hdu 6077 Time To Get Up【大模擬】
Time To Get Up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1226 Accepted Submission(s): 852
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
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題意:
給你一個時鐘表,輸出
分析:
就是一個大模擬,我將每個數分成8個邊,構成一個8位的二進位制數,然後對映到每一個具體的數
程式碼:
#include<bits/stdc++.h>
#define ll long long
#define mod 1000000007
using namespace std;
char c[30][100];
int a[1000];
int b[10]={5,5,2,5,5};
int main()
{
int t,i,j,num;
a[0]=-1;
a[119]=0;
a[96]=1;
a[62]=2;
a[124]=3;
a[105]=4;
a[93]=5;
a[95]=6;
a[100]=7;
a[127]=8;
a[125]=9;
scanf("%d",&t);
while(t--)
{
//scanf("%c",c[0][0]);
for(i=0;i<7;i++)
scanf("%s",c[i]);
for(i=0,j=0;i<21;i+=b[j],j++)
{
num=0;
if(c[1][i]=='X')
num++;
if(c[4][i]=='X')
num+=(1<<1);
if(c[0][i+1]=='X')
num+=(1<<2);
if(c[3][i+1]=='X')
num+=(1<<3);
if(c[6][i+1]=='X')
num+=(1<<4);
if(c[1][i+3]=='X')
num+=(1<<5);
if(c[4][i+3]=='X')
num+=(1<<6);
if(num==119||a[num])
printf("%d",a[num]);
else
printf(":");
}
printf("\n");
}
}