HDU 4578 Transformation

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

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A more complex line-of-tree problem

It is strongly recommended that the segment tree be written with fewer students to add the Operation $1,2,3$ one by one.

Direct write three operation thought may be disorderly

The problem is to ask for interval $1$ to $3$ and how to update for 1 operation needs deduction

Assume that the interval length is $len-$-A $ sum1$ $sum 2$ and respectively, and the 3$

After the update is $ Sum1 ' $ $sum 2 ' $ $sum 3 ' $ then

$sum 1 ' = sum1 + c * len$

$sum 2 ' = sum2 + sum1 * C * 2 + c * C * len$

$sum 3 ' = sum3 + sum2 * C * 3 + sum1 * c * c * 3 + c * C * c * len$

The rest is okay, just be patient. Analyze the code volume only $2k+$

1#include <cstdio>2#include <cstring>3#include <cmath>4#include <algorithm>5 using namespacestd;6 Const intN =100010, mod =10007;7 intsum[3][n <<2], flag[3][n <<2];8 intN, M;9 voidUpdateintXintLintRintTlintTrintOintc);Ten voidBuildintXintTlintTR) One { Asum[0][X] = sum[1][X] = sum[2][X] =0; -flag[0][X] = flag[2][X] =0; -flag[1][X] =1; the     if(TL = =TR) -         return; -     intMid = (tl + TR) >>1; -Build (x <<1, TL, mid); +Build (x <<1|1, Mid +1, tr); - } + voidPushdown (intXintTlintTR) A { at     intMid = (tl + TR) >>1; -      for(inti =2; I >=0; --i) -         if(Flag[i][x] &&! (i = =1&& Flag[i][x] = =1)) -         { -Update (x <<1, TL, Mid, TL, Mid, I, flag[i][x]); -Update (x <<1|1, Mid +1, TR, Mid +1, TR, I, flag[i][x]); inFLAG[I][X] = (i = =1?1:0); -         } to } + voidUpdateintXintLintRintTlintTrintOintc) - { the     intMid = (tl + TR) >>1; *     if(L <= tl && TR <=R) $     {Panax Notoginseng         if(O = =0) -         { theflag[0][X] = (flag[0][X] + c)%MoD; +sum[2][X] = (sum[2][X] + sum[1][X] * c *3%MoD A+ sum[0][X] * C% mod * c *3% mod + c * C% mod * C% mod * the(Tr-tl +1)% MoD)%MoD; +sum[1][X] = (sum[1][X] + sum[0][X] * c *2% mod + c * C% mod * -(Tr-tl +1)% MoD)%MoD; $sum[0][X] = (sum[0][X] + c * (Tr-tl +1)% MoD)%MoD; $         } -         Else if(O = =1) -         { theflag[0][X] = flag[0][X] * C%MoD; -flag[1][X] = flag[1][X] * C%MoD;Wuyisum[0][X] = sum[0][X] * C%MoD; thesum[1][X] = sum[1][X] * C% mod * C%MoD; -sum[2][X] = sum[2][X] * C% mod * C% mod * C%MoD; Wu         } -         Else About         { $flag[0][X] =0; -flag[1][X] =1; -flag[2][X] =C; -sum[0][X] = (TR-TL +1) * C%MoD; Asum[1][X] = (TR-TL +1) * C% mod * C%MoD; +sum[2][X] = (TR-TL +1* C% mod * C% mod * C%MoD; the         } -         return; $     } the pushdown (x, TL, TR); the     if(L <=mid) theUpdate (x <<1, L, R, TL, Mid, O, c); the     if(Mid <R) -Update (x <<1|1, L, R, Mid +1, TR, o, c); in      for(inti =0; I <3; ++i) theSUM[I][X] = (sum[i][x <<1] + sum[i][x <<1|1]) %MoD; the } About intQueryintXintLintRintTlintTrintp) the { the     if(L <= tl && TR <=R) the         returnSum[p][x]; + pushdown (x, TL, TR); -     intMid = (tl + TR) >>1, re=0; the     if(L <=mid)BayiRe + = query (x <<1, L, R, TL, Mid, p); the     if(Mid <R) theRe + = query (x <<1|1, L, R, Mid +1, TR, p); -     returnRe%MoD; - } the intMain () the { the      while(SCANF ("%d%d", &n, &m), N) the     { -Build1,1, n); the         into, x, y, C; the          while(m--) the         {94scanf"%d%d%d%d", &o, &x, &y, &c); the             if(O! =4) theUpdate1, X, Y,1, N, O-1, c); the             Else98printf"%d\n", Query (1, X, Y,1, N, C-1)); About         } -     }101     return 0;102}

HDU 4578 Transformation