Given a hash table of size N, we can define a hash function (. 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 (≤), 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

题目要求是给出了hashing的结果，而且是用的linear probing线性探测解决冲突，要求给出初始序列。负数直接跳过，序列中的数都是非负数。所以遍历给出序列，如果说当前位置的s[i] % n等于i那么就是说不存在冲突，如果不相等就是存在冲突的，经过线性探测才存在当前位置，所以从s[i] % n位置到i之前的数都是比是s[i]先进行操作的，如此我们可以找到序列中数的顺序，进行拓扑排序，如果存在多种可能要求输出最小的，所以用优先队列。
代码：
#include <cstdio>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

int n,s[1000],l[1000],flag;
vector<int> v[1000];
struct cmp {
    bool operator ()(const int &a,const int &b) {
        return s[a] > s[b];
    }
};
int main() {
    scanf("%d",&n);
    for(int i = 0;i < n;i ++) {
        scanf("%d",&s[i]);
    }
    priority_queue<int,vector<int>,cmp> q;
    for(int i = 0;i < n;i ++) {
        if(s[i] < 0) continue;
        if(s[i] % n != i) {
            int d = s[i] % n > i ? i + n : i;
            for(int j = s[i] % n;j < d;j ++) {
                if(s[j % n] < 0) continue;
                v[j % n].push_back(i);
                l[i] ++;
            }
        }
        else q.push(i);
    }
    while(!q.empty()) {
        int temp = q.top();
        q.pop();
        if(flag) putchar(' ');
        else flag = 1;
        printf("%d",s[temp]);
        for(int i = 0;i < v[temp].size();i ++) {
            int d = v[temp][i];
            l[d] --;
            if(!l[d]) q.push(d);
        }
    }
    return 0;
}