HDU P4578 Transformation

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. 

                                                         --by HDUoj

http://acm.hdu.edu.cn/showproblem.php?pid=4578

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因为p<4,所以，直接线段树维护区间和，平方和，立方和，三棵树，打上乘法，加法，更改标记，注意down的顺序；

所以，会线段树的话，这题主要考代码能力......和信仰。

记得及时取模。

代码如下：

 #include<cstdio>
 #define mod 10007
 using namespace std;

 long long tree[][];
 long long mark1[],mark2[],mark3[];

 int n,m,L,R;
 long long a,b;

 void work();
 void build(int ,int ,int );
 void up(int );
 void down(int ,int ,int );
 void change(int ,int ,int );
 long long sum(int ,int ,int );

 int main()
 {
     while()
     {
         scanf("%d%d",&n,&m);
         if(n==&&m==)
             return ;
         work();
     }
 }

 void work()
 {
     int i;
     long long ans=;
     build(,n,);
     for(i=;i<=m;i++)
     {
         scanf("%d%d%d%d",&b,&L,&R,&a);
         if(b==)
         {
                 ans=sum(,n,);
                 printf("%lld\n",ans%mod);
         }
         else
             change(,n,);
     }
 }

 void build(int l,int r,int nu)
 {
     tree[][nu]=tree[][nu]=tree[][nu]=mark1[nu]=mark3[nu]=;
     mark2[nu]=;
     if(l==r)
         return ;
     int mid=(l+r)>>;
     build(l,mid,nu<<);
     build(mid+,r,nu<<|);
 }

 void up(int nu)
 {
     tree[][nu]=(tree[][nu<<]+tree[][nu<<|])%mod;
     tree[][nu]=(tree[][nu<<]+tree[][nu<<|])%mod;
     tree[][nu]=(tree[][nu<<]+tree[][nu<<|])%mod;
 }

 void down(int l,int r,int nu)
 {
     int mid=(l+r)>>;
     if(mark3[nu])
     {
         tree[][nu<<]=tree[][nu<<]=tree[][nu<<]=;
         mark1[nu<<]=;mark2[nu<<]=;
         mark3[nu<<]=mark3[nu];
         tree[][nu<<|]=tree[][nu<<|]=tree[][nu<<|]=;
         mark1[nu<<|]=;mark2[nu<<|]=;
         mark3[nu<<|]=mark3[nu];
     }
     tree[][nu<<]=(tree[][nu<<]*mark2[nu]*mark2[nu]*mark2[nu])%mod;
     tree[][nu<<]=(tree[][nu<<]*mark2[nu]*mark2[nu])%mod;
     tree[][nu<<]=(tree[][nu<<]*mark2[nu])%mod;
     tree[][nu<<]=(tree[][nu<<]+*tree[][nu<<]*mark1[nu]+*tree[][nu<<]*mark1[nu]*mark1[nu]+(mid-l+)*mark1[nu]*mark1[nu]*mark1[nu])%mod;
     tree[][nu<<]=(tree[][nu<<]+*mark1[nu]*tree[][nu<<]+(mid-l+)*mark1[nu]*mark1[nu])%mod;
     tree[][nu<<]=(tree[][nu<<]+mark1[nu]*(mid-l+))%mod;
     mark2[nu<<]=(mark2[nu<<]*mark2[nu])%mod;
     mark1[nu<<]=(mark1[nu<<]*mark2[nu]+mark1[nu])%mod;
     tree[][nu<<|]=(tree[][nu<<|]*mark2[nu]*mark2[nu]*mark2[nu])%mod;
     tree[][nu<<|]=(tree[][nu<<|]*mark2[nu]*mark2[nu])%mod;
     tree[][nu<<|]=(tree[][nu<<|]*mark2[nu])%mod;
     tree[][nu<<|]=(tree[][nu<<|]+*tree[][nu<<|]*mark1[nu]+*tree[][nu<<|]*mark1[nu]*mark1[nu]+(r-mid)*mark1[nu]*mark1[nu]*mark1[nu])%mod;
     tree[][nu<<|]=(tree[][nu<<|]+*mark1[nu]*tree[][nu<<|]+(r-mid)*mark1[nu]*mark1[nu])%mod;
     tree[][nu<<|]=(tree[][nu<<|]+mark1[nu]*(r-mid))%mod;
     mark2[nu<<|]=(mark2[nu<<|]*mark2[nu])%mod;
     mark1[nu<<|]=(mark1[nu<<|]*mark2[nu]+mark1[nu])%mod;
     mark1[nu]=mark3[nu]=;mark2[nu]=;
 }

 void change(int l,int r,int nu)
 {
     if(L<=l&&r<=R)
     {
         if(b==)
         {
             mark1[nu]=a;
             mark2[nu]=;
             mark3[nu]=;
             tree[][nu]=a*(r-l+)%mod;
             tree[][nu]=(a*a*(r-l+))%mod;
             tree[][nu]=(a*a*a*(r-l+))%mod;
         }
         if(b==)
         {
             mark1[nu]=(mark1[nu]*a)%mod;
             mark2[nu]=(mark2[nu]*a)%mod;
             tree[][nu]=(tree[][nu]*a)%mod;
             tree[][nu]=(tree[][nu]*a*a)%mod;
             tree[][nu]=(tree[][nu]*a*a*a)%mod;
         }
         if(b==)
         {
             mark1[nu]=(mark1[nu]+a)%mod;
             tree[][nu]=(tree[][nu]+*tree[][nu]*a+*tree[][nu]*a*a+(r-l+)*a*a*a)%mod;
             tree[][nu]=(tree[][nu]+*a*tree[][nu]+(r-l+)*a*a)%mod;
             tree[][nu]=(tree[][nu]+a*(r-l+))%mod;
         }
         return;
     }
     down(l,r,nu);
     int mid=(l+r)>>;
     if(L<=mid)
         change(l,mid,nu<<);
     if(R>=mid+)
         change(mid+,r,nu<<|);
     up(nu);
 }

 long long sum(int l,int r,int nu)
 {
     long long su=;
     if(L<=l&&r<=R)
         return tree[a][nu];
     down(l,r,nu);
     int mid=(l+r)>>;
     if(L<=mid)
         su+=sum(l,mid,nu<<);
     if(R>=mid+)
         su+=sum(mid+,r,nu<<|);
     su=su%mod;
     return su;
 }

祝AC哟；