hdu4578(线段树)

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

题意：n个数，初始值为0，4种操作：

1。将某个区间所有值加上另一个值；

2。将区间所有值都乘上另一个值；

3。将区间所有值置为某个值；

4。查询区间中所有值的p次方和。

详细分析：http://www.cnblogs.com/GBRgbr/archive/2013/08/13/3254442.html

#pragma comment(linker,"/STACK:102400000,102400000")
#include <cstdio>
#include <cstring>
#include <string>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cstdlib>
#include <stack>
#include <vector>
#include <set>
#include <map>
#define LL long long
#define mod 10007
#define inf 0x3f3f3f3f
#define N 100010
#define FILL(a,b) (memset(a,b,sizeof(a)))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
using namespace std;
LL sum1[N<<2],sum2[N<<2],sum3[N<<2];
LL mul[N<<2],add[N<<2];
void Pushup(int rt)
{
    sum1[rt]=(sum1[rt<<1]+sum1[rt<<1|1])%mod;
    sum2[rt]=(sum2[rt<<1]+sum2[rt<<1|1])%mod;
    sum3[rt]=(sum3[rt<<1]+sum3[rt<<1|1])%mod;
}
void Pushdown(int rt,int len)
{
    if(mul[rt]!=1)
    {
        mul[rt<<1]=mul[rt<<1]*mul[rt]%mod;
        mul[rt<<1|1]=mul[rt<<1|1]*mul[rt]%mod;
        add[rt<<1]=add[rt<<1]*mul[rt]%mod;
        add[rt<<1|1]=add[rt<<1|1]*mul[rt]%mod;
        LL u=mul[rt]*mul[rt]%mod,v=mul[rt]*mul[rt]*mul[rt]%mod;
        sum1[rt<<1]=sum1[rt<<1]*mul[rt]%mod;
        sum1[rt<<1|1]=sum1[rt<<1|1]*mul[rt]%mod;
        sum2[rt<<1]=sum2[rt<<1]*u%mod;
        sum2[rt<<1|1]=sum2[rt<<1|1]*u%mod;
        sum3[rt<<1]=sum3[rt<<1]*v%mod;
        sum3[rt<<1|1]=sum3[rt<<1|1]*v%mod;
        mul[rt]=1;
    }
    if(add[rt]!=0)
    {
        add[rt<<1]=(add[rt<<1]+add[rt])%mod;
        add[rt<<1|1]=(add[rt<<1|1]+add[rt])%mod;
        LL u=add[rt]*add[rt]%mod,v=add[rt]*add[rt]*add[rt]%mod;
        LL t1=sum1[rt<<1],t2=sum1[rt<<1|1];
        LL t3=sum2[rt<<1],t4=sum2[rt<<1|1];
        sum1[rt<<1]=(sum1[rt<<1]+(len-(len>>1))*add[rt])%mod;
        sum1[rt<<1|1]=(sum1[rt<<1|1]+(len>>1)*add[rt])%mod;
        sum2[rt<<1]=(sum2[rt<<1]+(len-(len>>1))*u+t1*add[rt]*2)%mod;
        sum2[rt<<1|1]=(sum2[rt<<1|1]+(len>>1)*u+t2*add[rt]*2)%mod;
        sum3[rt<<1]=(sum3[rt<<1]+(len-(len>>1))*v+t3*add[rt]*3+3*u*t1)%mod;
        sum3[rt<<1|1]=(sum3[rt<<1|1]+(len>>1)*v+t4*add[rt]*3+3*u*t2)%mod;
        add[rt]=0;
    }
}

void build(int l,int r,int rt)
{
    sum1[rt]=sum2[rt]=sum3[rt]=0;
    mul[rt]=1;add[rt]=0;
    if(l==r)return;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
void update(int L,int R,LL c,int l,int r,int rt,int op)
{
    if(L<=l&&r<=R)
    {
        LL u=c*c%mod,v=c*c*c%mod;
        LL len=r-l+1;
        if(op==1)
        {
            LL t1=sum1[rt],t2=sum2[rt];
            add[rt]+=c;add[rt]%=mod;
            sum1[rt]=(sum1[rt]+len*c)%mod;
            sum2[rt]=(sum2[rt]+2*t1*c+len*u)%mod;
            sum3[rt]=(sum3[rt]+len*v+3*c*t2+3*u*t1)%mod;
        }
        else if(op==2)
        {
            mul[rt]*=c;mul[rt]%=mod;
            add[rt]*=c;add[rt]%=mod;
            sum1[rt]=sum1[rt]*c%mod;
            sum2[rt]=sum2[rt]*u%mod;
            sum3[rt]=sum3[rt]*v%mod;
        }
        else
        {
            add[rt]=c;mul[rt]=0;
            sum1[rt]=len*c%mod;
            sum2[rt]=len*u%mod;
            sum3[rt]=len*v%mod;
        }
        return;
    }
    Pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(L<=m)update(L,R,c,lson,op);
    if(m<R)update(L,R,c,rson,op);
    Pushup(rt);
}
LL query(int L,int R,int l,int r,int rt,int op)
{
    if(L<=l&&r<=R)
    {
        if(op==1)return sum1[rt];
        else if(op==2)return sum2[rt];
        else return sum3[rt];
    }
    Pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    LL res=0;
    if(L<=m)res+=query(L,R,lson,op);
    if(m<R)res+=query(L,R,rson,op);
    return res%mod;
}

int main()
{
    int n,m;
    int op,x,y,c;
    while(scanf("%d%d",&n,&m)>0)
    {
        if(m+n==0)break;
        build(1,n,1);
        while(m--)
        {
            scanf("%d%d%d%d",&op,&x,&y,&c);
            if(op==4)
            {
                printf("%I64d
",query(x,y,1,n,1,c));
            }
            else
            {
                update(x,y,c,1,n,1,op);
               // printf("%d
",sum1[1]);
            }
        }
    }
}