【POJ2104/2761】K-th Number
Description
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
Input
The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output
Sample Input
7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3
Sample Output
5
6
3
Hint
Source
#include <iostream>
#include <cstdio>
#include <algorithm>
#define N 100000
using namespace std;
int sum[],ls[],rs[],root[N+],hash[N+],a[N+],num[N+];
int n,m,cnt,size;
int findpos(int x)
{
int l=,r=cnt,mid;
while (l<=r)
{
mid=(l+r)>>;
if (hash[mid]<x) l=mid+;
else r=mid-;
}
return l;
} void updata(int l,int r,int x,int &y,int v)
{
y=++size; sum[y]=sum[x]+;
if (l==r) return;
ls[y]=ls[x]; rs[y]=rs[x];
int mid=(l+r)>>;
if (v<=mid) updata(l,mid,ls[x],ls[y],v);
else updata(mid+,r,rs[x],rs[y],v);
} int query(int l,int r,int x,int y,int k)
{
if (l==r) return l;
int mid=(l+r)>>;
if (sum[ls[y]]-sum[ls[x]]>=k) return query(l,mid,ls[x],ls[y],k);
else return query(mid+,r,rs[x],rs[y],k-(sum[ls[y]]-sum[ls[x]]));
} int main()
{
scanf("%d%d",&n,&m);
for (int i=;i<=n;i++)
{
scanf("%d",&a[i]);
num[i]=a[i];
}
sort(num+,num+n+);
hash[++cnt]=num[];
for (int i=;i<=n;i++)
if (num[i]!=num[i-])
hash[++cnt]=num[i];
for (int i=;i<=n;i++)
{
int t=findpos(a[i]);
updata(,cnt,root[i-],root[i],t);
}
for (int i=;i<=m;i++)
{
int a,b,x,re;
scanf("%d%d%d",&a,&b,&x);
re=query(,cnt,root[a-],root[b],x);
printf("%d\n",hash[re]);
}
return ;
}
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