<<1].ad=N[rt<<1].ad+m%mod;
 48         N[(rt<<1)|1].ad=N[(rt<<1)|1].ad+m%mod;
 49         N[rt<<1].a[3]=N[rt<<1].a[3]+m%mod*m%mod*m%mod*(N[rt<<1].r-N[rt<<1].l+1)%mod+3*m*N[rt<<1].a[2]%mod+3*m%mod*m%mod*N[rt<<1].a[1]%mod;
 50         N[rt<<1].a[3]%=mod;
 51         N[rt<<1].a[2]=N[rt<<1].a[2]+m%mod*m%mod*(N[rt<<1].r-N[rt<<1].l+1)%mod+2*m*N[rt<<1].a[1]%mod;
 52         N[rt<<1].a[2]%=mod;
 53         N[rt<<1].a[1]=N[rt<<1].a[1]+m*(N[rt<<1].r-N[rt<<1].l+1)%mod;
 54         N[rt<<1].a[1]%=mod;
 55         N[(rt<<1)|1].a[3]=N[(rt<<1)|1].a[3]%mod+m%mod*m%mod*m%mod*(N[(rt<<1)|1].r-N[(rt<<1)|1].l+1)%mod+3*m*N[(rt<<1)|1].a[2]%mod+3*m%mod*m%mod*N[(rt<<1)|1].a[1]%mod;
 56         N[(rt<<1)|1].a[3]%=mod;
 57         N[(rt<<1)|1].a[2]=N[(rt<<1)|1].a[2]%mod+m%mod*m%mod*(N[(rt<<1)|1].r-N[(rt<<1)|1].l+1)%mod+2*m*N[(rt<<1)|1].a[1]%mod;
 58         N[(rt<<1)|1].a[2]%=mod;
 59         N[(rt<<1)|1].a[1]=N[(rt<<1)|1].a[1]+m*(N[(rt<<1)|1].r-N[(rt<<1)|1].l+1)%mod;
 60         N[(rt<<1)|1].a[1]%=mod;
 61     }
 62 }
 63 void build(int L,int R,int rt){
 64     N[rt].l=L;
 65     N[rt].r=R;
 66     N[rt].mu=1;
 67     N[rt].ad=0;
 68     if(L==R){
 69         N[rt].a[1]=0;
 70         N[rt].a[2]=0;
 71         N[rt].a[3]=0;
 72         return;
 73     }
 74     int mid=(L+R)/2;
 75     build(L,mid,rt<<1);
 76     build(mid+1,R,(rt<<1)|1);
 77     pushup(rt);
 78 }
 79 void update(int L,int R,int rt,int l1,int r1,ll add,ll mul){
 80     N[rt].mu%=mod;
 81     N[rt].ad%=mod;
 82     N[rt].a[1]%=mod;
 83     N[rt].a[2]%=mod;
 84     N[rt].a[3]%=mod;
 85     if(l1<=L&&r1>=R){
 86     if(mul!=1){
 87         ll m=mul;
 88         N[rt].ad=N[rt].ad*m%mod;
 89         N[rt].mu=N[rt].mu*m%mod;
 90         N[rt].a[1]=N[rt].a[1]*m%mod;
 91         N[rt].a[2]=N[rt].a[2]*m%mod*m%mod;
 92         N[rt].a[3]=N[rt].a[3]*m%mod*m%mod*m%mod;
 93     }
 94     if(add!=0){
 95         ll m=add;
 96         N[rt].ad=N[rt].ad+m%mod;
 97         N[rt].a[3]=N[rt].a[3]+m%mod*m%mod*m%mod*(N[rt].r-N[rt].l+1)%mod+3*m*N[rt].a[2]%mod+3*m%mod*m%mod*N[rt].a[1]%mod;
 98         N[rt].a[3]%=mod;
 99         N[rt].a[2]=N[rt].a[2]+m%mod*m%mod*(N[rt].r-N[rt].l+1)%mod+2*m%mod*N[rt].a[1]%mod;
100         N[rt].a[2]%=mod;
101         N[rt].a[1]=N[rt].a[1]+m*(N[rt].r-N[rt].l+1)%mod;
102         N[rt].a[1]%=mod;
103     }
104         return;
105     }
106     pushdown(rt);
107     int mid=(L+R)/2;
108     if(mid>=l1){
109         update(L,mid,rt<<1,l1,r1,add,mul);
110     }
111     if(mid<r1){
112         update(mid+1,R,(rt<<1)|1,l1,r1,add,mul);
113     }
114     pushup(rt);
115 }
116 ll query(int L,int R,int rt,int l1,int r1,int p){
117     N[rt].mu%=mod;
118     N[rt].ad%=mod;
119     N[rt].a[1]%=mod;
120     N[rt].a[2]%=mod;
121     N[rt].a[3]%=mod;
122     if(l1<=L&&r1>=R){
123         return N[rt].a[p]%mod;
124     }
125     int mid=(L+R)/2;
126     ll ak=0;
127     pushdown(rt);
128     if(mid>=l1){
129         ak+=query(L,mid,rt<<1,l1,r1,p);
130     }
131     if(mid<r1){
132         ak+=query(mid+1,R,(rt<<1)|1,l1,r1,p);
133     }
134 
135 
136     return ak%mod;
137 }
138 int main(){
139     int n,m;
140     scai(n);
141     scai(m);
142     while(n||m){
143         build(1,n,1);
144         forp(m){
145             ll a,b,c,d;
146             scal(a);
147             scal(b);
148             scal(c);
149             scal(d);
150             if(a==1){
151                 update(1,n,1,b,c,d,1);
152             }
153             else if(a==2){
154                 update(1,n,1,b,c,0,d);
155             }
156             else if(a==3){
157                 update(1,n,1,b,c,d,0);
158             }
159             else{
160                 prl(query(1,n,1,b,c,d));
161             }
162         }
163         scai(n);
164         scai(m);
165     }
166     return 0;
167 }

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[hdoj4578][多延迟标记的线段树]

标签：sub hide c++ 区间 help led his 更新 line

原文地址：https://www.cnblogs.com/MekakuCityActor/p/10804864.html

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