Time To Get Up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others) Total Submission(s): 478 Accepted Submission(s): 382
Problem Description
Little Q's clock is alarming! It's time to get up now! However, after reading the time on the clock, Little Q lies down and starts sleeping again. Well, he has
5
alarms, and it's just the first one, he can continue sleeping for a while. Little Q's clock uses a standard 7-segment LCD display for all digits, plus two small segments for the '':'', and shows all times in a 24-hour format. The '':'' segments are on at all times.
Your job is to help Little Q read the time shown on his clock.
Input
The first line of the input contains an integer
T(1≤T≤1440)
, denoting the number of test cases. In each test case, there is an
7×21
ASCII image of the clock screen. All digit segments are represented by two characters, and each colon segment is represented by one character. The character ''X'' indicates a segment that is on while ''.'' indicates anything else. See the sample input for details.
Output
For each test case, print a single line containing a string
t
in the format of
HH:MM
, where
t(00:00≤t≤23:59)
, denoting the time shown on the clock.
Sample Input
1
.XX...XX.....XX...XX.
X..X....X......X.X..X
X..X....X.X....X.X..X
......XX.....XX...XX.
X..X.X....X....X.X..X
X..X.X.........X.X..X
.XX...XX.....XX...XX.
Sample Output
02:38
解析：看到这题首先想到的就是把时钟模拟出来吧，模拟时钟上数字显示的点数，来判断这个数字是什么。相信大佬们也都是这样想的，我写的代码在杭电上A了，但是。。。在我们学校自己的网站上提交却直接Wa掉了（我也很绝望啊！），至今还未找出错误，暂且先列出来，欢迎大家来吐槽错误。
首先先看一下借（偷偷借的）的大佬的代码。
C++ Code
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#include
<iostream>
#include
<cstdio>
#include
<cstring>
using
namespace
std;
const
char
num[
10
][
7
][
5
] = {
".XX."
,
"X..X"
,
"X..X"
,
"...."
,
"X..X"
,
"X..X"
,
".XX."
,
"...."
,
"...X"
,
"...X"
,
"...."
,
"...X"
,
"...X"
,
"...."
,
".XX."
,
"...X"
,
"...X"
,
".XX."
,
"X..."
,
"X..."
,
".XX."
,
".XX."
,
"...X"
,
"...X"
,
".XX."
,
"...X"
,
"...X"
,
".XX."
,
"...."
,
"X..X"
,
"X..X"
,
".XX."
,
"...X"
,
"...X"
,
"...."
,
".XX."
,
"X..."
,
"X..."
,
".XX."
,
"...X"
,
"...X"
,
".XX."
,
".XX."
,
"X..."
,
"X..."
,
".XX."
,
"X..X"
,
"X..X"
,
".XX."
,
".XX."
,
"...X"
,
"...X"
,
"...."
,
"...X"
,
"...X"
,
"...."
,
".XX."
,
"X..X"
,
"X..X"
,
".XX."
,
"X..X"
,
"X..X"
,
".XX."
,
".XX."
,
"X..X"
,
"X..X"
,
".XX."
,
"...X"
,
"...X"
,
".XX."
