2017杭电多校第四场 1011 Time To Get Up（模拟）HDU 6077

000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 27    Accepted Submission(s): 25

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..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

 

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题意：读数

解题思路：就模拟。。

#include<cstdio>#include<iostream>#include<cstring>using namespace std;int f[10][9]={{1,1,1,1,1,1,0},{0,0,1,1,0,0,0},{0,1,1,0,1,1,1},{0,1,1,1,1,0,1},{1,0,1,1,0,0,1},{1,1,0,1,1,0,1},{1,1,0,1,1,1,1},{0,1,1,1,0,0,0},{1,1,1,1,1,1,1},{1,1,1,1,1,0,1}};char c[10][25];int pan(int a[]){for(int i=0;i<10;i++){int flag=0;for(int j=0;j<7;j++){//printf("a[%d]=%d f[%d][%d]=%d\n",j,a[j],i,j,f[i][j]);if(a[j]!=f[i][j]){flag=1;break;}}if(flag==0)return i;}return 10;}int main(){int t;scanf("%d",&t);while(t--){for(int i=0;i<7;i++){scanf("%s",c[i]);}int h1,h2,m1,m2;int a[10];memset(a,0,sizeof(a));if(c[0][1]=='X'&&c[0][2]=='X')a[1]=1;if(c[3][1]=='X'&&c[3][2]=='X')a[6]=1;if(c[6][1]=='X'&&c[6][2]=='X')a[4]=1;if(c[1][0]=='X'&&c[2][0]=='X')a[0]=1;if(c[4][0]=='X'&&c[5][0]=='X')a[5]=1;if(c[1][3]=='X'&&c[2][3]=='X')a[2]=1;if(c[4][3]=='X'&&c[5][3]=='X')a[3]=1;//for(int i=0;i<7;i++)cout<<a[i]<<" ";h1=pan(a);//cout<<h1<<endl;memset(a,0,sizeof(a));if(c[0][6]=='X'&&c[0][7]=='X')a[1]=1;if(c[3][6]=='X'&&c[3][7]=='X')a[6]=1;if(c[6][6]=='X'&&c[6][7]=='X')a[4]=1;if(c[1][5]=='X'&&c[2][5]=='X')a[0]=1;if(c[4][5]=='X'&&c[5][5]=='X')a[5]=1;if(c[1][8]=='X'&&c[2][8]=='X')a[2]=1;if(c[4][8]=='X'&&c[5][8]=='X')a[3]=1;//for(int i=0;i<7;i++)cout<<a[i]<<" ";h2=pan(a);//cout<<h2<<endl;memset(a,0,sizeof(a));if(c[0][13]=='X'&&c[0][14]=='X')a[1]=1;if(c[3][13]=='X'&&c[3][14]=='X')a[6]=1;if(c[6][13]=='X'&&c[6][14]=='X')a[4]=1;if(c[1][12]=='X'&&c[2][12]=='X')a[0]=1;if(c[4][12]=='X'&&c[5][12]=='X')a[5]=1;if(c[1][15]=='X'&&c[2][15]=='X')a[2]=1;if(c[4][15]=='X'&&c[5][15]=='X')a[3]=1;//for(int i=0;i<7;i++)cout<<a[i]<<" ";m1=pan(a);//cout<<m1<<endl;memset(a,0,sizeof(a));if(c[0][18]=='X'&&c[0][19]=='X')a[1]=1;if(c[3][18]=='X'&&c[3][19]=='X')a[6]=1;if(c[6][18]=='X'&&c[6][19]=='X')a[4]=1;if(c[1][17]=='X'&&c[2][17]=='X')a[0]=1;if(c[4][17]=='X'&&c[5][17]=='X')a[5]=1;if(c[1][20]=='X'&&c[2][20]=='X')a[2]=1;if(c[4][20]=='X'&&c[5][20]=='X')a[3]=1;m2=pan(a);printf("%d%d:%d%d\n",h1,h2,m1,m2);}return 0;}