"Bzoj [1878" [Sdoi2009]hh's Necklace

DESCRIPTIONHH has a chain of beautiful shells. HH believes that different shells will bring good luck, so after each walk, he will take out a shell and think about what they mean. HH keeps collecting new shells, so his necklace is getting longer. One day, he suddenly raised a question: how many different shells are included in a certain piece of shell? The question is difficult to answer ... Because the necklace is too long. So he had to ask the wise you to solve the problem. Input first line: An integer n that represents the length of the necklace. Second line: n integers representing the number of shells in the necklace (integers numbered 0 to 1000000, in turn). Line three: An integer m that indicates the number of HH queries. Next m line: two integers per line, L and R (1≤l≤r≤n), indicating the interval of the inquiry. OUTPUTM lines, one integer per line, in turn, to ask the corresponding answer. Sample Input6
1 2 3 4 3 5
3
1 2
3 5
2 6
Sample Output2
2
4
HINT

For 20% of data, n≤100,m≤1000;
For 40% of data, n≤3000,m≤200000;
For 100% of data, n≤50000,m≤200000.

MO Team Water over

1#include <cstdio>2#include <cmath>3#include <algorithm>4 using namespacestd;5 structnode{intNo,l,r,ans;} b[200010];6 intsum[1000010],a[50010],bel[50010];7 intn,m,ans,i;8 BOOLCMP1 (node A,node b) {9     if(BEL[A.L]==BEL[B.L])returna.r<B.R;Ten     returna.l<B.L; One } A   - BOOLCMP2 (node A,node b) { -     returna.no<b.no; the } -   - voidUpdata (intWzintAddintk) { -     if(add==-1&&sum[a[wz]]==1) ans--; +     if(add==1&&sum[a[wz]]==0) ans++; -sum[a[wz]]+=add; + } A   at intMain () { -scanf"%d",&n); -      for(intI=1; i<=n;i++) scanf ("%d",&a[i]); -scanf"%d",&m); -      for(intI=1; i<=m;i++){ -scanf"%d%d",&b[i].l,&B[I].R); inb[i].no=i; -     } to     intblock=sqrt (n); +      for(intI=1; i<=n;i++) bel[i]= (i/block) +1; -Sort (b +1, b+m+1, CMP1); the     intL=1, r=0; *      for(intI=1; i<=m;i++){ $          for(; r<b[i].r;r++) Updata (r+1,1, i);Panax Notoginseng          for(; r>b[i].r;r--) Updata (r,-1, i); -          for(; l>b[i].l;l--) Updata (L-1,1, i); the          for(; l<b[i].l;l++) Updata (l,-1, i); +b[i].ans=ans; A     } theSort (b +1, b+m+1, CMP2); +      for(intI=1; i<=m;i++) printf ("%d\n", B[i].ans); -}

"Bzoj [1878" [Sdoi2009]hh's Necklace