Dlutoj #1394 Magic Questions

Transmission Door

Time Limit:3 Sec Memory limit:128 MB

Description

Alice likes playing games. So she'll take part in the movements of M within N days, and each game was represented in an integer between 1 and M. Roo Mmates has Q magic questions:how many different kinds of games does Alice participate between Lth day and Rth Day (includ ing Lth Day and Rth day)?

Input

You'll be given a number of cases; Each case contains blocks of several lines. The first line contains 2 numbers of N and M. The second line contains n numbers implying the game numbers this Alice take part in within n days. The third line contains a number of Q. Then Q lines is entered. Each line contain the numbers of L and R.

1≤n,m,q≤100000

Output

There should is Q output lines per test case containing Q answers required.

Sample INPUT5 3 1 2 3 2 2 3 1 4 2 4 1 5Sample Output3 2 3HINT

This is the school game of this year's K, a classic topic, but the scene did not a.

Online can be used by the Chairman tree, not yet. There is a clever use of the array of the offline solution, relatively good writing.

The main points are:

1. Sort the query by the right endpoint from small to large.

2. Record the position of each number in the last occurrence. When the number reappears, it removes the count from its last occurrence position.

Implementation:

The subject is a double pointer.

#include <bits/stdc++.h>using namespacestd;Const intN (1e5+5);intN, M, Q, A[n], pos[n], bit[n], ans[n];voidAddintXintv) {     for(; x<=n; Bit[x]+=v, x+=x&-x);}intSumintx) {    intres=0;  for(; x; res+=bit[x], x-=x&-x); returnRes;}structp{intL, R, id; P (intLintRintID): L (L), R (r), ID (ID) {} P () {}; BOOL operator< (ConstP&AMP;B)Const{returnr<B.R;}} P[n];intMain () {//Ios::sync_with_stdio (false);     for(; ~SCANF ("%d%d", &n, &m); ){         for(intI=1; i<=n; i++) scanf ("%d", A +i); scanf ("%d", &P);  for(intL, R, i=0; i<q; i++) scanf ("%d%d", &l, &r), p[i]={L, R, i}; Sort (p, p+Q);//Error-pronememset (bit,0,sizeof(bit)); memset (POS,0,sizeof(POS));  for(intI=1, j=0K j<q&&i<=n; ){//Error-prone             for(; i<=p[j].r; i++){                if(Pos[a[i]]) Add (Pos[a[i]],-1); Pos[a[i]]=i; Add (i,1); }             for(k=j; k<q&&p[k].r==p[j].r; k++)//Error-proneAns[p[k].id]=sum (P[K].R)-sum (p[k].l-1); J=K; }         for(intI=0; i<q; i++) printf ("%d\n", Ans[i]);//Error-prone    }    return 0;}

Dlutoj #1394 Magic Questions