BZOJ3809 Gty's two-cornered sister sequence

Finally did the BZ on the latest topic 2333

The question is ... Team, then query the tree-like array.

The result t, eh eh eh eh%>_<%, how can this!

Another way to find him: Hzwer's Pieces

NN, is to divide the color into n blocks, then the word modified O (1), the word query O (n/sz + 2 * sz)

SZ = sqrt (N/2) is best (theoretically), actually SZ = sqrt (n) is not slow at all .... do you want to try Sz = log (n) next time ?

1 /**************************************************************2 problem:38093 User:rausen4 language:c++5 result:accepted6 time:26248 Ms7 memory:25428 KB8 ****************************************************************/9  Ten#include <cstdio> One#include <cmath> A#include <algorithm> -   - using namespacestd; the Const intN =100005; - Const intM =1000005; - Const intSqrt_n = -; -   + intN, Q, sz; - intA[n]; + intB[sqrt_n], w[n], cnt[n]; A intAns[m]; at   - structData { -     intL, R, a, B, W; -       -InlineBOOL operator< (ConstData &x)Const { -         returnW[l] = = W[x.l]? R < X.r:w[l] <W[X.L]; in     } - } Q[m]; to   +InlineintRead () { -     intx =0; the     CharCH =GetChar (); *      while(Ch <'0'||'9'<ch) $CH =GetChar ();Panax Notoginseng      while('0'<= CH && Ch <='9') { -x = x *Ten+ CH-'0'; theCH =GetChar (); +     } A     returnx; the } +   - intQueryintXinty) { $     intres =0, I; $     if(W[x] = =W[y]) { -          for(i = x; i <= y; + +)i) -             if(Cnt[i]) + +Res; the}Else { -          for(i = w[x] +1; i < w[y]; ++i)WuyiRes + =B[i]; the          for(i = w[x] * SZ; i >= x;--i) -             if(Cnt[i]) + +Res; Wu          for(i = (W[y]-1) * sz +1; I <= y; ++i) -             if(Cnt[i]) + +Res; About     } $     returnRes; - } -   -InlinevoidAddintCintdel) { A     if(!cnt[c]) + +B[w[c]]; +CNT[C] + =del; the     if(!cnt[c])--B[w[c]]; - } $   the intMain () { the     intI, L, R; then = Read (), Q = Read (), Sz = (int) sqrt (N/2); the      for(i =1; I <= N; ++i) -A[i] = Read (), w[i] = (int) (I-1)/sz +1; in      for(i =1; I <= Q; ++i) { theQ[I].L = Read (), Q[I].R = Read (), q[i].a = Read (), q[i].b =read (); theQ[i]. W =i; About     } theSort (q +1, q + q +1); the       the      for(i = L =1, r =0; I <= Q; ++i) { +          for(; l < Q[I].L;) Add (a[l++],-1); -          for(; l > q[i].l;) Add (A[--l],1); the          for(; R < Q[I].R;) Add (A[++r],1);Bayi          for(; r > q[i].r;) Add (a[r--],-1); theAns[q[i]. W] =query (Q[I].A, q[i].b); the     } -      for(i =1; I <= Q; ++i) -printf"%d\n", Ans[i]); the     return 0; the}

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(P.S. Rank 2 is good, Orz rank 1 is not the MO team . ）

[Email protected] 2014/12/21 23:13

Wow, rank 1 god Ben wangyisong1996 take the initiative to find Konjac Konjac play ~ very happy AH

God Ben said he used the card! Often! Number! Mo Team, time unexpectedly only half, simply to kneel >_<

Another: Sz = log (n) is dead, T is miserable

Why write output optimization instead of slower = = unscientific

BZOJ3809 Gty's two-cornered sister sequence