public Program(string inputFile, string outputFile)
{
io = new IOHelper(inputFile, outputFile, Encoding.Default);
//n==6 -- 1 3 5 5 3 1 + 2 4 6 4 2 + 6
//n==7 -- 1 3 5 7 5 3 1 + 2 4 6 6 4 2 + 7
int n = io.NextInt();
for (int i = 1; i <= n; i += 2)
{
io.Write(i + " ");
}
for (int i = n - (n % 2 == 0 ? 1 : 2); i >= 1; i -= 2)
{
io.Write(i + " ");
}
for (int i = 2; i <= n; i += 2)
{
io.Write(i + " ");
}
for (int i = n - (n % 2 == 0 ? 2 : 1); i >= 1; i -= 2)
{
io.Write(i + " ");
}
io.WriteLine(n);
io.Dispose();
}
public Program(string inputFile, string outputFile)
{
io = new IOHelper(inputFile, outputFile, Encoding.Default);
//直接思路：枚举1~m的lcm，对每个lcm枚举约数，然后约数的bin里都要，复杂度msqrt(m)
//那么现在反转，枚举约数，再枚举这个数是哪些可能的lcm值的约数
//根据调和级数公式，n/1+n/2+n/3+...n/n=nlogn
int n=io.NextInt(),m = io.NextInt();
List<int>[] bin = new List<int>[m + 1];
for (int i = 0; i <= m; i++) bin[i] = new List<int>();
for (int i = 1; i <= n; i++) {
int x = io.NextInt();
if (x <= m) bin[x].Add(i);
}
int[] cnt = new int[m + 1];
for (int i = 1; i <= m; i++) {
if (bin[i].Count == 0) continue;
for (int j = i; j <= m; j += i) {
cnt[j] += bin[i].Count;
}
}
int maxElement=cnt.Max();
for (int i = 1; i <= m; i++) {
if (cnt[i] == maxElement) {
io.WriteLine(i+" "+maxElement);
List<int> ans=new List<int>();
for (int j = 1; j <= Math.Sqrt(i); j++) {
if (i % j == 0) {
ans.AddRange(bin[j]);
if (j * j != i) ans.AddRange(bin[i / j]);
}
}
ans.Sort();
foreach (int k in ans) {
io.Write(k + " ");
}
break;
}
}
io.Dispose();
}
public Program(string inputFile, string outputFile)
{
io = new IOHelper(inputFile, outputFile, Encoding.Default);
//n==6 -- 1 3 5 5 3 1 + 2 4 6 4 2 + 6
//n==7 -- 1 3 5 7 5 3 1 + 2 4 6 6 4 2 + 7
int n = io.NextInt();
for (int i = 1; i <= n; i += 2)
io.Write(i + " ");
for (int i = n - (n % 2 == 0 ? 1 : 2); i >= 1; i -= 2)
io.Write(i + " ");
for (int i = 2; i <= n; i += 2)
io.Write(i + " ");
for (int i = n - (n % 2 == 0 ? 2 : 1); i >= 1; i -= 2)
io.Write(i + " ");
io.WriteLine(n);
io.Dispose();
}
public Program(string inputFile, string outputFile)
{
io = new IOHelper(inputFile, outputFile, Encoding.Default);
//直接思路：枚举1~m的lcm，对每个lcm枚举约数，然后约数的bin里都要，复杂度msqrt(m)
//那么现在反转，枚举约数，再枚举这个数是哪些可能的lcm值的约数
//根据调和级数公式，n/1+n/2+n/3+...n/n=nlogn
int n = io.NextInt(), m = io.NextInt();
List <int>[] bin = new List <int> [m + 1];
for (int i = 0; i <= m; i++)
{
bin[i] = new List <int>();
}
for (int i = 1; i <= n; i++)
{
int x = io.NextInt();
if (x <= m)
{
bin[x].Add(i);
}
}
int[] cnt = new int[m + 1];
for (int i = 1; i <= m; i++)
{
if (bin[i].Count == 0)
{
continue;
}
for (int j = i; j <= m; j += i)
{
cnt[j] += bin[i].Count;
}
}
int maxElement = cnt.Max();
for (int i = 1; i <= m; i++)
{
if (cnt[i] == maxElement)
{
io.WriteLine(i + " " + maxElement);
List <int> ans = new List <int>();
for (int j = 1; j <= Math.Sqrt(i); j++)
{
if (i % j == 0)
{
ans.AddRange(bin[j]);
if (j * j != i)
{
ans.AddRange(bin[i / j]);
}
}
}
ans.Sort();
foreach (int k in ans)
{
io.Write(k + " ");
}
break;
}
}
io.Dispose();
}
