You are given a sequence of n integers a1 , a2 , ... , an in non-decreasing order. In addition to that, you are 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 input consists of several test cases. Each test case starts with a line containing two integers n and q (1 ≤ n, q ≤ 100000). The next line contains n integers a1 , ... , an (-100000 ≤ ai ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: ai ≤ ai+1. The following q lines contain one query each, consisting of two integers i and j (1 ≤ i ≤ j ≤ n), which indicate the boundary indices for the
query.
The last test case is followed by a line containing a single 0.
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.
1 #include <iostream> 2 #include <string> 3 #include <cstring> 4 #include <cmath> 5 #include <queue> 6 #include <cstdio> 7 #define MAX 100050 8 #define INF 0X7FFFFFFFF 9 #define LL long long 10 #define mem(a,b) memset(a,b,sizeof(a)) 11 #define L(x) x<<1 12 #define R(x) x<<1|1 13 using namespace std; 14 int data[MAX],vis[MAX],maxx; 15 int n,i,j,k,q; 16 struct node 17 { 18 int le,ri,mid,value; 19 }tree[4*MAX]; 20 21 struct point 22 { 23 int st,en; 24 }cnt[MAX]; 25 void up(int num) 26 { 27 tree[num].value=max(tree[L(num)].value,tree[R(num)].value); 28 } 29 void build(int le,int ri,int root) 30 { 31 tree[root].le=le; 32 tree[root].ri=ri; 33 tree[root].mid=(le+ri)>>1; 34 if(le==ri) 35 { 36 tree[root].value=cnt[le].en-cnt[le].st+1; 37 return; 38 } 39 build(le,tree[root].mid,L(root)); 40 build(tree[root].mid+1,ri,R(root)); 41 up(root); 42 } 43 44 int query(int le,int ri,int root) 45 { 46 if(le<=tree[root].le&&tree[root].ri<=ri) 47 return tree[root].value; 48 if(le>tree[root].mid) 49 return query(le,ri,R(root)); 50 else 51 if(ri<=tree[root].mid) 52 return query(le,ri,L(root)); 53 else 54 { 55 return max(query(le,tree[root].mid,L(root)),query(tree[root].mid+1,ri,R(root))); 56 } 57 } 58 59 int main() 60 { 61 62 while(scanf("%d",&n)==1) 63 { 64 if(n==0) 65 break; 66 scanf("%d",&q); 67 for(i=1;i<=n;i++) 68 scanf("%d",&data[i]); 69 memset(vis,0,sizeof(vis)); 70 int t=1; 71 vis[1]=t; 72 cnt[t].st=1; 73 if(n==1) 74 cnt[t].en=1; 75 for(i=2;i<=n;i++) 76 { 77 if(data[i]!=data[i-1]) 78 { 79 cnt[t].en=i-1; 80 t++; 81 cnt[t].st=i; 82 vis[i]=t; 83 } 84 else 85 vis[i]=t; 86 } 87 vis[n]=t; 88 cnt[t].en=n; 89 build(1,t,1); 90 int a,b; 91 for(i=1;i<=q;i++) 92 { 93 scanf("%d%d",&a,&b); 94 if(vis[a]==vis[b]) 95 { 96 printf("%d ",b-a+1); 97 continue; 98 } 99 if(vis[b]-vis[a]==1) 100 { 101 printf("%d ",max(cnt[vis[a]].en-a+1,b-cnt[vis[b]].st+1)); 102 continue; 103 } 104 int p=max(cnt[vis[a]].en-a+1,b-cnt[vis[b]].st+1); 105 printf("%d ",max(p,query(vis[a]+1,vis[b]-1,1))); 106 } 107 } 108 }