顯然最短迴圈節長度=i-next[i]，則相當於給定next陣列構造字串。然後按照kmp的過程模擬即可。雖然這看起來是一個染色問題，但是由圖的特殊性，如果next=0只要貪心地選最小的就可以了，然而並不會證。

#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<'0'||c>'9')) c=getchar();return c;}
int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
int read()
{
    int x=0,f=1;char c=getchar();
    while (c<'0'||c>'9') {if
 
 (c=='-') f=-1;c=getchar();}
    while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
    return x*f;
}
int n,nxt[N];
bool flag[26];
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=1;i<=n;i++) nxt[i]=i-read();
    nxt[0]=-1;
    for (int i=1;i<=n;i++)
    {
        int j=nxt[i-1];memset(flag,0,sizeof(flag));
        while (~j&&j+1!=nxt[i]) flag[s[j+1]-'a']=1,j=nxt[j];
        if (j==-1) {for (int j=0;j<26;j++) if (!flag[j]) {s[i]=j+'a';break;}}
        else s[i]=s[j+1];
    }
    printf("%s",s+1);
    return 0;
}