显然最短循环节长度=i-next[i],则相当于给定next数组构造字符串。然后按照kmp的过程模拟即可。虽然这看起来是一个染色问题,但是由图的特殊性,如果next=0只要贪心地选最小的就可以了,稍微想一下容易证明。
#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<algorithm>
#include<cassert>
using namespace std;
#define ll long long
#define N 100010
char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<''||c>'')) c=getchar();return c;}
int gcd(int n,int m){return m==?n:gcd(m,n%m);}
int read()
{
int x=,f=;char c=getchar();
while (c<''||c>'') {if (c=='-') f=-;c=getchar();}
while (c>=''&&c<='') x=(x<<)+(x<<)+(c^),c=getchar();
return x*f;
}
int n,nxt[N];
bool flag[];
char s[N];
int main()
{
#ifndef ONLINE_JUDGE
freopen("bzoj4974.in","r",stdin);
freopen("bzoj4974.out","w",stdout);
const char LL[]="%I64d\n";
#else
const char LL[]="%lld\n";
#endif
n=read();
for (int i=;i<=n;i++) nxt[i]=i-read();
nxt[]=-;
for (int i=;i<=n;i++)
{
int j=nxt[i-];memset(flag,,sizeof(flag));
while (~j&&j+!=nxt[i]) flag[s[j+]-'a']=,j=nxt[j];
if (j==-) {for (int j=;j<;j++) if (!flag[j]) {s[i]=j+'a';break;}}
else s[i]=s[j+];
}
printf("%s",s+);
return ;
}