hdu 2404 Permutation Recovery

Problem Description

Professor Permula gave a number of permutations of the n integers 1, 2, ..., n to her students. For each integer i (1 <= i <= n), she asks the students to write down the number of integers greater than i that appears before i in the given permutation. This number is denoted ai. For example, if n = 8 and the permutation is 2,7,3,5,4,1,8,6, then a1 = 5 because there are 5 numbers (2, 7, 3, 5, 4) greater than 1 appearing before it. Similarly, a4 = 2 because there are 2 numbers (7, 5) greater than 4 appearing before it.

John, one of the students in the class, is studying for the final exams now. He found out that he has lost the assignment questions. He only has the answers (the ai's) but not the original permutation. Can you help him determine the original permutation, so he can review how to obtain the answers?

 

Input

The input consists of a number of test cases. Each test case starts with a line containing the integer n (n <= 500). The next n lines give the values of a1, ..., an. The input ends with n = 0.

 

Output

For each test case, print a line specifying the original permutation. Adjacent elements of a permutation should be separated by a comma. Note that some cases may require you to print lines containing more than 80 characters.

 

Sample Input

8501212001098765432100

 

Sample Output

2,7,3,5,4,1,8,610,9,8,7,6,5,4,3,2,1
已知某一排列序列的逆序列，如何恢复原来的排列序列，先考虑当全部数组都为空时，如果1的逆序数为m,那么1显然会在m+1位置上，顺序处理1,2.......k,（递增序），假设此时k的逆序数为m',那么k会出现在什么位置呢？显然1,2,,,,k-1都无法对k的逆序数产生影响，则1,2,,,,k-1占据的位置可以直接忽视，只统计那些可以对逆序数产生影响的位置（即当前依然为空的位置），从左往右遍历数组，如果遇到已被(1,2,,,,k-1)占据的数据时，直接跳过，如果遇到空位置，则统计空位置数量，直至扫描至某一位index,空数据位置数量刚好达到m',则k的位置就是index+1
#include <iostream>#include <string>using namespace std;int reserve[1000];int per[1000];int n;void permulation(){int i, j;memset(per,0,sizeof(per));for ( i = 1; i <= n; ++i )//枚举每个数应该在哪个位置{for ( j = 1; j <= n; ++j ){if(per[j] == 0){if ( reserve[i] != 0 )//统计空白格子--reserve[i];elsebreak;}}per[j] = i;}}int main(){int i;while(cin >> n && n != 0){for(i=1; i <= n; ++i)cin >> reserve[i];permulation();cout << per[1];for(i = 2; i <= n; ++i)cout << ',' << per[i];cout << endl;}return 0;}