poj 2104 K-th Number

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int N_MAX = 100000 + 4,M_MAX=5000+1;
const int ST_SIZE = (1<<18)-1;
int N, M;
int A[N_MAX];
int l[M_MAX], r[M_MAX], k[M_MAX];
int nums[N_MAX];//nums是排好序的数列，对于其中每一个数进行二分搜索，判断是否符合要求
vector<int>dat[ST_SIZE];

void init(int k,int l,int r) {//节点k,对应区间[l,r)
    if (r - l == 1) {
        dat[k].push_back(A[l]);//节点处的数列只有一个值
    }
    else {
        int k1 = 2 * k + 1, k2 = 2 * k + 2;
        init(k1,l,(l+r)/2);
        init(k2,(l+r)/2,r);
        dat[k].resize(r-l);//先预置空间，才可以进行归并
        merge(dat[k1].begin(), dat[k1].end(),dat[k2].begin(),dat[k2].end(),dat[k].begin());
    }
}

int query(int a,int b,int x,int k,int l,int r) {//查找区间[a,b)中小于等于x的数的个数
    if (l >= b || r <= a) {
        return 0;
    }
    else if (a <= l&&r <= b) {
        return upper_bound(dat[k].begin(),dat[k].end(),x)-dat[k].begin();
    }
    else {
        int lc = query(a, b, x, 2 * k + 1, l, (l + r) / 2);
        int rc = query(a, b, x, 2 * k + 2, (l + r) / 2, r);
        return lc + rc;
    }
}


int main() {
    scanf("%d%d",&N,&M);
    for (int i = 0; i < N; i++) {
        scanf("%d",&A[i]);
        nums[i] = A[i];
    }
    sort(nums,nums+N);
    for (int i = 0; i < M; i++) {
        scanf("%d%d%d",&l[i],&r[i],&k[i]);
        l[i]--, r[i]--;
    }
    init(0,0,N);
    
    for (int i = 0; i < M; i++) {
        int L = l[i], R = r[i]+1, K = k[i];//!!!!!+1别忘
        int lb = -1, ub = N - 1;
        while (ub - lb > 1) {
            int mid = (lb + ub) >> 1;
            int kk = query(L, R, nums[mid], 0, 0, N);
            if (kk >= K)ub = mid;
            else lb = mid;
        }
        printf("%d\n",nums[ub]);
    }
    return 0;
}