hdu 1806 Frequent values（二分+线段树）

Frequent values

Time Limit: 1000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1050 Accepted Submission(s): 375

Problem Description

You are given a sequence of n integers a₁ , a₂ , ... , a_n 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 a_i , ... , a_j .

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 , ... , a_n(-100000 ≤ a_i ≤ 100000, for each i ∈ {1, ..., n}) separated by spaces. You can assume that for each i ∈ {1, ..., n-1}: a_i ≤ a_i+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 10 102 31 105 100

Sample Output

143
样例：
其余你应该知道怎么做了！
#include <iostream>#include <cstdio>#include <vector>#include <cstring>#include <cstdlib>using namespace std;const int maxn = 100010;int v[maxn] , sum[maxn];struct tree{int l , r , Max;tree(int a = 0 , int b = 0 , int c = 0){l = a , r = b , Max = c;}}a[4*maxn];int n , q , cnt;void build(int l , int r , int k){a[k].l = l;a[k].r = r;if(l == r){a[k].Max = v[l];}else{int mid = (l+r)/2;build(l , mid , 2*k);build(mid+1 , r , 2*k+1);a[k].Max = max(a[2*k].Max , a[2*k+1].Max);}}int querry(int l , int r , int k){if(l <= a[k].l && a[k].r <= r){return a[k].Max;}else{int mid = (a[k].l+a[k].r)/2;if(mid >= r) return querry(l , r , 2*k);else if(mid < l) return querry(l , r , 2*k+1);else return max(querry(l , mid , 2*k) , querry(mid+1 , r , 2*k+1));}}void initial(){cnt = 1;for(int i = 0; i < maxn; i++){v[i] = 0;sum[i] = 0;}}void readcase(){cin >> q;int num , pre;cin >> pre;v[1] = 1;for(int i = 1; i < n; i++){scanf("%d" , &num);if(num == pre) v[cnt]++;else{cnt++;v[cnt]++;pre = num;}}for(int i = 1; i <= cnt; i++){sum[i] = sum[i-1]+v[i];}}int binary_search(int k){int l = 1 , r = cnt;while(l < r){int mid = (l+r)/2;if(sum[mid] >= k){r = mid;}else{l = mid+1;}}return l;}void computing(){build(1 , cnt , 1);int i , j , ans;while(q--){//cin >> i >> j;scanf("%d%d" , &i ,&j);int L = binary_search(i);int R = binary_search(j);if(L == R) printf("%d\n" , j-i+1);else{ans = max(sum[L]-i+1 , j-sum[R-1]);if(L+1 <= R-1) ans = max(ans , querry(L+1 , R-1 , 1));printf("%d\n" , ans);}}}int main(){while(cin >> n && n){initial();readcase();computing();}return 0;}