此题很神奇请看大神题解。点这里
我的代码
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <cstring>
#include <iostream>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define N 65
#define mo 500000
#define ll long long
int n,next[N],jy,jy1,len;
ll f[N],k,ans[N],er[N];
ll hash(int len,int i,int j)
{
ll x=0,y=0;
fo(k,i-j+1,len) x=x*2+ans[k];
fo(k,1,len-i+j) y=y*2+ans[k];
return x==y;
}
ll hzj()
{
memset(f,0,sizeof(f));
fo(i,1,n)
{
if (i<len && next[i]==0) f[i]=1;
if (i<len) continue;
f[i]=er[(i-len)];
fo(j,1,i/2)
{
ll x;
if (j>=len) x=er[(i-2*j)];
if (len>j && len<=i-j) x=er[(i-2*j-len+j)];
if (len>i-j) x=hash(len,i,j);
f[i]-=x*f[j];
}
}
return f[n];
}
void kmp()
{
if (len==1) return;
while (jy!=0 && ans[len]!=ans[jy+1]) jy=next[jy];
if (ans[len]==ans[jy+1]) jy++;
next[len]=jy;
}
int main()
{
freopen("word.in","r",stdin);freopen("word.out","w",stdout);
er[0]=1;fo(i,1,63) er[i]=er[i-1]*2;
int ak;scanf("%d",&ak);
while (ak-->0)
{
scanf("%d%lld",&n,&k);jy=len=0;
memset(next,0,sizeof(next));
ll lh=hzj();printf("%lld\n",lh);
for(len=1;len<=n;len++)
{
ans[len]=0;jy1=jy;
kmp();
ll lh=hzj();
if (lh<k)
{
jy=jy1;ans[len]=1;k-=lh;kmp();
}
}
fo(i,1,n) printf("%c",ans[i]==0?'a':'b');
printf("\n");
}
fclose(stdin);fclose(stdout);
}