POJ-3368 Frequent values

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 nintegers 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 10
2 3
1 10
5 10
0

Sample Output

1
4
3

// RMQ在线查询#include <cstdio>
#include <algorithm>
#include <cmath>#define MAX 100010using namespace std;int maxsum[MAX][20];
int f[MAX];
int a[MAX];
int n, q;void RMQ( int n ) { //预处理->O(nlogn)for( int i = 1; i <= n; i++ ) {maxsum[i][0] = f[i];}for( int j = 1; j < 20; ++j )for( int i = 1; i <= n; ++i )if( i + (1 << j) - 1 <= n ) {maxsum[i][j] = max( maxsum[i][j - 1], maxsum[i + (1 << (j - 1))][j - 1] );}
}int query( int left, int right ) {if( left > right ) return 0;int k = (int)(log( right - left + 1.0 ) / log( 2.0 ));int maxres = max( maxsum[left][k], maxsum[right - ( 1 << k ) + 1][k] );return maxres;
}int main() {while( scanf( "%d", &n ) ) {memset( a, 0, sizeof( a ) );memset( f, 0, sizeof( f ) );memset( maxsum, 0, sizeof( maxsum ) );if( n == 0 ) break;scanf( "%d", &q );for( int i = 1; i <= n; i++ ) {scanf( "%d", &a[i] );if( i == 1 ) f[1] = 1;else if( a[i] == a[i - 1] ) f[i] = f[i - 1] + 1;else f[i] = 1;}RMQ( n );while( q-- ) {int left, right;scanf( "%d%d", &left, &right );// 先寻找前半段满足连续相等值的位置，因为前半段可能会产生影响int t = left;while( t <= right && a[t] == a[t - 1] ) t++;int ans = query( t, right );    // 计算后半段ans = max( ans, t - left );     // 选两者最大值printf( "%d\n", ans );}}return 0;
}

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