toj 2807 Number Sort

 

2807.   Number Sort

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Time Limit: 10.0 Seconds   Memory Limit: 65536K
Total Runs: 469   Accepted Runs: 160

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Given you N positive integers, your job is to output these N numbers in the ascending order in every digit (if two numbers in some digit are the same, you should output the smaller one first). For example, there are five numbers 1 11 111 1111 11111. You should output five lines as sample output. If a number doesn't have the i^th digit, you can consider its i^th digit zero.

Input

The input contains several test cases. Each case contains two lines. The first line contains a positive integer N (N ≤ 1000), indicating the length of the sequence. The second line gives N numbers (each number will not exceed 2147483647). If N = 0 it signals the end of the input and isn't considered to be processed.

Output

Each case contains several lines. The first line is indicated as sample output. The i + 1 line contains these N numbers in ascending order in the i^th digit.

Sample Input

5
1 11 111 1111 11111
7
19 28 37 46 55 55 45
0

Sample Output

Case 1:
1 11 111 1111 11111
1 11 111 1111 11111
1 11 111 1111 11111
1 11 111 1111 11111
1 11 111 1111 11111
Case 2:
45 55 55 46 37 28 19
19 28 37 45 46 55 55

Problem Setter: mmatch@TJU

Source: TJU Programming Contest 2007 Preliminary

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//

#include <iostream>
#include <algorithm>
#include <cmath>
#define MAX 1002
using namespace std;
int data[MAX],ddd[MAX];
struct node
{
    int num;
    int dd;
    int da;
}s[MAX];
bool comp(node a,node b)
{
    if(a.dd!=b.dd)
        return a.dd<b.dd;
    else
        return a.da<b.da;
}
int n;
int main()
{
    int i,MM,j,zzz=1,k;
    while(scanf("%d",&n)!=EOF && n)
    {
        MM=-1;
        for(i=0;i<n;i++)
        {
            scanf("%d",&data[i]);
            ddd[i]=data[i];
            if(data[i]>MM)
                MM=data[i];
        }
        int nn=0;
        while(MM!=0)
        {
            MM=MM/10;
            nn++;
        }
        int cs=10;
        printf("Case %d:\n",zzz++);
        for(i=0;i<nn;i++)
        {
            for(j=0;j<n;j++)
            {
                s[j].num=j;
                s[j].da=data[j];
                s[j].dd=ddd[j]%cs;
                ddd[j]=ddd[j]/10;
            }
            sort(s,s+n,comp);
            printf("%d",data[s[0].num]);
            for(k=1;k<n;k++)
            {
                printf(" %d",data[s[k].num]);
            }
            printf("\n");
        }
    }
    return 0;
}