};
char
tm[
10
][
25
];
char
n[
7
][
5
];
int
check() {
for
(
int
i=
0
; i<
10
; i++) {
bool
flag =
true
;
for
(
int
j=
0
; j<
7
; j++) {
if
(strcmp(n[j], num[i][j]) !=
0
) { flag =
false
;
break
; } }
if
(flag) {
return
i; } }
return
-
1
; }
int
main() {
int
t; scanf(
"%d"
, &t);
while
(t >
0
) { t--;
for
(
int
i=
0
; i<
7
; i++) scanf(
"%s"
, tm[i]);
int
a, b, c, d;
for
(
int
i=
0
; i<
7
; i++) { strncpy(n[i],tm[i],
4
); } a = check();
for
(
int
i=
0
; i<
7
; i++) { strncpy(n[i],tm[i]+
5
,
4
); } b = check();
for
(
int
i=
0
; i<
7
; i++) { strncpy(n[i], tm[i]+
12
,
4
); } c = check();
for
(
int
i=
0
; i<
7
; i++) { strncpy(n[i], tm[i]+
17
,
4
); } d = check(); printf(
"%d%d:%d%d\n"
, a,b,c,d); }
return
0
; }
简洁明了；
在这里再厚着脸皮亮一下自己小渣渣的代码：
C++ Code
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///找到电子时钟的显示规律；
#include
<iostream>
#include
<cstdio>
using
namespace
std;
int
main() {
char
str[
7
][
21
];
int
o,l;
int
a[
4
]; scanf(
"%d"
,&o); getchar();
while
(o--) {
for
(
int
i=
0
;i<
7
;i++)
///时钟的前两个数字
{
for
(
int
j=
0
;j<
21
;j++) { str[i][j]=getchar(); } getchar(); } l=
0
;
for
(
int
p=
0
;p<
6
;p+=
5
) {
if
(str[
3
][
1
+p]==
'.'
) {
if
(str[
6
][
1
+p]==
'X'
) { a[l]=
0
; l++;
continue
; }
else
if
(str[
0
][
1
+p]==
'.'
) { a[l]=
1
; l++;
continue
; }
else
{ a[l]=
7
; l++;
continue
; } }
else
if
(str[
4
][
3
+p]==
'.'
) { a[l]=
2
; l++;
continue
; }
else
if
(str[
1
][
0
+p]==
'.'
) { a[l]=
3
; l++;
continue
; }
else
if
(str[
0
][
1
+p]==
'.'
) { a[l]=
4
; l++;
continue
; }
else
if
(str[
1
][
3
+p]==
'.'
) {
if
(str[
4
][
0
+p]==
'.'
) { a[l]=
5
; l++;
continue
; }
else
{ a[l]=
6
; l++;
continue
; } }
else
if
(str[
4
][
0
+p]==
'X'
) { a[l]=
8
; l++;
continue
; }
else
{ a[l]=
9
; l++;
continue
; } }
for
(
int
p=
12
;p<
19
;p+=
5
)
///时钟的后两个数字
{
if
(str[
3
][
1
+p]==
'.'
) {
if
(str[
6
][
1
+p]==
'X'
) { a[l]=
0
; l++;
continue
; }
else
if
(str[
0
][
1
+p]==
'.'
) { a[l]=
1
; l++;
continue
; }
else
{ a[l]=
7
; l++;
continue
; } }
else
if
(str[
4
][
3
+p]==
'.'
) { a[l]=
2
; l++;
continue
; }
else
if
(str[
1
][
0
+p]==
'.'
) { a[l]=
3
; l++;
continue
; }
else
if
(str[
0
][
1
+p]==
'.'
) { a[l]=
4
; l++;
continue
; }
else
if
(str[
1
][
3
+p]==
'.'
) {
if
(str[
4
][
0
+p]==
'.'
) { a[l]=
5
; l++;
continue
; }
else
{ a[l]=
6
; l++;
continue
; } }
else
if
(str[
4
][
0
+p]==
'X'
) { a[l]=
8
; l++;
continue
; }
else
{ a[l]=
9
; l++;
continue
; } } printf(
"%d%d:%d%d\n"
,a[
0
],a[
1
],a[
2
],a[
3
]); }
return
0
; }
过程看起来比较繁琐，但是。。也确实是有那么一点麻烦，但是真正写的倒是不多，通过每一个数字或几个数字在时钟上的显示点不同来判断数字，上下有分了两个部分（前两个数字和后两个数字，后两个数字直接复制前面的改一下起始位置即可），自我感觉还是比较好理解的；应该还有一些错误我没有考虑到的欢迎吐槽。
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