Kth number

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

Total Submission(s): 37 Accepted Submission(s): 25

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）

 

Recommend

zty

 

/*----------------------------------------------File: Date: 2017/6/9 21:40:22Author: LyuCheng----------------------------------------------*/#include 
     
      #define MAXN 100005using namespace std;int t,n;int q,l,r,k;/************************************划分树模板************************************/int a[MAXN];       //原数组int sorted[MAXN];  //排序好的数组//是一棵树，但把同一层的放在一个数组里。int num[20][MAXN];   //num[i] 表示i前面有多少个点进入左孩子int val[20][MAXN];   //20层,每一层元素排放，0层就是原数组void build(int l,int r,int ceng){    if(l==r) return ;    int mid=(l+r)/2,isame=mid-l+1;  //isame保存有多少和sorted[mid]一样大的数进入左孩子    for(int i=l;i<=r;i++) if(val[ceng][i]
      
       0)        {            val[ceng+1][ln++]=val[ceng][i];            num[ceng][i]++;            if(val[ceng][i]==sorted[mid]) isame--;        }        else        {            val[ceng+1][rn++]=val[ceng][i];        }    }    build(l,mid,ceng+1);    build(mid+1,r,ceng+1);}int look(int ceng,int sl,int sr,int l,int r,int k){    if(sl==sr) return val[ceng][sl];    int ly;  //ly 表示l 前面有多少元素进入左孩子    if(l==sl) ly=0;  //和左端点重合时    else ly=num[ceng][l-1];    int tolef=num[ceng][r]-ly;  //这一层l到r之间进入左子树的有tolef个    if(tolef>=k)    {        return look(ceng+1,sl,(sl+sr)/2,sl+ly,sl+num[ceng][r]-1,k);    }    else    {        // l-sl 表示l前面有多少数，再减ly 表示这些数中去右子树的有多少个        int lr = (sl+sr)/2 + 1 + (l-sl-ly);  //l-r 去右边的开头位置        // r-l+1 表示l到r有多少数，减去去左边的，剩下是去右边的，去右边1个，下标就是lr，所以减1        return look(ceng+1,(sl+sr)/2+1,sr,lr,lr+r-l+1-tolef-1,k-tolef);    }}/************************************划分树模板************************************/int main(int argc, char *argv[]){        // freopen("in.txt","r",stdin);    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&q);        for(int i=1;i<=n;i++){            scanf("%d",&val[0][i]);            sorted[i]=val[0][i];        }        sort(sorted+1,sorted+n+1);        build(1,n,0);        while(q--){            scanf("%d%d%d",&l,&r,&k);            printf("%d\n",look(0,1,n,l,r,k));        }    }    return 0;}