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代码：
#include
#include
using namespace std;
const int MAXN = 1e5+10;
struct T{
int L,R;
int sum;
T(){
sum = 0;
}
}Tree[MAXN*20];//这类题尽量开大点，否则很容易RE，甚至可能T。
struct V{
int value,id;
bool operator < (const struct V &b)const{
return value < b.value;
}
}board[MAXN];
int root[MAXN];
int Rank[MAXN];
int tot;
void Updata(int num,int &rt,int left,int right){
Tree[++tot] = Tree[rt];
rt = tot;
Tree[rt].sum++;
if(left == right)return ;
int mid = left + (right-left)/2;
if(num <= mid)Updata(num,Tree[rt].L,left,mid);
else Updata(num,Tree[rt].R,mid+1,right);
}
int Query(int ql,int qr,int k,int left,int right){
int t = Tree[Tree[qr].L].sum - Tree[Tree[ql].L].sum;
if(left == right)return left;
int mid = left + (right-left)/2;
if(k <= t)return Query(Tree[ql].L,Tree[qr].L,k,left,mid);
else return Query(Tree[ql].R,Tree[qr].R,k-t,mid+1,right);
}
inline void init(){
tot = 0;
root[0] = 0;
Tree[0].L = Tree[0].R = Tree[0].sum = 0;
}
int main(){
int T;
scanf("%d",&T);
int N,M;
while(T--){
scanf("%d %d",&N,&M);
init();
for(int i=1 ; i<=N ; ++i){
scanf("%d",&board[i].value);
board[i].id = i;
}
sort(board+1,board+1+N);
for(int i=1 ; i<=N ; ++i)Rank[board[i].id] = i;
for(int i=1 ; i<=N ; ++i){
root[i] = root[i-1];
Updata(Rank[i],root[i],1,N);
}
int left,right,k;
for(int i=1 ; i<=M ; ++i){
scanf("%d %d %d",&left,&right,&k);
printf("%d\n",board[Query(root[left-1],root[right],k,1,N)].value);
}
}
return 0;
}