Given a hash table of size N, we can define a hash function H(x) = x%N. Suppose that the linear probing is used to solve collisions, we can easily obtain the status of the hash table with a given sequence of input numbers.

However, now you are asked to solve the reversed problem: reconstruct the input sequence from the given status of the hash table. Whenever there are multiple choices, the smallest number is always taken.

Input Specification:

Each input file contains one test case. For each test case, the first line contains a positive integer N (<=1000), which is the size of the hash table. The next line contains N integers, separated by a space. A negative integer represents an empty cell in the hash table. It is guaranteed that all the non-negative integers are distinct in the table.

Output Specification:

For each test case, print a line that contains the input sequence, with the numbers separated by a space. Notice that there must be no extra space at the end of each line.

Sample Input:

11

33 1 13 12 34 38 27 22 32 -1 21

Sample Output:

1 13 12 21 33 34 38 27 22 32

 #include<cstdio>

 #include<algorithm>

 #include<iostream>

 #include<cstring>

 #include<queue>

 using namespace std;

 struct node{

     bool exist;

     node(){

         exist=false;

     }

 };

 node h1[];

 priority_queue< pair<int,int>,vector<pair<int ,int> >,greater<pair<int,int> > > pq;//记录值和放的位置，按pair的第一个元素升序排列

 pair<int,bool> input[];//标记元素有没有进入队列

 queue<int> q;

 int main(){

     //freopen("D:\\INPUT.txt","r",stdin);

     int n,num,i,count=;

     scanf("%d",&n);

     for(i=;i<n;i++){

         scanf("%d",&num);

         input[i]=make_pair(num,false);

         if(num>=){//记录不等于负数的元素个数

             count++;

         }

     }

     int j,k,post;

     for(i=;i<count;i++){//复杂度n^2：扫数列count次，每次将当前满足条件的最小元素进入队列q

         for(j=;j<n;j++){//遍历数列的每个元素

             if(!input[j].second&&input[j].first>=){//寻找没有进入队列并且不等于-1的元素

                 post=input[j].first%n;//元素本来应该在哈希表中的位置

                 if(post==j){//本来要放的位置和现在在哈希表中的位置相同

                     pq.push(make_pair(input[j].first,j));

                     //h1[post].exist=true;//标记

                     input[j].second=true;

                     continue;

                 }

                 if(post<j){//本来要放的位置在现在的位置前面

                     for(k=post;k<j;k++){

                         if(!h1[k].exist){//前面没有元素填充

                             break;

                         }

                     }

                     if(k==j){//可以进队pq

                         pq.push(make_pair(input[j].first,j));

                         //h1[j].exist=true;

                         input[j].second=true;

                     }

                 }

                 else{//post>j

                     for(k=post;k<n;k++){//先由post开始向后遍历

                         if(!h1[k].exist){

                             break;

                         }

                     }

                     if(k==n){//满足条件后，由0开始向j遍历

                         for(k=;k<j;k++){

                             if(!h1[k].exist){

                                 break;

                             }

                         }

                         if(k==j){

                             pq.push(make_pair(input[j].first,j));

                             //h1[j].exist=true;

                             input[j].second=true;

                         }

                     }

                 }

             }

         }

         pair<int,int> top=pq.top();

         pq.pop();

         q.push(top.first);//找到本次满足条件：在队列pq且最小的元素，进入输出队列q

         h1[top.second].exist=true;//意味着top.second位置已经放了元素

     }

     printf("%d",q.front());//对输出队列q进行输出

     q.pop();

     while(!q.empty()){

         printf(" %d",q.front());

         q.pop();

     }

     return ;

 }