HDU 2665(主席树，无修改第k小）

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）

说实话  我不知道我该怎么去解释 因为这就是主席树的板子

无区间修改的主席树 不解释

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cctype>
 #include<cmath>
 #include<cstring>
 #include<map>
 #include<stack>
 #include<set>
 #include<vector>
 #include<algorithm>
 #include<string.h>
 #define ll long long
 #define LL unsigned long long
 using namespace std;
 const int INF=0x3f3f3f3f;
 const double eps=0.0000000001;
 const int N=+;
 struct node{
     int left,right;
     int val;
 }tree[N*];
 int cnt;
 int a[N],b[N];
 int root[N];
 int build(int l,int r){
     int pos=++cnt;
     tree[pos].val=;
     if(l==r)return pos;
     int mid=(l+r)>>;
     tree[pos].left=build(l,mid);
     tree[pos].right=build(mid+,r);
     return pos;
 }
 void update(int pre,int &now,int x,int l,int r){
     tree[++cnt]=tree[pre];
     now=cnt;
     tree[now].val++;
     if(l==r)return;
     int mid=(l+r)>>;
     if(x<=mid){
         update(tree[pre].left,tree[now].left,x,l,mid);
     }
     else{
         update(tree[pre].right,tree[now].right,x,mid+,r);
     }

 }
 int query(int L,int R,int k,int l,int r){
     if(l==r) return l;
     int mid=(l+r)>>;
     int sum=tree[tree[R].left].val-tree[tree[L].left].val;
     if(k<=sum){
         return query(tree[L].left,tree[R].left,k,l,mid);
     }
     else{
         return query(tree[L].right,tree[R].right,k-sum,mid+,r);
     }

 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--){
         memset(root,,sizeof(root));
         memset(a,,sizeof(a));
         memset(b,,sizeof(b));
         int n,m;
         scanf("%d%d",&n,&m);
         cnt=;
         for(int i=;i<=n;i++) {
             scanf("%d",&a[i]);
             b[i]=a[i];
         }
         sort(b+,b++n);
         int tt = unique(b+,b++n)-b-;
           root[0]=build(1,tt);
         for(int i=;i<=n;i++) {
             a[i]=lower_bound(b+,b++tt,a[i])-b;//二分找到a[i]的位置
             update(root[i-],root[i],a[i],,tt);//root[i-1]表示上一个版本的线段树
         }
         for(int i=;i<=m;i++) {
             int l,r,k;
             scanf("%d%d%d",&l,&r,&k);
             int ans=query(root[l-],root[r],k,,tt);
             printf("%d\n", b[ans]);
         }
     }
     return ;
 }