You're given a permutation aa of length nn (1 le n le 10^51≤n≤105).
For each i in [1,n]i∈[1,n], construct a sequence s_isi by the following rules:
- s_i[1]=isi[1]=i;
- The length of s_isi is nn, and for each j in [2, n]j∈[2,n], s_i[j] le s_i[j-1]si[j]≤si[j−1];
- First, we must choose all the possible elements of s_isi from permutation aa. If the index of s_i[j]si[j] in permutation aa is pos[j]pos[j], for each j ge 2j≥2, |pos[j]-pos[j-1]|le k∣pos[j]−pos[j−1]∣≤k (1 le k le 10^51≤k≤105). And for each s_isi, every element of s_isi must occur in aa at most once.
- After we choose all possible elements for s_isi, if the length of s_isi is smaller than nn, the value of every undetermined element of s_isi is 00;
- For each s_isi, we must make its weight high enough.
Consider two sequences C = [c_1, c_2, ... c_n]C=[c1,c2,...cn] and D=[d_1, d_2, ..., d_n]D=[d1,d2,...,dn], we say the weight of CC is higher thanthat of DD if and only if there exists an integer kk such that 1 le k le n1≤k≤n, c_i=d_ici=di for all 1 le i < k1≤i<k, and c_k > d_kck>dk.
If for each i in [1,n]i∈[1,n], c_i=d_ici=di, the weight of CC is equal to the weight of DD.
For each i in [1,n]i∈[1,n], print the number of non-zero elements of s_isi separated by a space.
It's guaranteed that there is only one possible answer.
Input
There are multiple test cases.
The first line contains one integer T(1 le T le 20)T(1≤T≤20), denoting the number of test cases.
Each test case contains two lines, the first line contains two integers nn and kk (1 le n,k le 10^51≤n,k≤105), the second line contains nn distinct integers a_1, a_2, ..., a_na1,a2,...,an (1 le a_i le n1≤ai≤n) separated by a space, which is the permutation aa.
Output
For each test case, print one line consists of nn integers |s_1|, |s_2|, ..., |s_n|∣s1∣,∣s2∣,...,∣sn∣ separated by a space.
|s_i|∣si∣ is the number of non-zero elements of sequence s_isi.
There is no space at the end of the line.
样例输入复制
代码语言：javascript复制2
3 1
3 2 1
7 2
3 1 4 6 2 5 7样例输出复制
代码语言：javascript复制1 2 3
1 1 2 3 2 3 3这题是队友出的，这个题好像线段树能做，但是我们是暴力加贪心做的，前几遍是枚举的区间位置纯暴力，后来是枚举左右区间长度，对于每个元素的下一个元素都是唯一固定的。之后记忆化搜索就可以求出每个序列。
代码语言：javascript复制#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string.h>
#include<math.h>
#include<vector>
#include<queue>
#include<map>
using namespace std;
#define LL long long
#define MAXN 100100
int a[MAXN],pos[MAXN];
int s[MAXN],oo[MAXN];
int main()
{
int t;
scanf("%d",&t);
int n,k,l,r,maxx;
while(t--){
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i ){
scanf("%d",a i);
pos[a[i]]=i;
s[i]=1;
}
s[0]=0;
for(int i=2;i<=n;i ){
int j=i-1;
while(j>0){
if(abs(pos[i]-pos[j])<=k){
s[i] =s[j];
break;
}
j--;
}
}
for(int i=1;i<=n;i ){printf("%d",s[i]);if(i<n)printf(" ");}
printf("n");
}
}
