Kth number

Time Limit: 15000/5000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7616 Accepted Submission(s): 2402

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

题意：

给你一串数字，然后是m次查询，每次查找[s,t]区间的第k大值 /*划分树学习

/*模板题，hdu2665*/#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          
           #include 
           
            #include 
            using namespace std;typedef long long ll;typedef long double ld;using namespace std;const int maxn = 100010;int tree[20][maxn];int sorted[maxn];int toleft[20][maxn];void build(int l,int r,int dep) //模拟快排 并记录左树中比i小的个数{ if(l == r) return; int mid = (l+r)>>1; int same = mid-l+1;//可能有很多值与中间那个相等，但不一定被分到左边 for(int i = l;i <= r;i++) { if(tree[dep][i] < sorted[mid]) same--; } int lpos = l; int rpos = mid+1; for(int i = l;i <= r;i++) { if(tree[dep][i] < sorted[mid]) tree[dep+1][lpos++] = tree[dep][i]; else if(tree[dep][i] == sorted[mid] && same > 0) { tree[dep+1][lpos++] = tree[dep][i]; same --; } else tree[dep+1][rpos++] = tree[dep][i]; toleft[dep][i] = toleft[dep][l-1] + lpos -l; } build(l,mid,dep+1); build(mid+1,r,dep+1); } int query(int L,int R,int l,int r,int dep,int k) { if(l == r) return tree[dep][l]; int mid = (L+R)>>1; int cnt = toleft[dep][r]-toleft[dep][l-1]; //所查找区间放在左树中的个数 if(cnt >= k) { //得到l左边放到左子树的个数，加上L即是开始位置 int lpos = L+toleft[dep][l-1]-toleft[dep][L-1]; int rpos = lpos+cnt-1; return query(L,mid,lpos,rpos,dep+1,k); } else { //R-r可以得出后面空出了多少位置 int rpos = r+toleft[dep][R]-toleft[dep][r]; int lpos = rpos-(r-l-cnt); return query(mid+1,R,lpos,rpos,dep+1,k-cnt); } } int main() { int T,n,m; scanf("%d",&T); while(T--) { scanf("%d%d",&n,&m); memset(tree,0,sizeof(tree)); for(int i = 1;i <= n;i++) { scanf("%d",&sorted[i]); tree[0][i] = sorted[i]; } sort(sorted+1,sorted+n+1); build(1,n,0); int l,r,k; while(m--) { scanf("%d%d%d",&l,&r,&k); printf("%d\n",query(1,n,l,r,0,k)); } } return 0; }