HDU6077 思维（水）

Time To Get Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 407    Accepted Submission(s): 319

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 has5 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 stringt 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..XX..X....X.X....X.X..X......XX.....XX...XX.X..X.X....X....X.X..XX..X.X.........X.X..X.XX...XX.....XX...XX.

 

Sample Output

02:38

 

Source

2017 Multi-University Training Contest - Team 4

比较耗时间的方法：

#include<cstdio>#include<iostream>#include<cstring>#include<string>using namespace std;char str[10][30];char map[10][7][10]={                   {".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.",}};int get_num(int s){     int i,j,k;      for(k=0;k<10;k++)  {    for(i=0;i<7;i++){for(j=s;j<s+4;j++){if(map[k][i][j-s]!=str[i][j])break;}if(j!=s+4)        break;}if(i==7&&j==s+4){//cout<<k<<endl; return k;}   }}int main(){int t;int h1,h2,m1,m2;scanf("%d",&t);while(t--){for(int i=0;i<7;i++){scanf("%s",str[i]);}        h1=get_num(0);        h2=get_num(5);        m1=get_num(12);        m2=get_num(17);        printf("%d%d:%d%d\n",h1,h2,m1,m2);}} 

找规律特点的便捷的方法：

#include <stdio.h>#include <string.h>#include <iostream>#include <algorithm>using namespace std;typedef long long LL;const int MOD = 1e9 + 7;const int MAXN = 5e5 + 10;char str[30][30];int check(int s) {if (str[0][s + 1] == 'X') {if (str[1][s] == 'X') {if (str[1][s + 3] == 'X') {if (str[3][s + 1] == 'X') {if (str[4][s] == 'X') return 8;else return 9;} else {return 0;}} else {if (str[4][s] == 'X') return 6;else return 5;}} else {if (str[4][s] == 'X') return 2;else if (str[3][s+1] == 'X') return 3;else return 7;}} else {if (str[1][s] == 'X') return 4;else return 1;}}int main(){int T;for (scanf("%d", &T); T--; ) {for (int i = 0; i < 7; i++) {scanf("%s", str[i]);}printf("%d%d:%d%d\n", check(0), check(5), check(12), check(17));int h1 = check(0);}return 0;}