hdu 4578 Transformation(线段树维护区间次方和)
Transformation
Time Limit: 15000/8000 MS (Java/Others) Memory Limit: 65535/65536 K (Java/Others)Total Submission(s): 1378 Accepted Submission(s): 303
Problem 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.
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.
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
Source
2013ACM-ICPC杭州赛区全国邀请赛
题意：
对区间内的数进行各种操作。
1，把[x,y]区间上的每个数加上c。
2，把[x,y]区间上的每个数乘上c.
3，把[x,y]区间上的数置成c。
4,询问区间[x,y]上的每个数的p次方的和。
思路：
一看就是线段树。开始被次方和吓到了。以前从来没遇到这种类型的。但是再看p的范围只有1到3。于是想到了每个结点维护三个和即sum[0],sum[1],sum[2]分别表示区间内数1次方2次方3次方的和。但是怎么更新结点这几个值呢？
想到线段树都是从儿子结点获得信息然后跟新父亲结点信息的。考虑对一个区间内的数加上一个数c后对这三个值有什么影响。首先考虑一个数c的情况。一次方就不说了直接加上c*区间长度len就行了。
二次方。一个数开始为a。后来为a+c。则平方后为a*a+2*a*c+c*c。比原来多了2*a*c+c*c。而对于整个区间增加
2*sum[0]*c+len*c*c。三次方增加3*sum[1]*c+3*sum[0]*c*c+len*c*c*c。
根据题目的操作在每个结点加三个标记。
same[k]表示区间内数值是否一样。如果相同在用一个变量value[k]存该值。
addv[k]表示区间内的数加的值。
mul[k]表示区间内的数乘的值。
但是想到这里又卡住了。因为在addv[k]和mul[k]是冲突的。因为当这两个标记时不知道是先加的还是先乘的。于是想到标记递归往下传知道不冲突或到了叶子结点停止。比赛时没敢写。觉得这样可能更新到叶子结点效率太低，结果下来一写2sA了。
只是要注意到取模的问题。因为算次方很容易超的。
详细见代码：
#include
#include
#include
using namespace std;
const int maxn=100010,mod=10007;
int addv[maxn<<2],mul[maxn<<2],value[maxn<<2],sum[maxn<<2][3];
bool same[maxn<<2];
int n,m;
void small(int k)//对各sum取模
{
int i;
for(i=0;i<3;i++)
if(sum[k][i]>=mod)
sum[k][i]%=mod;
}
void btree(int L,int R,int k)//建树
{
int ls,rs,mid;
addv[k]=0;
mul[k]=1;
same[k]=true;
value[k]=0;
sum[k][0]=sum[k][1]=sum[k][2]=0;
if(L==R)
return;
ls=k<<1;
rs=k<<1|1;
mid=(L+R)>>1;
btree(L,mid,ls);
btree(mid+1,R,rs);
}
void pushdown(int L,int R,int k)//根据是否互斥下传标记
{
int ls,rs,mid,lenl,lenr;
ls=k<<1;
rs=k<<1|1;
mid=(L+R)>>1;
lenl=mid-L+1;
lenr=R-mid;
if(same[k])
{
same[ls]=same[rs]=true;
value[ls]=value[rs]=value[k]%mod;
sum[ls][0]=value[ls]*lenl%mod;
sum[ls][1]=sum[ls][0]*value[ls]%mod;
sum[ls][2]=sum[ls][1]*value[ls]%mod;
sum[rs][0]=value[rs]*lenr%mod;
sum[rs][1]=sum[rs][0]*value[rs]%mod;
sum[rs][2]=sum[rs][1]*value[rs]%mod;
small(ls);
small(rs);
addv[ls]=addv[rs]=0;
mul[ls]=mul[rs]=1;
same[k]=false;
}
if(addv[k]!=0)
{
if(same[ls])
{
value[ls]+=addv[k]%mod;
value[ls]%=mod;
sum[ls][0]=value[ls]*lenl%mod;
sum[ls][1]=sum[ls][0]*value[ls]%mod;
sum[ls][2]=sum[ls][1]*value[ls]%mod;
small(ls);
}
else if(mul[ls]!=1)
{
pushdown(L,mid,ls);
mul[ls]=1;
