HDU 5273 Dylans loves sequence (interval inverse logarithm-mo team algorithm)

n<=1000,q<=100000, in order to find the logarithm of the interval, from [l,r] can obviously log (n) time to move to [l-1,r],[l+1,r],[l,r-1],[l,r+1], then you can use MO team offline

The complexity is probably O (n*sqrt (n) *log2 (n)), but can be violent to raise points, and then back to the statistics, then O (1) answer, but N is no longer able to solve the big, even if the n<=1e5 can quickly get answers, open-O optimization, 1e6 can also be quickly answered

#include <bits/stdc++.h>using namespace std;const int Maxn=1e5+5;const int maxm=1e5+5;/* number range is larger, discretized to [1,n] just a few There are 1000, q queries [l,r] between the reverse logarithm if [l,r] to seek [l,r+1] or [l,r-1], as long as statistics a[i] contribution is good a[i] contribution can be obtained by querying the tree array log2 (1000) time to get this property is suitable for the MO team algorithm, the overall The Complexity O (n*sqrt (n) *log2 (n) +q) can be quickly answered even if n<=1e5, can also quickly get answers */void Jia (int x) {printf ("Add%d\n", x);} void Jian (int x) {printf ("--%d\n", x);}    void read (int &res) {char c=getchar ();    while (!isdigit (c)) C=getchar ();    res=0; while (IsDigit (c)) res=res*10+c-' 0 ', C=getchar ();} int a[maxn],b[maxn],c[maxn],d[maxn];int n,q,block;int ql[maxm],qr[maxm],qq[maxm],ans[maxm];inline int lowbit (int x) { Return x& (-X);} BOOL Cmp1 (int I,int j) {return a[i]<a[j];}    void Add (int x,int val) {//if (val>0) Jia (x); else jian (x); while (x<=n) c[x]+=val,x+=lowbit (x);}    int query (int x) {int res=0;    while (x>0) res+=c[x],x-=lowbit (x); return res;}    BOOL Cmp2 (int i,int j) {int q1=ql[i]/block;    int q2=ql[j]/block; if (Q1!=Q2) return q1<q2;    if (Qr[i]!=qr[j]) return qr[i]<qr[j]; return qr[i]<qr[j];}    int main () {freopen ("in", "R", stdin);    Freopen ("WA", "w", stdout);    Read (n); read (q);    BLOCK=SQRT (n);    for (int i=1;i<=n;i++) read (A[i]), b[i]=i;    Sort (B+1,B+1+N,CMP1);    int tot=0,last=-1;        for (int i=1;i<=n;i++) {if (a[b[i]]==last) D[b[i]]=tot;    else D[b[i]]=++tot,last=a[b[i]];    } for (int i=0;i<q;i++) read (Ql[i]), read (Qr[i]), qq[i]=i;    Sort (QQ,QQ+Q,CMP2);    int l=1,r=1,res=0;    Add (d[1],1);        for (int i=0;i<q;i++) {int l=ql[qq[i]],r=qr[qq[i]];            while (R<r) {res+=r-l+1-query (d[r+1]);            Add (d[r+1],1);        r++;            } while (R>r) {res-=r-l+1-query (d[r]);            Add (d[r],-1);        r--;            } while (L<l) {res-=query (d[l]-1);            Add (d[l],-1);        l++;            } while (L>l) {res+=query (d[l-1]-1);            Add (d[l-1],1);        l--; } ANs[qq[i]]=res;    } for (int i=0;i<q;i++) printf ("%d\n", Ans[i]); return 0;}

　　

HDU 5273 Dylans loves sequence (interval inverse logarithm-mo team algorithm)