HDU P4578 Transformation
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.
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;
#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;
}
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