addv[ls]=addv[k]%mod;
sum[ls][2]+=3*sum[ls][1]%mod*addv[k]%mod+3*sum[ls][0]%mod*addv[k]%mod*addv[k]%mod+lenl%mod*addv[k]%mod*addv[k]%mod*addv[k]%mod;
sum[ls][1]+=lenl%mod*addv[k]%mod*addv[k]%mod+2*sum[ls][0]%mod*addv[k]%mod;
sum[ls][0]+=lenl%mod*addv[k]%mod;
small(ls);
}
else
{
addv[ls]+=addv[k]%mod;
addv[ls]%=mod;
sum[ls][2]+=3*sum[ls][1]%mod*addv[k]%mod+3*sum[ls][0]%mod*addv[k]%mod*addv[k]%mod+lenl%mod*addv[k]%mod*addv[k]%mod*addv[k]%mod;
sum[ls][1]+=lenl%mod*addv[k]%mod*addv[k]%mod+2*sum[ls][0]%mod*addv[k]%mod;
sum[ls][0]+=lenl%mod*addv[k]%mod;
small(ls);
}
if(same[rs])
{
value[rs]+=addv[k]%mod;
value[rs]%=mod;
sum[rs][0]=value[rs]*lenr%mod;
sum[rs][1]=sum[rs][0]*value[rs]%mod;
sum[rs][2]=sum[rs][1]*value[rs]%mod;
small(rs);
}
else if(mul[rs]!=1)
{
pushdown(mid+1,R,rs);
mul[rs]=1;
addv[rs]=addv[k]%mod;
sum[rs][2]+=3*sum[rs][1]%mod*addv[k]%mod+3*sum[rs][0]%mod*addv[k]%mod*addv[k]%mod+lenr%mod*addv[k]%mod*addv[k]%mod*addv[k]%mod;
sum[rs][1]+=lenr%mod*addv[k]%mod*addv[k]%mod+2*sum[rs][0]%mod*addv[k]%mod;
sum[rs][0]+=lenr*addv[k]%mod;
small(rs);
}
else
{
addv[rs]+=addv[k]%mod;
addv[rs]%=mod;
sum[rs][2]+=3*sum[rs][1]%mod*addv[k]%mod+3*sum[rs][0]%mod*addv[k]%mod*addv[k]%mod+lenr%mod*addv[k]%mod*addv[k]%mod*addv[k]%mod;
sum[rs][1]+=lenr%mod*addv[k]%mod*addv[k]%mod+2*sum[rs][0]%mod*addv[k]%mod;
sum[rs][0]+=lenr*addv[k]%mod;
small(rs);
}
addv[k]=0;
}
else if(mul[k]!=1)
{
if(same[ls])
{
value[ls]*=mul[k]%mod;
value[ls]%=mod;
sum[ls][0]=lenl%mod*value[ls]%mod;
sum[ls][1]=sum[ls][0]*value[ls]%mod;
sum[ls][2]=sum[ls][1]*value[ls]%mod;
small(ls);
}
else if(addv[ls]!=0)
{
pushdown(L,mid,ls);
mul[ls]=mul[k]%mod;
sum[ls][0]*=mul[k]%mod;
sum[ls][1]*=mul[k]%mod*mul[k]%mod;
sum[ls][2]*=mul[k]%mod*mul[k]%mod*mul[k]%mod;
small(ls);
}
else
{
mul[ls]*=mul[k]%mod;
mul[ls]%=mod;
sum[ls][0]*=mul[k]%mod;
sum[ls][1]*=mul[k]%mod*mul[k]%mod;
sum[ls][2]*=mul[k]%mod*mul[k]%mod*mul[k]%mod;
small(ls);
}
if(same[rs])
{
value[rs]*=mul[k]%mod;
value[rs]%=mod;
sum[rs][0]=lenr%mod*value[rs]%mod;
sum[rs][1]=sum[rs][0]*value[rs]%mod;
sum[rs][2]=sum[rs][1]*value[rs]%mod;
small(rs);
}
else if(addv[rs]!=0)
{
pushdown(mid+1,R,rs);
mul[rs]=mul[k]%mod;
sum[rs][0]*=mul[k]%mod;
sum[rs][1]*=mul[k]%mod*mul[k]%mod;
sum[rs][2]*=mul[k]%mod*mul[k]%mod*mul[k]%mod;
small(rs);
}
else
{
mul[rs]*=mul[k]%mod;
mul[rs]%=mod;
