本文已参与「新人创作礼」活动，一起开启掘金创作之路。
Problem Description
You are given an array aa of length nn. You must perform exactly kk times operations.
For each operation,
\bullet∙ First, you select two integers l,rl,r ((1\leq l\leq r \leq n1≤l≤r≤n)),
\bullet∙ Second, change aa to bb, satisfy :
\ \ \ \ \circ ∘ For each ii ((1\le i< l1≤i<l)) , b_i=a_ibi=ai;
\ \ \ \ \circ ∘ For each ii ((l\le i< rl≤i<r)) , b_i=a_{i+1}bi=ai+1;
\ \ \ \ \circ ∘ b_r=a_lbr=al
\ \ \ \ \circ ∘ For each ii ((r< i\le nr<i≤n)) , b_i=a_ibi=ai;
Find the lexicographically largest possible array after kk times operations.
Array xx is lexicographically greater than array yy if there exists an index ii (( 1\leq i\leq n1≤i≤n )) such that x_ixi >> y_iyi and for every j (1\leq j \lt i) ,j(1≤j<i), x_j=y_jxj=yj.
Input
The first line of the input contains one integer TT ((1\leq T\leq 1001≤T≤100 )) --- the number of test cases. Then TT test cases follow.
The first line of the test case contains two integers n,kn,k ((1\le n,k\le 3000001≤n,k≤300000))
The second line of the test case contains nn integers a_1,a_2,...,a_na1,a2,...,an((1\le a_i\le 3000001≤ai≤300000))
The sum of nn over all testcases doesn't exceed 10^{6}106.
The sum of kk over all testcases doesn't exceed 10^{6}106.
Output
For each testcase,one line contains nn integers ,a_1,a_2,...,a_na1,a2,...,an --- the lexicographically largest possible array after kk times operations.
Don‘t have spaces at the end of the line
Sample Input
2
7 3
1 4 2 1 4 2 4
5 2
4 3 5 4 5
Sample Output
4 4 2 4 2 1 1
5 4 5 4 3
题意：
给出长度为n的数组以及k次操作机会，每次操作可以选择一个区间[l, r]使区间内的点循环左移1位，问最终字典序最大的序列是什么。
分析：
选择区间[l, r]内的循环左移实际上等价于选择一个值a[i]将其拿出并放在后面任意一个位置上，贪心地想，要想字典序最大那么从前往后看每个位置都要尽可能大，所以遍历一遍数组，第i个位置能够通过小于等于k次的操作变成a[i]到a[i+k]中的某个值，显然第i个位置应该变成其中的最大值，所以这需要我们提前维护出区间最大值，这可以通过st表O(nlogn)处理出来，过程中被弹出的值用一个数组t2存下来，取到的最大值也用一个数组t1存下来，最后可以将这些弹出的值任意插入回t1，所以对t2降序排列，字典序最大也就是一个类似归并t1和t2的过程。
具体代码如下：
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
using namespace std;
int f[300005][20], a[300005], n, k, ans[300005];
bool cmp(int x, int y){
return x > y;
}
void st()
{
for(int i = 1; i <= n; i ++) f[i][0] = a[i];
for(int j = 1; j < 20; j ++)
for(int i = 1; i+(1<<j)-1 <= n; i ++)
f[i][j] = max(f[i][j-1],f[i+(1<<(j-1))][j-1]);//均分为两部分
}
int query(int l, int r)
{
int len = abs(r-l+1);//询问区间长度
int t = log(len)/log(2);//换底公式，得到以2为底的下取整对数
return max(f[l][t],f[r-(1<<t)+1][t]);
}
signed main()
{
int T;
cin >> T;
while(T--){
scanf("%d%d", &n, &k);
for(int i = 1; i <= n; i++)
scanf("%d", &a[i]);
st();
vector<int> temp1, temp2;
for(int i = 1; i <= n; i++){
int l = i, r = i+k;
if(r > n) r = n;
int mx = query(l, r);
temp1.push_back(mx);
while(a[i] != mx){
if(!k) break;
k--;
temp2.push_back(a[i]);
i++;
}
}
sort(temp2.begin(), temp2.end(), cmp);
int cnt = 0, i = 0, j = 0;
for(; i < temp1.size() && j < temp2.size();){
if(temp1[i] < temp2[j]){
ans[++cnt] = temp2[j];
j++;
}
else if(temp1[i] >= temp2[j]){
ans[++cnt] = temp1[i];
i++;
}
}
while(i < temp1.size()){
ans[++cnt] = temp1[i];
++i;
}
while(j < temp2.size()){
ans[++cnt] = temp2[j];
++j;
}
printf("%d", ans[1]);
for(int i = 2; i <= n; i++)
printf(" %d", ans[i]);
puts("");
}
return 0;
}
