Codeforces 220b-little Elephant and array offline tree array

This problem can is solve in simpler o(nsqrtn) solution, but I'll describe O( nlogn) one.

We'll solve this problem in offline. for eachx (0?≤? ) x? <? n)we should keep all the queries so end inx. Iterate thatxFrom 0 to n?-? 1. Also we need to keep some arrayDsuch .x  Dl? +? D l. +?1? +?...? +? D x Would be is the answer for query[l; X]. To keepDcorrect, before the processing all queries so end inx, we need to updateD. LetTBeing the current integer inA, i. E. Ax , and VectorPBe the list of indices of previous occurences ofT(0-based numeration of Vector). Then, if| P|? ≥? T , you need to add1 to DP[| P|? -? T] , because this position are now the first ( from right) that contains exactlyToccurences ofT. After, if| P|? ? T , you need to subtract2 from DP[| P|? -? T?-? 1] , in order to close current interval and cancel previous. Finally, if| P|? ? t? +?1, then you need additionally add1 to DP[| P|? -? T?-? 2] To cancel previous close of the interval.

#include <cstdio> #include <cstring> #include <algorithm> #include <string> #include < iostream> #include <iomanip> #include <cmath> #include <map> #include <set> #include < queue>using namespace std; #define LS (RT) Rt*2#define RS (RT) Rt*2+1#define ll long long#define ull unsigned long long#de Fine Rep (I,s,e) for (int. i=s;i<e;i++) #define REPE (i,s,e) for (int i=s;i<=e;i++) #define CL (A, B) memset (A,b,sizeof (A )) #define IN (s) freopen (S, "R", stdin) #define OUT (s) freopen (S, "w", stdout) const int MAXN = 1e5+100;int N,m;int a[maxn],c[    maxn],ans[maxn];struct query{int l,r,id; BOOL operator < (const Query &t) const {return R&LT;T.R;}} q[maxn];inline int lowbit (int x) {return x& (-X);}        void Add (int i, int v) {while (i<=n) {c[i]+=v;    I+=lowbit (i);    }}int sum (int x) {int ret=0;        while (x>0) {ret+=c[x];    X-=lowbit (x); } return ret;}    int main () {int sz; while (~SCANF ("%d%d", &n,&amP;m)) {vector<int>data[maxn];        CL (c,0);        for (int i=1;i<=n;i++) scanf ("%d", &a[i]);            for (int i=1;i<=m;i++) {scanf ("%d%d", &AMP;Q[I].L,&AMP;Q[I].R);        Q[i].id=i;        } sort (q+1,q+1+m);                for (int i=1,k=1;i<=n;i++) {if (a[i]<=n) {data[a[i]].push_back (i);                Sz=data[a[i]].size ();                    if (Sz>=a[i]) {Add (data[a[i]][sz-a[i]],1);                    if (Sz>a[i]) Add (data[a[i]][sz-a[i]-1],-2);                if (sz>a[i]+1) Add (data[a[i]][sz-a[i]-2],1); }} while (Q[k].r==i && k<=m) {ans[q[k].id]=sum (Q[K].R)-sum (q[k                ].L-1);            k++;    }} for (int i=1;i<=m;i++) printf ("%d\n", Ans[i]); } return 0;}

Code for debugging understanding and adding gaze

#include <cstdio> #include <cstring> #include <algorithm> #include <string> #include < iostream> #include <iomanip> #include <cmath> #include <map> #include <set> #include < queue>using namespace std; #define LS (RT) Rt*2#define RS (RT) Rt*2+1#define ll long long#define ull unsigned long long#de Fine Rep (I,s,e) for (int. i=s;i<e;i++) #define REPE (i,s,e) for (int i=s;i<=e;i++) #define CL (A, B) memset (A,b,sizeof (A )) #define IN (s) freopen (S, "R", stdin) #define OUT (s) freopen (S, "w", stdout) const int MAXN = 1e5+100;int N,m;int a[maxn],c[    maxn],ans[maxn];struct query{int l,r,id; BOOL operator < (const Query &t) const {return R&LT;T.R;}} q[maxn];inline int lowbit (int x) {return x& (-X);}        void Add (int i, int v) {while (i<=n) {c[i]+=v;    I+=lowbit (i);    }}int sum (int x) {int ret=0;        while (x>0) {ret+=c[x];    X-=lowbit (x); } return ret;}    int main () {int sz; while (~SCANF ("%d%d", &n,&amP;m)) {vector<int>data[maxn];        CL (c,0);        for (int i=1;i<=n;i++) scanf ("%d", &a[i]);            for (int i=1;i<=m;i++) {scanf ("%d%d", &AMP;Q[I].L,&AMP;Q[I].R);        Q[i].id=i;        } sort (q+1,q+1+m);                for (int i=1,k=1;i<=n;i++) {if (a[i]<=n) {data[a[i]].push_back (i);                Sz=data[a[i]].size ();                    if (Sz>=a[i]) {Add (data[a[i]][sz-a[i]],1),///From right to left A[i] times A[i] position +1                    if (Sz>a[i]) Add (data[a[i]][sz-a[i]-1],-2); From right to left a[i]+1 A[i] Position-2,//due to sz==a[i], this position has been added 1.                    This time I was read.                    The position of a[i] from right to left A[i] is also + 1. Then the position of the query a[i]+1 appears a[i] to I. The answer is -2+1+1=0,//query a[i] The position of a[i] to I, the answer is 1 if (sz>a[i]+1) Add (Data[a[i]][sz-a[i]-2],                    1);  From right to left a[i]+2 times appears a[i] position +1, before being + 1-2, so becomes 0                  These three lines of code are maintained, from the current I to the left number, a[i] times appear a[i] position always 1//a[i]+1 occurrences A[i] position always 1, a[i]+2 and many other times the position is always 0, which                The query results for the interval with I as the right endpoint are all right}} while (Q[k].r==i && k<=m) {                printf ("#i =%d#\n", i);                for (int j=0;j<=n;j++) printf ("c[%d]=%d\n", J,c[j]);                Ans[q[k].id]=sum (Q[K].R)-sum (Q[K].L-1);            k++;    }} for (int i=1;i<=m;i++) printf ("%d\n", Ans[i]); } return 0;}

Codeforces 220b-little Elephant and array offline tree array