sum[rs][0]*=mul[k]%mod;
sum[rs][1]*=mul[k]%mod*mul[k]%mod;
sum[rs][2]*=mul[k]%mod*mul[k]%mod*mul[k]%mod;
small(rs);
}
mul[k]=1;
}
}
void update(int L,int R,int l,int r,int k,int op,int c)//更新
{
int ls,rs,mid,len;
len=R-L+1;
c%=mod;
len%=mod;
if(l==L&&r==R)
{
if(op==1)
{
if(same[k])
{
value[k]+=c;
value[k]%=mod;
sum[k][0]=len*value[k]%mod;
sum[k][1]=sum[k][0]*value[k]%mod;
sum[k][2]=sum[k][1]*value[k]%mod;
}
else if(mul[k]!=1)
{
pushdown(L,R,k);
addv[k]=c;
sum[k][2]+=3*sum[k][1]%mod*c%mod+3*sum[k][0]%mod*c%mod*c%mod+len*c%mod*c%mod*c%mod;
sum[k][1]+=len*c%mod*c%mod+2*sum[k][0]%mod*c%mod;
sum[k][0]+=len*c%mod;
}
else
{
addv[k]+=c;
addv[k]%=mod;
sum[k][2]+=3*sum[k][1]%mod*c%mod+3*sum[k][0]%mod*c%mod*c%mod+len*c%mod*c%mod*c%mod;
sum[k][1]+=len*c%mod*c%mod+2*sum[k][0]%mod*c%mod;
sum[k][0]+=len*c%mod;
}
}
else if(op==2)
{
if(same[k])
{
value[k]*=c;
value[k]%=mod;
sum[k][0]=value[k]*len%mod;
sum[k][1]=sum[k][0]*value[k]%mod;
sum[k][2]=sum[k][1]*value[k]%mod;
}
else if(addv[k]!=0)
{
pushdown(L,R,k);
mul[k]=c;
sum[k][0]*=c;
sum[k][1]*=c*c%mod;
sum[k][2]*=c*c%mod*c%mod;
}
else
{
mul[k]*=c%mod;
mul[k]%=mod;
sum[k][0]*=c%mod;
sum[k][1]*=c%mod*c%mod;
sum[k][2]*=c%mod*c%mod*c%mod;
}
}
else if(op==3)
{
same[k]=true;
value[k]=c;
addv[k]=0;
mul[k]=1;
sum[k][0]=len*c%mod;
sum[k][1]=sum[k][0]*c%mod;
sum[k][2]=sum[k][1]*c%mod;
}
small(k);
return;
}
ls=k<<1;
rs=k<<1|1;
mid=(L+R)>>1;
pushdown(L,R,k);
if(l>mid)
update(mid+1,R,l,r,rs,op,c);
else if(r<=mid)
update(L,mid,l,r,ls,op,c);
else
{
update(L,mid,l,mid,ls,op,c);
update(mid+1,R,mid+1,r,rs,op,c);
}
sum[k][0]=sum[ls][0]+sum[rs][0];
sum[k][1]=sum[ls][1]+sum[rs][1];
sum[k][2]=sum[ls][2]+sum[rs][2];
small(k);
//printf("%d->%d %d %d %d\n",L,R,sum[k][0],sum[k][1],sum[k][2]);
}
int qu(int L,int R,int l,int r,int k,int p)
{
int ls,rs,mid;
if(l==L&&r==R)
return sum[k][p];
ls=k<<1;
rs=k<<1|1;
mid=(L+R)>>1;
pushdown(L,R,k);
if(l>mid)
return qu(mid+1,R,l,r,rs,p);
else if(r<=mid)
return qu(L,mid,l,r,ls,p);
else
return (qu(L,mid,l,mid,ls,p)+qu(mid+1,R,mid+1,r,rs,p))%mod;
}
int main()
{
int com,i,x,y,c;
while(scanf("%d%d",&n,&m),n||m)
{
btree(1,n,1);
for(i=1; i<=m; i++)
{
scanf("%d%d%d%d",&com,&x,&y,&c);
if(com!=4)
update(1,n,x,y,1,com,c);
else
printf("%d\n",qu(1,n,x,y,1,c-1));
}
}
return 0;
}
哎。。。长的我自己都不想看了。。。
下传标记应该有更高效的方法的。有空再研究。。。