#寒假1# R - Frequent values

题目

https://vjudge.net/problem/POJ-3368
就是：求一段区间中出现次数最多的数字有几个

思路

这样，查询的时候，先求两个端点的max（要减去之前&之后的个数），再求其他区间的max，之后取max

代码

# include<iostream>
# include<algorithm>
#include <cstring>
#include <cmath>

#define MAXX 100005
using namespace std;
//typedef long long ll;
int a[MAXX], leftt[MAXX], rightt[MAXX];
int pos[MAXX], f[MAXX][50];
int cnt;

void init()
{
    memset(f, 0, sizeof f);
    for (int i = 0; i <= cnt; ++i)
    {
        f[i][0] = rightt[i] - leftt[i] + 1;
    }
    for (int j = 1; (1 << j) <= cnt; ++j)
    {
        int l = 1 << j;
        for (int i = 0; i + l - 1 <= cnt; ++i)
        {
            f[i][j] = max(f[i][j - 1], f[i + (l >> 1)][j - 1]);
            // cout<<f[i][j]<<' ';
        }
        // cout<<endl;
    }
}

int query(int ll, int rr)
{
    int k = log(rr - ll + 1.0) / log(2.0);
    return max(f[ll][k], f[rr - (1 << k) + 1][k]);
}

int main()
{
    int n, q;
    while (cin >> n >> q && n)
    {
        memset(a, 0, sizeof a);
        memset(leftt, 0, sizeof leftt);
        memset(rightt, 0, sizeof rightt);
        a[0] = MAXX;
        cnt = 0;
        for (int i = 1; i <= n; ++i)
        {
            cin >> a[i];
            if (a[i] != a[i - 1])
            {
                cnt++;
                pos[i] = cnt;
                leftt[cnt] = rightt[cnt] = i;
            } else
            {
                pos[i] = cnt;
                rightt[cnt]++;
            }
        }
        /*for (int i = 1; i <= cnt; ++i)
        {
            cout<<leftt[i]<<' ';
        }
        cout<<endl;
        for (int i = 1; i <= cnt; ++i)
        {
            cout<<rightt[i]<<' ';
        }*/
        init();
        int ll, rr;
        int ans;
        while (q--)
        {
            scanf("%d%d", &ll, &rr);
            if (pos[ll] == pos[rr])
                ans = rr - ll + 1;
            else if (pos[rr] - pos[ll] == 1)
                ans = max(rightt[pos[ll]] - ll + 1, rr - leftt[pos[rr]] + 1);  
                //憨憨本憨。。这里的[]括号括错了。。。。改了一个又忘记改下面的了。。。。
            else
            {
                ans = max(rightt[pos[ll]] - ll + 1, rr - leftt[pos[rr]] + 1);
                ans = max(ans, query(pos[ll] + 1, pos[rr] - 1));
            }
            cout << ans << endl;
        }
    }
    return 0;
}