codepeng

Roman numerals are based on seven symbols: I = 1, V = 5, X = 10, L = 50, C = 100, D = 500 and M = 1000.

Symbols are iterated to produce multiples of the decimal (1, 10, 100, 1,000) values, with V, L, D substituted for a multiple of five, and the iteration continuing: I "1", II "2", III "3", V "5", VI "6", VII "7", etc., and the same for other bases: X "10", XX "20", XXX "30", L "50", LXXX "80"; CC "200", DCC "700", etc. At the fourth iteration, a subtractive principle is employed, with the base placed before the higher base: IV for "4", IX for "9", XL for "40", XC for "90", CD for "400", CM for "900".

The basic multiples of Roman numerals thus follow a pattern:

A practical way to write a Roman number is to consider the modern Arabic numeral system, and separately convert the thousands, hundreds, tens, and ones as given in the chart above. So, for instance, 1234 may be thought of as "one thousand and two hundreds and three tens and four", obtaining M (one thousand) + CC (two hundreds) + XXX (thirty) + IV (four), for MCCXXXIV. Thus eleven is XI (ten and one), 29 is XXIX (twenty and nine), and 2011 is MMXI (two thousand and ten and one). Note that the subtractive principle is not extended beyond the chart: for example, IL is not used for 49, rather this should be written as forty (XL) and nine (IX), or XLIX.

Given a list of numbers, you are to rearrange them so that if we write them as Roman numbers, they are in lexicographical order.

Input

There are multiple test cases. The first line of input is an integer T ≈ 100 indicating the number of test cases.

Each test case starts with an integer 1 ≤ n ≤ 10000. Then n numbers 0 < ai < 4000.

Output

For each test case, output the n numbers in specified order.

Sample Input

3
3
1 2 3
7
1 5 10 50 100 500 1000
11
4 5 6 7 8 9 10 11 12 13 14

Sample Output

1 2 3
100 500 1 50 1000 5 10
4 9 5 6 7 8 10 11 12 13 14

代碼：

 1#include<stdio.h>
 2#include<string.h>
 3#include<algorithm>
 4using namespace std;
 5char strHundreds[10][5]={"C","CC","CCC","CD","D","DC","DCC","DCCC","CM"};
 6char strTens[10][5]={"X","XX","XXX","XL","L","LX","LXX","LXXX","XC"};
 7char strOnes[10][5]={"I","II","III","IV","V","VI","VII","VIII","IX"};
 8char strThousands[4][4]={"M","MM","MMM"};
 9typedef struct node
10{
11    char str[20];
12    int flg;
13}NODE;
14NODE data[10005];
15int num[10005]={0};
16int cmp(NODE b, NODE c)
17{
18    if(strcmp(b.str,c.str)<0)
19        return 1;
20    else
21        return 0;
22}
23int main()
24{
25    int i,T,n,num_temp_a;
26    char tmp[20]="\0";
27    scanf("%d",&T);
28    while(T--)
29    {
30        scanf("%d",&n);
31        for(i=0;i<n;i++)
32        {
33            scanf("%d",&num[i]);
34            num_temp_a=num[i];
35            if(num_temp_a/1000!=0)
36            {
37                strcat(tmp,strThousands[num_temp_a/1000-1]);
38                num_temp_a=num_temp_a%1000;
39            }
40            if(num_temp_a/100!=0)
41            {
42                strcat(tmp,strHundreds[num_temp_a/100-1]);
43                num_temp_a=num_temp_a%100;
44            }
45            if(num_temp_a/10!=0)
46            {
47                strcat(tmp,strTens[num_temp_a/10-1]);
48                num_temp_a=num_temp_a%10;
49            }
50            if(num_temp_a!=0)
51            {
52                strcat(tmp,strOnes[num_temp_a-1]);
53            }
54            strcpy(data[i].str,tmp);
55            data[i].flg=i;
56            memset(tmp,0,sizeof(tmp));
57        }
58        sort(data,data+n,cmp);
59        printf("%d",num[data[0].flg]);
60        for(i=1;i<n;i++)
61            printf(" %d",num[data[i].flg]);
62        printf("\n");
63    }
64    return 0;
65}
66