K-th Number 【POJ - 2104】【可持久化线段树】

题目链接

因为这道题没有删除修改之类的，所以很多人会用离散化之后的线段树来做，但是实际上（可能是我懒得去做离散化这个操作了），然后就是直接写可持久化线段树，区间的长度就是int的从最小到最大的长度，然后记录的是size，我们最后直接返回的是对应的位置即可。

---

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <string>
 #include <cstring>
 #include <algorithm>
 #include <limits>
 #include <vector>
 #include <stack>
 #include <queue>
 #include <set>
 #include <map>
 #define lowbit(x) ( x&(-x) )
 #define pi 3.141592653589793
 #define e 2.718281828459045
 #define INF 0x3f3f3f3f
 #define HalF (l + r)>>1
 #define lsn rt<<1
 #define rsn rt<<1|1
 #define Lson lsn, l, mid
 #define Rson rsn, mid+1, r
 #define QL Lson, ql, qr
 #define QR Rson, ql, qr
 #define myself rt, l, r
 using namespace std;
 typedef unsigned long long ull;
 typedef long long ll;
 const int maxN = 1e5 + , _UP = , _DOWN = -;
 int N, M, root[maxN], tot, siz[ * maxN], lc[ * maxN], rc[ * maxN];
 inline void insert(int &rt, int old, ll l, ll r, int qx)
 {
     rt = ++tot;
     siz[rt] = siz[old] + ; lc[rt] = lc[old];   rc[rt] = rc[old];
     if(l == r) return;
     ll mid = HalF;
     if(qx <= mid) insert(lc[rt], lc[old], l, mid, qx);
     else insert(rc[rt], rc[old], mid + , r, qx);
 }
 inline ll query(int now, int old, ll l, ll r, int k)
 {
     if(l == r) return l;
     ll mid = HalF;
     if(k <= siz[lc[now]] - siz[lc[old]]) return query(lc[now], lc[old], l, mid, k);
     else return query(rc[now], rc[old], mid + , r, k - siz[lc[now]] + siz[lc[old]]);
 }
 inline void init()
 {
     for(int i=; i<=tot; i++) root[i] = ;
     tot = ;
 }
 int main()
 {
     while(scanf("%d%d", &N, &M) != EOF)
     {
         init();
         for(int i=, val; i<=N; i++)
         {
             scanf("%d", &val);
             insert(root[i], root[i - ], _DOWN, _UP, val);
         }
         int l, r, k;
         while(M--)
         {
             scanf("%d%d%d", &l, &r, &k);
             printf("%lld\n", query(root[r], root[l-], _DOWN, _UP, k));
         }
     }
     return ;
 }