Little Daniel loves to play with strings! He always finds different ways to have fun with strings! Knowing that, his friend Kinan decided to test his skills so he gave him a string S and asked him Q questions of the form:
If all distinct substrings of string S were sorted lexicographically, which one will be the K-th smallest?
After knowing the huge number of questions Kinan will ask, Daniel figured out that he can't do this alone. Daniel, of course, knows your exceptional programming skills, so he asked you to write him a program which given S will answer Kinan's questions.
Example:
S = "aaa" (without quotes)
substrings of S are "a" , "a" , "a" , "aa" , "aa" , "aaa". The sorted list of substrings will be:
"a", "aa", "aaa".
Input
In the first line there is Kinan's string S (with length no more than 90000 characters). It contains only small letters of English alphabet. The second line contains a single integer Q (Q <= 500) , the number of questions Daniel will be asked. In the next Q lines a single integer K is given (0 < K < 2^31).
Output
Output consists of Q lines, the i-th contains a string which is the answer to the i-th asked question.
Example
Input: aaa 2 2 3 Output: aa aaa
Edited: Some input file contains garbage at the end. Do not process them.
对串创建SAM，然后，拓扑一下，算出每个状态，如果继续往后边走，能够生成多少种不同的字符串。然后每次查询的时候，就按照字典顺序贪心着找
//#include<cstdio>
//#include<cstring>
//#include<string>
//#include<iostream>
//#include<algorithm>
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=1e5+10;//字符串长度
const int maxc=26;
char s[maxn];
struct Suffix_Automaton {//打*的不一定用到
int next[maxn<<1][maxc]; //状态转移(尾部加一个字符的下一个状态)
int len[maxn<<1]; //最长子串的长度(该节点子串数量=len[x]-len[link[x]])
int link[maxn<<1]; //后缀链接(最短串前部减少一个字符所到达的状态)
// int cnt[maxn<<1]; //被后缀连接的数(*）
int id; //结点编号
int last; //最后结点
int endpos[maxn<<1]; // endpos数（一类子串的数量）
int a[maxn];
int b[maxn<<1];
int sum[maxn<<1];//该节点后面所形成的子串的总数
// int dp[maxn<<1];
// int ans[maxn<<1];
// ll d[maxn<<1]; //d[i]表示从状态i出发，不同的子串的数目,即不同的路径数
void init() { //初始化
for(int i=1; i<=id; i++){ //常规初始化
link[i] = len[i] = 0;
memset(next[i],0,sizeof(next[i]));
endpos[i]=0;
a[i]=0;
b[i]=0;
}
// for(int i=1;i<=id;i++) {//非常规初始化
// d[i]=0;
// }
last = id = 1; //1表示root起始点 空集
}
//SAM建图
void add(int c) { //插入字符，为字符ascll码值
int x = ++id; //创建一个新结点x;
len[x] = len[last] + 1; // 长度等于最后一个结点+1
endpos[x] = 1; //接受结点子串除后缀连接还需加一
int p; //第一个有C转移的结点;
for (p = last; p && !next[p][c]; p = link[p])
next[p][c] = x;//沿着后缀连接 将所有没有字符c转移的节点直接指向新结点
if (!p){ //全部都没有c的转移 直接将新结点后缀连接到起点
link[x] = 1;
// cnt[1]++;
}
else {
int q = next[p][c]; //p通过c转移到的结点
if (len[p] + 1 == len[q]){//pq是连续的
link[x] = q;
// cnt[q]++; //将新结点后缀连接指向q即可,q结点的被后缀连接数+1
}
else {
int nq = ++id; //不连续 需要复制一份q结点
len[nq] = len[p] + 1; //令nq与p连续
link[nq] = link[q]; //因后面link[q]改变此处不加cnt
memcpy(next[nq], next[q], sizeof(next[q])); //复制q的信息给nq
for (; p&&next[p][c] == q; p = link[p])
