"Tree array interval revision interval sum" Codevs 1082 segment Tree Exercise 3

http://codevs.cn/problem/1082/

"AC"

1#include <bits/stdc++.h>2 using namespacestd;3typedefLong Longll;4 Const intmaxn=2e5+2;5 intN;6 ll A[MAXN];7 ll C1[MAXN];8 ll C2[MAXN];9 intLowbit (intx)Ten { One     returnx&-x; A } - voidAdd (ll *c,intk,ll val) - { the      while(k<=N) { -c[k]+=Val; -k+=Lowbit (k); -     } + } -ll query (ll *c,intk) + { All ans=0; at      while(k) -     { -ans+=C[k]; -k-=Lowbit (k); -     } -     returnans; in } -ll solve (intx) to { +ll ans=0; -ans+=x*query (c1,x); theans-=query (c2,x); *     returnans; $ }Panax Notoginsengll solve (intXinty) - { the     returnSolve (y)-solve (x1);  + } A intMain () the { +      while(~SCANF ("%d",&N)) -     { $a[0]=0; $          for(intI=1; i<=n;i++) -         { -scanf"%i64d",&a[i]); the         //cout<<a[i]<<endl; -Add (c1,i,a[i]-a[i-1]);WuyiAdd (C2,i, (i-1) * (a[i]-a[i-1])); the         }  -         intQ; Wuscanf"%d",&q); -         intTP; About          while(q--) $         { -scanf"%d",&tp); -             if(tp==1) -             { A                 intx,y;ll Val; +scanf"%d%d%i64d",&x,&y,&val);  the Add (c1,x,val); -Add (c1,y+1,-val); $Add (c2,x,1ll* (X-1)*val); theAdd (c2,y+1, -1ll*y*val); the             } the             Else the             { -                 intx, y; inscanf"%d%d",&x,&y); thell ans=solve (x, y); theprintf"%i64d\n", ans); About             } the         } the     } the     return 0; +}

View Code

Principle

The principle is to use a poor fractional group, reproduced from http://www.cnblogs.com/boceng/p/7222751.html

The time complexity of the tree array is O (Mlogn), which is better than the segment tree and less written.

The Divine Ox is introduced in the differential fractional group, to maintain the differential array ks[i] = A[i]-a[i-1]; can easily get a[i] = Ks[1] + ks[2] + ... + ks[i]; That is, the first I item and, for convenience to remember as Sigma (KS, i), can already see the shadow of the tree array, so to find the interval and the resulting

A[1] + a[2] +.. + A[n] = Sigma (KS, 1) + Sigma (KS, 2) + ... + sigma (KS, N);

= N*ks[1] + (n-1) *ks[2] + ... + 2*ks[n-1] + 1*ks[n];

= N (ks[1] + ks[2] +...+ ks[n])-(0*ks[1] + 1*ks[2] + ... + (n-1) *ks[n]);

So you can get Sum[n] =n * Sigma (KS, N)-(0*ks[1] + 1*ks[2] + ... + (n-1) *ks[n]);

Make jk[i] = (i-1) * ks[i];

Then sum[n] = n * Sigma (KS, N)-Sigma (JK, N);

Then it constructs two tree-like arrays;

1 intLowbit (intk) {2     returnK &-K;3 }4 voidAddintNint*c,intKintva) {5      while(k <=N) {6C[k] + =va;7K + =Lowbit (k);8     }9 }Ten  One //------------------------------------- A  -  for(i =1; I <= N; ++i) { -Add (n, C1, I, jk[i]-jk[i-1]); theAdd (n, c2, I, (i-1) * (jk[i]-jk[i-1])); -}

View Code

Then make the query sum

1 intSigmaint*c,intk) {2     intsum =0;3      while(k) {4Sum + =C[k];5K-=Lowbit (k);6     }7     returnsum;8 }9 intGetsum (intSintt) {Ten     return(T*sigma (c1, T)-sigma (C2, T))-((s1) *sigma (C1, S-1)-sigma (C2, S-1)); One}

View Code

When you make a single point query, only two parameters are passed in to that point.

In the interval update, the god bull through two maintenance C1, two times C2 get, but I speculate a few cases, can not explain the reasons for doing so well,

1 void Update (intint int va) {2    Add (C1, S, VA); 3     Add (c1, t+1,-va); 4     Add (C2, S, va* (S-1)); 5     Add (c2, t+1,-va*t); 6 }

View Code

"Tree array interval revision interval sum" Codevs 1082 segment Tree Exercise 3