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 ;
}

转载于:https://www.cnblogs.com/forever4444/archive/2009/05/14/1456960.html