Kth number

                                                Time Limit: 15000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                        Total Submission(s): 10682    Accepted Submission(s): 3268

Problem Description

Give you a sequence and ask you the kth big number of a inteval.

Input

The first line is the number of the test cases.
For
each test case, the first line contain two integer n and m (n, m <=
100000), indicates the number of integers in the sequence and the number
of the quaere.
The second line contains n integers, describe the sequence.
Each of following m lines contains three integers s, t, k.
[s, t] indicates the interval and k indicates the kth big number in interval [s, t]

Output

For each test case, output m lines. Each line contains the kth big number.

Sample Input

1

10 1

1 4 2 3 5 6 7 8 9 0

1 3 2

Sample Output

2

Source

HDU男生专场公开赛——赶在女生之前先过节（From WHU）

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 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cctype>
 5 #include<cmath>
 6 #include<cstring>
 7 #include<map>
 8 #include<stack>
 9 #include<set>
10 #include<vector>
11 #include<algorithm>
12 #include<string.h>
13 #define ll long long
14 #define LL unsigned long long
15 using namespace std;
16 const int INF=0x3f3f3f3f;
17 const double eps=0.0000000001;
18 const int N=100000+10;
19 struct node{
20     int left,right;
21     int val;
22 }tree[N*40];
23 int cnt;
24 int a[N],b[N];
25 int root[N];
26 int build(int l,int r){
27     int pos=++cnt;
28     tree[pos].val=0;
29     if(l==r)return pos;
30     int mid=(l+r)>>1;
31     tree[pos].left=build(l,mid);
32     tree[pos].right=build(mid+1,r);
33     return pos;
34 }
35 void update(int pre,int &now,int x,int l,int r){
36     tree[++cnt]=tree[pre];
37     now=cnt;
38     tree[now].val++;
39     if(l==r)return;
40     int mid=(l+r)>>1;
41     if(x<=mid){
42         update(tree[pre].left,tree[now].left,x,l,mid);
43     }
44     else{
45         update(tree[pre].right,tree[now].right,x,mid+1,r);
46     }
47
48 }
49 int query(int L,int R,int k,int l,int r){
50     if(l==r) return l;
51     int mid=(l+r)>>1;
52     int sum=tree[tree[R].left].val-tree[tree[L].left].val;
53     if(k<=sum){
54         return query(tree[L].left,tree[R].left,k,l,mid);
55     }
56     else{
57         return query(tree[L].right,tree[R].right,k-sum,mid+1,r);
58     }
59
60 }
61 int main()
62 {
63     int t;
64     scanf("%d",&t);
65     while(t--){
66         memset(root,0,sizeof(root));
67         memset(a,0,sizeof(a));
68         memset(b,0,sizeof(b));
69         int n,m;
70         scanf("%d%d",&n,&m);
71         cnt=0;
72         for(int i=1;i<=n;i++) {
73             scanf("%d",&a[i]);
74             b[i]=a[i];
75         }
76         sort(b+1,b+1+n);
77         int tt = unique(b+1,b+1+n)-b-1;
78         for(int i=1;i<=n;i++) {
79             a[i]=lower_bound(b+1,b+1+tt,a[i])-b;//二分找到a[i]的位置
81             update(root[i-1],root[i],a[i],1,tt);//root[i-1]表示上一个版本的线段树
82         }
83         for(int i=1;i<=m;i++) {
84             int l,r,k;
85             scanf("%d%d%d",&l,&r,&k);
86             int ans=query(root[l-1],root[r],k,1,tt); //ans是第k个数的位置
87             printf("%d\n", b[ans]);//因为询问的是哪个数,所以要b[ans]
88         }
89     }
90     return 0;
91 }