Shell Sort is a generalization of Insertion Sort to arrange a list of n elements A.
1 insertionSort(A, n, g)
2 for i = g to n-1
3 v = A[i]
4 j = i - g
5 while j >= 0 && A[j] > v
6 A[j+g] = A[j]
7 j = j - g
8 cnt++
9 A[j+g] = v
10
11 shellSort(A, n)
12 cnt = 0
13 m = ?
14 G[] = {?, ?,…, ?}
15 for i = 0 to m-1
16 insertionSort(A, n, G[i])
A function shellSort(A, n) performs a function insertionSort(A, n, g), which considers every g-th elements. Beginning with large values of g, it repeats the insertion sort with smaller g.
Your task is to complete the above program by filling ?. Write a program which reads an integer n and a sequence A, and prints m, Gi(i=0,1,…,m−1) in the pseudo code and the sequence A in ascending order. The output of your program must meet the following requirements:
1≤m≤100
0≤Gi≤n
cnt does not exceed ⌈n1.5⌉
Input
In the first line, an integer n is given. In the following n lines, Ai(i=0,1,…,n−1) are given for each line.
Output
In the first line, print an integer m. In the second line, print m integers Gi(i=0,1,…,m−1) separated by single space character in a line.
In the third line, print cnt in a line. In the following n lines, print Ai(i=0,1,…,n−1) respectively.
This problem has multiple solutions and the judge will be performed by a special validator.
Constraints
1≤n≤1,000,000
0≤Ai≤109
Sample Input 1
5
5
1
4
3
2
Sample Output 1
2
4 1
3
1
2
3
4
5
Sample Input 2
3
3
2
1
Sample Output 2
1
1
3
1
2
3
code
/*
^....0
^ .1 ^1^
.. 01
1.^ 1.0
^ 1 ^ ^0.1
1 ^ ^..^
0. ^ 0^
.0 1 .^
.1 ^0 .........001^
.1 1. .111100....01^
00 ^ 11^ ^1. .1^
1.^ ^0 0^
.^ ^0..1
.1 1..^
1 .0 ^ ^
^ 00. ^^0.^
1 ^ 0 ^^110.^
0. 0 ^ ^^^10.01
^^ 010^ 1 1 ^^^1110.1
0001 10 0 ^ 1.1 ^^^1111110
0^ 10 . 01 ^^ ^^ ^^^1111^1.^ ^^^
10 10^ 0^ ^^111^^^0.1^ 1....^
11 0 ^^11^^^ 0.. ....1^ ^ ^
1. 0^ ^11^^^ ^ 1 111^ ^ 0.
10 00 11 ^^^^^ 1 0 1.
0^ ^0 ^0 ^^^^ 0 0.
0^ 1.0 .^ ^^^^ 1 1 .0
^.^ ^^ 0^ ^1 ^^^^ 0. ^.1
1 ^ 11 1. ^^^ ^ ^ ..^
^..^ ^1 ^.^ ^^^ .0 ^.0
0..^ ^0 01 ^^^ .. 0..^
1 .. .1 ^.^ ^^^ 1 ^ ^0001
^ 1. 00 0. ^^^ ^.0 ^.1
. 0^. ^.^ ^.^ ^^^ ..0.0
1 .^^. .^ 1001 ^^ ^^^ . 1^
. ^ ^. 11 0. 1 ^ ^^ 0.
0 ^. 0 ^0 1 ^^^ 0.
0.^ 1. 0^ 0 .1 ^^^ ..
.1 1. 00 . .1 ^^^ ..
1 1. ^. 0 .^ ^^ ..
0. 1. .^ . 0 .
.1 1. 01 . . ^ 0
^.^ 00 ^0 1. ^ 1 1
.0 00 . ^^^^^^ .
.^ 00 01 ..
1. 00 10 1 ^
^.1 00 ^. ^^^ .1
.. 00 .1 1..01 ..
1.1 00 1. ..^ 10
^ 1^ 00 ^.1 0 1 1
.1 00 00 ^ 1 ^
. 00 ^.^ 10^ ^^
1.1 00 00 10^
..^ 1. ^. 1.
0 1 ^. 00 00 .^
^ ^. ^ 1 00 ^0000^ ^ 01
1 0 ^. 00.0^ ^00000 1.00.1 11
. 1 0 1^^0.01 ^^^ 01
.^ ^ 1 1^^ ^.^
1 1 0.
.. 1 ^
1 1
^ ^ .0
1 ^ 1
.. 1.1 ^0.0
^ 0 1..01^^100000..0^
1 1 ^ 1 ^^1111^ ^^
0 ^ ^ 1 1000^
.1 ^.^ . 00
.. 1.1 0. 0
1. . 1. .^
1. 1 1. ^0
^ . ^.1 00 01
^.0 001. .^
*/
// Virtual_Judge —— Shell Sort Aizu - ALDS1_2_D .cpp created by VB_KoKing on 2019,04,28,19.
/* Procedural objectives：
Variables required by the program:
Procedural thinking：
Functions required by the program:
*/
/* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first hurricane that passed through your ear is you,
The first star you see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE！！！
*/
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
using namespace std;
vector<int> G;
long long cnt=0;
int l,A[1000000],n;
//指定了间隔g的插入排序
void insert_ion_sort(int A[],int n,int g)
{
for (int i = g; i < n; i++) {
int v=A[i],j=i-g;
while (j>=0&&A[j]>v)
{
A[j+g]=A[j];
j-=g;
cnt++;
}
A[j+g]=v;
}
}
void shell_sort(int A[],int n)
{
//生成数列G={1,4,13,40,121,……}
int h=1;
while (true)
{
if (h>n) break;
G.push_back(h);
h=3*h+1;
}
for (int i = G.size()-1; i >=0 ; i--)
//按逆序制定G[i]=g
insert_ion_sort(A,n,G[i]);
}
int main()
{
cin>>n;
for (int i = 0; i < n; i++)
scanf("%d",&A[i]);
shell_sort(A,n);
cout<<G.size()<<endl;
for (int i = G.size()-1; i >=0 ; i--) {
printf("%d",G[i]);
if (i) printf(" ");
}
cout<<endl<<cnt<<endl;
for (int i = 0; i < n; i++)
printf("%d
",A[i]);
return 0;
}