2807.
Number Sort
Time Limit: 10.0 Seconds
Memory Limit: 65536K
Total Runs: 469
Accepted Runs: 160
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 ith digit, you can consider its ith 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
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