next[p][c] = nq; //沿着后缀连接 将所有通过c转移为q的改为nq
link[q] = link[x] = nq; //将x和q后缀连接改为nq
//cnt[nq] += 2; // nq增加两个后缀连接
}
}
last = x; //更新最后处理的结点
}
/*
ll getSubNum() { //求不相同子串数量
ll ans = 0;
for (int i = 2; i <= id; i++)
ans += len[i]-len[link[i]]; //一状态子串数量等于len[i]-len[link[i]]
return ans;
}
*/
void getTP(int &Len){//对sam的节点按照len，从小到大排序重新标号，即给定节点的拓扑序
for(int i=1;i<=id;i++) a[len[i]]++;
for(int i=1;i<=Len;i++) a[i]+=a[i-1];
for(int i=1;i<=id;i++) b[a[len[i]]--]=i;
}
void getendpos(){//求每类子串的数量 ,即endpos集合的大小
for(int i=id;i>=1;i--){ //按拓扑序遍历
int e=b[i];
endpos[link[e]]+=endpos[e];
}
}
/*
void get_len_max(int Len){//求长度为i的出现次数最多的子串的出现次数
for(int i=1;i<=id;i++) dp[len[i]]=max(dp[len[i]],endpos[i]);
for(int i=Len-1;i>=1;i--) dp[i]=max(dp[i],dp[i+1]);
for(int i=1;i<=Len;i++) printf("%d\n",dp[i]);
}
*/
void getsum(){
for(int i=id;i>=1;i--){
int &e=b[i];
sum[e]=1;
for(int j=0;j<26;j++){
sum[e]+=sum[next[e][j]];
}
}
}
void getmink(int k){
int now=1,p;
string s="";
while(k){
for(int i=0;i<26;i++){
if(next[now][i]&&k){
p=next[now][i];
if(sum[p]<k) k-=sum[p];
else{
k--;
now=p;
s+=(char)(i+'a');
break;
}
}
}
}
cout<<s<<endl;
}
/*
void LCS1(char s[],int Len){//求两个串的最长公共子串
int ans=0,cnt=0;
int now=1;
char base='a';
for(int i=0;i<Len;i++){
int c=s[i]-base;
if(next[now][c]){
cnt++;
now=next[now][c];
}
else{
while(now&&!next[now][c]) now=link[now];
if(!now) cnt=0,now=1;
else cnt=len[now]+1,now=next[now][c];
}
ans=max(ans,cnt);
}
printf("%d\n",ans);
}
*/
/*
void init_ans(){
for(int i=1;i<=id;i++) ans[i]=len[i];
}
void LCS2(char s[],int Len){ //求多个串的最长公共子串
for(int i=1;i<=id;i++) dp[i]=0;
int cnt=0;
int now=1;
char base='a';
for(int i=0;i<Len;i++){
int c=s[i]-base;
if(next[now][c]){
cnt++;
now=next[now][c];
}
else{
while(now&&!next[now][c]) now=link[now];
if(!now) cnt=0,now=1;
else cnt=len[now]+1,now=next[now][c];
}
dp[now]=max(dp[now],cnt);
}
for(int i=id;i>=1;i--){
int e=b[i];
dp[link[e]]=max(dp[link[e]],min(dp[e],len[link[e]]));
}
for(int i=1;i<=id;i++) ans[i]=min(ans[i],dp[i]);
}
void get_LCS2_ans(){
int cnt=0;
for(int i=1;i<=id;i++) cnt=max(cnt,ans[i]);
printf("%d\n",cnt);
}
*/
/*
void solve1(){ //求出现次数为k的子串种数
ll ans=0;
for(int i=1;i<=id;i++){
if(endpos[i]==K){
ans+=len[i]-len[link[i]];
}
}
printf("%lld\n",ans);
}
*/
/*
void solve1(){//求出现次数A<=K<=B的子串种数
for(int i=id;i>1;i--){
int v=b[i];
if(endpos[v]>=A&&endpos[v]<=B) d[v]++;
for(int j=0;j<26;j++){
if(next[v][j]) d[v]+=d[next[v][j]];
}
}
ll ans=0;
for(int i=0;i<26;i++){
if(next[1][i]) ans+=d[next[1][i]];
}
printf("%lld\n",ans);
}
*/
/*
void solve2(){//求出现次数>=k的子串的最大长度
int ans=0;
for(int i=1;i<=id;i++){
if(endpos[i]>=K){
ans=max(ans,len[i]);
}
}
printf("%d\n",ans);
}*/
} sam;
int main(){
scanf("%s",s);
int len=strlen(s);
sam.init();
for(int i=0;i<len;i++) sam.add(s[i]-'a');
sam.getTP(len);
sam.getsum();
int T;
scanf("%d",&T);
while(T--){
int K;
scanf("%d",&K);
sam.getmink(K);
}
return 0;
}