Simple POJ 3368 frequent values,rmq.

Description

You are given a sequence of n integers a1, A2, ..., an in non-decreasing order. In addition to, given several queries consisting of indices I and J (1≤i≤j≤n ). For each query, determine the most frequent value among the integers ai, ..., AJ.

The question is what is the highest frequency within the range.

With rmq words, the statistics of a number and the previous frequency, and then ask the time is to find the maximum value of two parts, the first part is the same number as X, the second part is the first and x different from the part of Y, with RMQ for the second part.

Then unable to spit groove himself, Logn array opened small, the result has been WA, really enough silly.

The code is as follows:

//The ━━━━━━ of gods and Beasts ━━━━━━//┏┓┏┓//┏┛┻━━━━━━━┛┻┓//┃┃//┃━┃//████━████┃//┃┃//┃┻┃//┃┃//┗━┓┏━┛//┃┃//┃┃//┃┗━━━┓//┃┣┓//┃┏┛//┗┓┓┏━━━━━┳┓┏┛//┃┫┫┃┫┫//┗┻┛┗┻┛////━━━━━━ Feel the ━━━━━━ of Meng Meng//author:whywhy//Created time:2015 July 17 Friday 17:11 45 seconds//File name:3368.cpp#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>#include<vector>#include<queue>#include<Set>#include<map>#include<string>#include<math.h>#include<stdlib.h>#include<time.h>using namespacestd;Const intmaxn=100005;intdp[maxn][ -];intLOGN[MAXN];//!!!voidInitintNintnum[]) {logn[0]=-1;  for(intI=1; i<=n;++i) {Logn[i]=logn[i-1]+ ((i& (i-1))==0); dp[i][0]=Num[i]; }     for(intj=1; j<=logn[n];++j) for(intI=1; i+ (1&LT;&LT;J)-1<=n;++i) dp[i][j]=max (dp[i][j-1],dp[i+ (1<< (J-1))][j-1]);}intRMQ (intAintb) {    if(a>b)return 0; intk=logn[b-a+1]; returnMax (dp[a][k],dp[b-(1&LT;&LT;K) +1][k]);}intNUM[MAXN];intREM[MAXN],WEI[MAXN];intn,q;intMain () {//freopen ("In.txt", "R", stdin); //freopen ("OUT.txt", "w", stdout);        intb;  while(~SCANF ("%d", &n) &&N) {scanf ("%d",&p); rem[1]=1;  for(intI=1; i<=n;++i) scanf ("%d",&Num[i]);  for(intI=2; i<=n;++i)if(num[i]==num[i-1]) Rem[i]=rem[i-1]+1; ElseRem[i]=1; Wei[n]=o;  for(intj=m-1; i>=1;--i)if(num[i]==num[i+1]) Wei[i]=wei[i+1]; ElseWei[i]=i;        Init (N,REM);  while(q--) {scanf ("%d%d",&a,&b); printf ("%d\n", Max (RMQ (wei[a]+1, b), Min (wei[a],b)-a+1)); }    }        return 0;}

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Simple POJ 3368 frequent values,rmq.