Rokua P3374 "template" Tree array 1 (CDQ Division)

Title Description

Title, you need to perform the following two operations, known as a sequence:

1. Add a number to the X

2. Find out the amount of each number in a range and

Input/output format

Input format:

The first line contains two integers n, M, respectively, indicating the number of numbers and the total number of operations.

The second line contains n space-delimited integers, where the I number represents the initial value of item I of a sequence.

The next m row contains 3 integers for each row, indicating an operation, as follows:

Action 1: Format: 1 x k Meaning: Add x number plus K

Action 2: Format: 2 x y meaning: the and of each number in the output interval [x, y]

Output format:

The output contains several line integers, which is the result of all Operation 2.

Input and Output sample input sample #: Copy

5 51 5 4 2 31 1 32 2 51 3-11 4 22 1 4

Output Example # #: Replication

1416

Description

Time limit: 1000ms,128m

Data size:

For 30% data: n<=8,m<=10

For 70% data: n<=10000,m<=10000

For 100% data: n<=500000,m<=500000

Sample Description:

Therefore, the output results 14, 16

Maintenance of two-dimensional partial order by CDQ

The first dimension is used to sort out

The second dimension is treated by CDQ

Slow to die.

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <ctime>5#include <cstdlib>6 using namespacestd;7 #defineLS T[now].ch[0]8 #defineRS T[now].ch[1]9 Const intmaxn=2*1e6+Ten;TenInlineCharNC () One { A     Static CharBuf[maxn],*p1=buf,*p2=buf; -     returnp1==p2&& (p2= (p1=buf) +fread (buf,1, Maxn,stdin), P1==P2)? eof:*p1++; - } theInlineintRead () - { -     CharC=NC ();intx=0, f=1; -      while(c<'0'|| C>'9'){if(c=='-') f=-1; c=NC ();} +      while(c>='0'&&c<='9') {x=x*Ten+c-'0', c=NC ();} -     returnx*F; + } A intn,m; at structnode - { -     intIdx//The first few questions -     intVal//the modified value -     intType//type of Operation -     BOOL operator< (ConstNode &a)Const { in         returnidx==a.idx?type<a.type:idx<A.idx;} - }Q[MAXN]; to intqidx=0;//Number of Operations + intaidx=0;//Number of Queries - intANS[MAXN]; the node TMP[MAXN]; * voidCDQ (intLintR) $ {Panax Notoginseng     if(r-l<=1)return ; -     intMid= (l+r) >>1; CDQ (L,mid); CDQ (mid,r); the     intsum=0; +     intp=l,q=mid,o=0; A      while(p<mid&&q<R) the     { +         if(q[p]<Q[q]) -         { $             if(q[p].type==1) sum+=Q[p].val; $tmp[o++]=q[p++]; -         } -         Else the         { -             if(q[q].type==2) ans[q[q].val]-=sum;Wuyi             Else if(q[q].type==3) ans[q[q].val]+=sum; thetmp[o++]=q[q++]; -         } Wu     } -      while(P<mid) tmp[o++]=q[p++]; About      while(q<R) $     { -         if(q[q].type==2) ans[q[q].val]-=sum; -         Else if(q[q].type==3) ans[q[q].val]+=sum; -tmp[o++]=q[q++]; A     } +      for(intI=0; i<o;i++) q[i+l]=Tmp[i]; the } - intMain () $ { the #ifdef WIN32 theFreopen ("a.in","R", stdin); the     #else the     #endif -N=read (); m=read (); in      for(intI=1; i<=n;i++) the     { theq[qidx].idx=i; AboutQ[qidx].type=1; theQ[qidx].val=read (); + +Qidx; the     } the      for(intI=0; i<m;i++) +     { -         intType=read (); theQ[qidx].type=type;Bayi         if(type==1) Q[qidx].idx=read (), q[qidx].val=read (); the         Else the         { -             intL=read (), r=read (); -q[qidx].idx=l-1; q[qidx].val=aidx;++Qidx; theQ[qidx].type=3; Q[qidx].idx=r; q[qidx].val=aidx;aidx++; the         } the++Qidx; the     } -CDQ (0, QIDX); the      for(intI=0; i<aidx;i++) printf ("%d\n", Ans[i]); the     return 0; the}

Rokua P3374 "template" Tree array 1 (CDQ Division)