51nod-1686 K-wide interval (binary + ruler)

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Section K Range

Base time limit: 1 seconds space limit: 131072 KB

Defines the number of times the value of an interval appears for its majority.
The number of n is given, and the value of K is the number of the value after sorting all the intervals.

Input

The first row two numbers n and K (1<=n<=100000,k<=n* (n-1)/2) The second row n number, 0<= each number <2^31

Output

A number represents the answer.

Input example

4 21 2) 3 2

Output example

2

Test instructions

Ideas:

First, the arrays are separated into [1,n] number, the attention is equal;
The second half of the answer T,check the interval of the number of regions greater than or equal to T, and (can be set to a range of more than or equal to the range of the T)
Check by using the ruler method, in order to enumerate the left end of each interval I, and then find the minimum interval right end point, with the left end of the legal interval of the scheme number is n (r-1);
Complexity of O (NLOGN)

AC Code:

//#include <bits/stdc++.h>#include <vector>#include<iostream>#include<queue>#include<cmath>#include<map>#include<cstring>#include<algorithm>#include<cstdio>using namespacestd;#defineRiep (n) for (int i=1;i<=n;i++)#defineRIOP (n) for (int i=0;i<n;i++)#defineRJEP (n) for (int j=1;j<=n;j++)#defineRJOP (n) for (int j=0;j<n;j++)#defineMST (SS,B) memset (ss,b,sizeof (ss));typedefLong Longll;template<classT>voidRead (t&num) {    CharCH;BOOLf=false;  for(Ch=getchar (); ch<'0'|| Ch>'9'; f= ch=='-', ch=GetChar ());  for(num=0; ch>='0'&&ch<='9'; num=num*Ten+ch-'0', ch=GetChar ()); F&& (num=-num);}intstk[ -], Tp;template<classT> Inlinevoidprint (T p) {if(!p) {Puts ("0");return; }  while(p) stk[++ TP] = p%Ten, p/=Ten;  while(TP) Putchar (stk[tp--] +'0'); Putchar ('\ n');}ConstLL mod=1e9+7;Const DoublePi=acos (-1.0);ConstLL inf=1e10;Const intn=1e5+ the;intn,k;intA[n],flag[n];structnode{intA,id,ans;} Po[n];intCMP (node X,node y) {if(X.A==Y.A)returnx.id<y.id; returnx.a<y.a;}intcmp1 (node X,node y) {returnx.id<y.id;}intCheckintx) {MST (flag,0); LL ans=0; intnum=0; intR=0; Riep (n) { while(1)        {            intpos=Po[r].ans; if(pos!=-1)if(flag[pos]>=x| | R>n) Break; R++; POS=Po[r].ans; Flag[pos]++; } ans=ans+n-(R1); Flag[po[i].ans]--; }    if(ans>=k)return 1; return 0;}intMain () {read (n); read (k); Riep (n) Read (PO[I].A), Po[i].id=i; Sort (PO+1, po+n+1, CMP); po[0].ans=po[0].a=-1; Riep (n) {if(po[i].a!=po[i-1].A) po[i].ans=i; Elsepo[i].ans=po[i-1].ans; } Sort (PO+1, po+n+1, CMP1); intL=1, r=N;  while(l<=r) {intMid= (l+r) >>1; if(Check (mid)) l=mid+1; Elser=mid-1; } print (L-1); return 0;}

51nod-1686 K-wide interval (binary + ruler)