二进制log2（t）+1 便是位数1
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80#include<iostream>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
int n;
int p;
int s=0;
int sol[999];
int ans=0;
int log2(int a)//log
{
switch (a)
{
case 1:
return 1;
case 2:
return 2;
case 4:
return 3;
case 8:
return 4;
case 16:
return 5;
case 32:
return 6;
case 64:
return 7;
case 128:
return 8;
case 256:
return 9;
case 512:
return 10;
case 1024:
return 11;
case 2048:
return 12;
case 4096:
return 13;
}
}
void print()
{
for(int i=1;i<=n;i++)
{
printf("%d ",log2(sol[i]));
}
printf("\n");
}
void dfs(int r,int rl,int ll)
{
if(r==p)
{
ans++;
if(ans<=3)
print();
return;
}
int pos=p&~(r|rl|ll);
while(pos!=0)
{
int t=pos&(-pos);
pos-=t;
sol[++s]=t;
dfs(r|t,(rl|t)<<1,(ll|t)>>1);
--s;
}
}
int main()
{
scanf("%d",&n);
p=(1<<n)-1;
dfs(0,0,0);
printf("%d",ans);
return 0;
}