Frequent values Description
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.
Input
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.
Output
For each query, print one line with one integer: The number of occurrences of the most frequent value within the given range.
Sample Input
10 3 -1 -1 1 1 1 1 3 10 10 10 2 3 1 10 5 10 0 Sample Output
1 4 3 Source
Ulm Local 2007
题意： 给你一个大小为n的非递减整型数组，有q次查询，问[L,R]中出现次数最多的数出现的次数
分析：“来自白书第三章” 由于是非递减的，所以相同的数会在一起出现，我们可以从左到右按数值将数组分段。count[i]表示第i段的数出现的次数，num[p],left[p],right[p]分别表示位置p所在的段的编号和左右端点的位置。 每次查询[L,R]的结果为一下3个部分的最大值：从L到L所在段的结束处的元素的个数即 right[L] - L + 1,从R所在段的开始处到R处的元素的个数即 R - lfet[R] +1 ,中间第num[L]+1段到第num[R]-1段的count的最大值。 特殊情况： 如果L,R在同一段，则答案为R-L+1
AC代码：
// UVa11235 Frequent Values // Rujia Liu #include<cstdio> #include<algorithm> #include<vector> using namespace std; const int maxn = 100000 + 5; const int maxlog = 20; // 区间最*大*值 struct RMQ { int d[maxn][maxlog]; void init(const vector<int>& A) { int n = A.size(); for(int i = 0; i < n; i++) d[i][0] = A[i]; for(int j = 1; (1<<j) <= n; j++) for(int i = 0; i + (1<<j) - 1 < n; i++) d[i][j] = max(d[i][j-1], d[i + (1<<(j-1))][j-1]); } int query(int L, int R) { int k = 0; while((1<<(k+1)) <= R-L+1) k++; // 如果2^(k+1)<=R-L+1，那么k还可以加1 return max(d[L][k], d[R-(1<<k)+1][k]); } }; int a[maxn], num[maxn], left[maxn], right[maxn]; RMQ rmq; int main() { int n, q; while(scanf("%d%d", &n, &q) == 2) { for(int i = 0; i < n; i++) scanf("%d", &a[i]); a[n] = a[n-1] + 1; // 哨兵 int start = -1; vector<int> count; for(int i = 0; i <= n; i++) { if(i == 0 || a[i] > a[i-1]) { // 新段开始 if(i > 0) { count.push_back(i - start); for(int j = start; j < i; j++) { num[j] = count.size() - 1; left[j] = start; right[j] = i-1; } } start = i; } } rmq.init(count); while(q--) { int L, R, ans; scanf("%d%d", &L, &R); L--; R--; if(num[L] == num[R]) ans = R-L+1; else { ans = max(R-left[R]+1, right[L]-L+1); if(num[L]+1 < num[R]) ans = max(ans, rmq.query(num[L]+1, num[R]-1)); } printf("%d\n", ans); } } return 0; }