POJ-2761-树状数组+第k小数+离散化

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原创最新推荐文章于 2026-07-10 12:19:21 发布 1.3k 阅读 AI 写同款· GEO 优化 ›

【MATLAB/Simulink】从模型到飞行：基于PX4与Rflysim的无人机MBD实战指南本文详细介绍了基于MATLAB/Simulink与PX4、Rflysim的无人机模型开发与飞行控制实战指南。通过MBD（基于模型设计）方法，从建模、仿真到代码生成与硬件部署，实现无人机开发的高效闭环。文章涵盖开发环境搭建、动力学建模、控制算法实现及测试验证全流程，为无人机开发者提供实用技巧与优化建议。阅读详情

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;

const int inf = 0x3f3f3f3f;
const int maxn = 100009;
const double esp = 1e-6;

int n, m, num;
int bef[maxn];
int aft[maxn];
int ans[50009];
int C[maxn];

struct Q
{
    int i, j, k, id;
}q[50009];

int cmp(Q a, Q b) //查询 先安起始位置从小到大排序；再按结束为止从小到大排序
{
    if(a.i==b.i) return a.j<b.j;
    else return a.i<b.i;
}

inline int lowbit(int x)
{
    return x&(-x);
}

void add(int i, int val)
{
    for(; i<=num; i+=lowbit(i))
        C[i]+=val;
}

int query(int i)
{
    int a = 0;
    for(; i>0; i-=lowbit(i))
        a += C[i];
    return a;
}

int find(int val) //二分查找离散化后的编号
{
    int l = 1, r = num, mid;
    while(1)
    {
        mid = (l+r)/2;
        if( aft[mid]==val ) return mid;
        else if( aft[mid]>val ) r = mid-1;
        else l = mid+1;
    }
}

int answer(int val) 
{
    int l=1, r=num, mid;
    while(l<r)
    {
        mid = (l+r)/2;
        int t = query(mid);
        if( t < val) l = mid+1;
        else r = mid;
    }
    return aft[l];
}

void solve()
{
    memset(C, 0, sizeof(C));
    int l = q[1].i, r = q[1].j;
    for(int i = l; i<=r; i++) 
    {
        add( find( bef[i] ), 1 );
    }
    ans[ q[1].id ] = answer( q[1].k );
    for(int i=2; i<=m; i++)
    {
        for(int j=q[i-1].i; j<q[i].i; j++)
            add( find(bef[j]), -1 );
        if( q[i-1].j<q[i].j )
        {
            for(int j=q[i-1].j+1; j<=q[i].j; j++)
                add( find(bef[j]), 1 );
        }
        else
        {
            for(int j=q[i-1].j; j>q[i].j; j--)
                add( find(bef[j]), -1 );
        }
        ans[ q[i].id ] = answer( q[i].k );
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i=1; i<=n; i++)
    {
        scanf("%d", &bef[i]);
        aft[i] = bef[i];
    }
    sort(aft+1, aft+n+1);
    num = 0;
    for(int i=1; i<=n; i++)
    if(aft[i]!=aft[i-1])  //离散化，去重
        aft[++num] = aft[i];

    for(int i=1; i<=m; i++)
    {
        scanf("%d%d%d", &q[i].i, &q[i].j, &q[i].k);
        q[i].id = i;
    }
    sort(q+1, q+m+1, cmp);
    solve();
    for(int i=1; i<=m; i++)
        printf("%d\n", ans[i]);
	return 0;
}