hdu4578Transformation（线段树多个lz标记）

这里以3次方来举例讲一下这题的做法，其它维类似。

如果要求某一个值的3次方那么sum = t^3,设t = x+y。那也就是sum = (x+y)^3.

假如我让每个数都加z t = x+y+z，我可以让新的y = y+z,这里发现新来的总会加在y上，那么可以给他一个延迟，slz，

那么新的值t = t + slz,再看如果让每个数都乘k，t = (x+y)*k = xk+yk.可以看出刚才的slz = slz*k; 另外发现系数会跟着变换，可以给他一个延迟，mlz。

那么一个数t = (mlz*x+slz)^3 = mlz^3*x^3+2*mlz^2*slz*x^2+2*mlz*slz^2*x+x^3.

如果求一段这样的和的话，会发现整体就是k1*原本x^3的和+k2*原本x^2的和+k1原本x^1的和 +（r-l+1)*x^3;

这样就可以分别保存3维的和。

把某一段的值设置为v的时候，可以使mlz =0 slz = v.

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 100000
 #define LL __int64
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 const int mod = ;
 int slz[N<<],mlz[N<<];//lz[N<<2];
 LL s[N<<][];
 int p[][];
 void init()
 {
     int i,j;
     for(i = ; i <=  ; i++)
     {
         p[i][] = i;
         for(j =  ; j <=  ; j++)
         p[i][j] = (p[i][j-]*i)%mod;
     }
 }
 void up(int w)
 {
     for(int i = ; i <=  ; i++)
     s[w][i] = (s[w<<][i]+s[w<<|][i])%mod;
 }
 void build(int l,int r,int w)
 {
     mlz[w] = ;
     slz[w] = ;
     //lz[w] = 0;
     if(l==r)
     {
         for(int i = ; i <=  ; i++)
         s[w][i] = ;
         return ;
     }
     int m = (l+r)>>;
     build(l,m,w<<);
     build(m+,r,w<<|);
     up(w);
 }
 void down(int w,int m)
 {

     if(mlz[w]!=||slz[w]!=)
     {
         mlz[w<<] = (mlz[w<<]*mlz[w])%mod;
         slz[w<<] = (mlz[w]*slz[w<<]+slz[w])%mod;
         mlz[w<<|] = (mlz[w<<|]*mlz[w])%mod;
         slz[w<<|] = (mlz[w]*slz[w<<|]+slz[w])%mod;

         int x1 ,x2,x3,y1,y2,y3;
         x1 = mlz[w]; y1 = slz[w];
         x2 = p[x1][]; y2 = p[y1][];
         x3 = p[x1][]; y3 = p[y1][];
         s[w<<][] = (x3*s[w<<][]+(*x2*y1)%mod*s[w<<][]+(*x1*y2)%mod*s[w<<][]+y3*(m-m/))%mod;
         s[w<<|][] = (x3*s[w<<|][]+(*x2*y1)%mod*s[w<<|][]+(*x1*y2)%mod*s[w<<|][]+y3*(m/))%mod;
         s[w<<][] = (x2*s[w<<][]+(*x1*y1)%mod*s[w<<][]+y2*(m-m/))%mod;
         s[w<<|][] = (x2*s[w<<|][]+(*x1*y1)%mod*s[w<<|][]+y2*(m/))%mod;
         s[w<<][] = (x1*s[w<<][]+y1*(m-m/))%mod;
         s[w<<|][] = (x1*s[w<<|][]+y1*(m/))%mod;
         mlz[w] = ;
         slz[w] = ;
     }
 }
 void update(int f,int c,int a,int b,int l,int r,int w)
 {
     if(a<=l&&b>=r)
     {
         if(f==)
         {

             mlz[w] = ;
             slz[w] = c;
             for(int i = ; i <=  ; i++)
             s[w][i] = (p[c][i]*(r-l+))%mod;

         }
         else if(f==)
         {

             slz[w]=(slz[w]+c)%mod;
             s[w][] = ((s[w][]+*s[w][]*c+*s[w][]*c*c)%mod+(LL)c*c*c*(r-l+))%mod;
             s[w][] = ((s[w][]+*c*s[w][])%mod+(LL)c*c*(r-l+))%mod;
             s[w][] = s[w][]+c*(r-l+); s[w][]%=mod;
         }
         else if(f==)
         {

                 mlz[w] *= c;mlz[w]%=mod;
                 slz[w] *= c;slz[w]%=mod;
            // }
             //cout<<l<<" "<<r<<" "<<s[w][2]<<endl;
             s[w][] = (s[w][]*c*c*c)%mod;
             s[w][] = (s[w][]*c*c)%mod;
             s[w][] = (c*s[w][])%mod;
             //cout<<l<<" "<<r<<" "<<s[w][2]<<endl;
         }
         return ;
     }
     down(w,r-l+);
     int m = (l+r)>>;
     if(a<=m)
     update(f,c,a,b,l,m,w<<);
     if(b>m) update(f,c,a,b,m+,r,w<<|);
     up(w);
 }
 LL query(int f,int a,int b,int l,int r,int w)
 {
     if(a<=l&&b>=r)
     {
         return s[w][f];
     }
     down(w,r-l+);
     int m = (l+r)>>;
     LL res=;
     if(a<=m) res+=query(f,a,b,l,m,w<<);
     res%=mod;
     if(b>m) res+=query(f,a,b,m+,r,w<<|);
     res%=mod;
     return res;
 }
 int main()
 {
     int n,m,i;
     int k,x,y,c;
     init();
     while(scanf("%d%d",&n,&m)&&n&&m)
     {
         build(,n,);
         for(i =  ;i <= m; i++)
         {
             scanf("%d%d%d%d",&k,&x,&y,&c);
             if(k<=)
                 update(k,c,x,y,,n,);
             else
                 printf("%I64d\n",(query(c,x,y,,n,))%mod);
         }
     }
     return ;
 }