HDU6077-Time To Get Up

Time To Get Up

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 117 Accepted Submission(s): 101

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

Source
2017 Multi-University Training Contest - Team 4

题目大意：给出时间的图像，打印出数字表示。
解题思路：模拟，观察数字的特征，哪里是实心的，哪里是空心的。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>#define mem(a,b) memset(a,b,sizeof(a))using namespace std;typedef long long LL;const LL INF=1e18;const int MAXN=1e3+100;const int MOD=1e9+7;char G[10][25];int judge(int i){    int l,r;    if(i==1) l=0,r=3;    else if(i==2) l=5,r=8;    else if(i==3) l=12,r=15;    else l=17,r=20;    if(G[1][l+1]!='X')    {        if(G[4][l+1]=='X') return 4;        else return 1;    }else    {        if(G[2][r]!='X')        {            if(G[5][l]=='X') return 6;            else return 5;        }        if(G[4][l+1]!='X')        {            if(G[2][l]=='X') return 0;            else  return 7;        }        if(G[2][l]=='X')        {            if(G[5][l]=='X') return 8;            else return 9;        }        if(G[5][l]=='X') return 2;        else return 3;    }}int main(){    int T;    scanf("%d",&T);    while(T--)    {        for(int i=1;i<=7;++i)        {            scanf("%s",&G[i]);        }        int h1=judge(1);        int h2=judge(2);        int h3=judge(3);        int h4=judge(4);        printf("%d%d:%d%d\n",h1,h2,h3,h4);    }    return 0;}/*.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.*/