HDU 6077 Time To Get Up 模拟
Time To Get Up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 277 Accepted Submission(s): 227
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.
![](http://acm.hdu.edu.cn/data/images/C762-1011-1.jpg)
Your job is to help Little Q read the time shown on his clock.
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.
![](http://acm.hdu.edu.cn/data/images/C762-1011-1.jpg)
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.
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
就是一条挺傻逼的模拟 然而当时多校还写错了一个地方wa了一发.....
#include <iostream> #include <cstring> #include <cstdio> #include <algorithm> #include <queue> #include <vector> #include <iomanip> #include <math.h> #include <map> using namespace std; #define FIN freopen("input.txt","r",stdin); #define FOUT freopen("output.txt","w",stdout); #define INF 0x3f3f3f3f #define INFLL 0x3f3f3f3f3f3f3f #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 typedef long long LL; typedef pair<int, int> PII; char mp[10][30]; int judge(int a, int b, int c, int d, int e, int f, int g) { if(a == 1 && b == 1 && c == 1 && d == 0 && e == 1 && f == 1 && g == 1) return 0; else if(a == 0 && b == 0 && c == 1 && d == 0 && e == 0 && f == 0 && g == 1) return 1; else if(a == 0 && b == 1 && c == 1 && d == 1 && e == 1 && f == 1 && g == 0) return 2; else if(a == 0 && b == 1 && c == 1 && d == 1 && e == 0 && f == 1 && g == 1) return 3; else if(a == 1 && b == 0 && c == 1 && d == 1 && e == 0 && f == 0 && g == 1) return 4; else if(a == 1 && b == 1 && c == 0 && d == 1 && e == 0 && f == 1 && g == 1) return 5; else if(a == 1 && b == 1 && c == 0 && d == 1 && e == 1 && f == 1 && g == 1) return 6; else if(a == 0 && b == 1 && c == 1 && d == 0 && e == 0 && f == 0 && g == 1) return 7; else if(a == 1 && b == 1 && c == 1 && d == 1 && e == 1 && f == 1 && g == 1) return 8; else if(a == 1 && b == 1 && c == 1 && d == 1 && e == 0 && f == 1 && g == 1) return 9; } int main() { //FIN int T; scanf("%d", &T); while(T--) { for(int i = 0; i < 7; i++) scanf("%s", mp[i]); //for(int i = 0; i < 7; i++) cout << mp[i] <<endl; int a = 0, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0; if(mp[1][0] == 'X') a = 1; if(mp[0][1] == 'X') b = 1; if(mp[1][3] == 'X') c = 1; if(mp[3][1] == 'X') d = 1; if(mp[4][0] == 'X') e = 1; if(mp[6][1] == 'X') f = 1; if(mp[4][3] == 'X') g = 1; int ans1 = judge(a, b, c, d, e, f, g); a = 0, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0; if(mp[1][5] == 'X') a = 1; if(mp[0][7] == 'X') b = 1; if(mp[1][8] == 'X') c = 1; if(mp[3][7] == 'X') d = 1; if(mp[4][5] == 'X') e = 1; if(mp[6][7] == 'X') f = 1; if(mp[4][8] == 'X') g = 1; //cout << mp[1][9] <<endl; int ans2 = judge(a, b, c, d, e, f, g); //cout << a<<" "<<b<<" "<<c<<" "<<d<<" "<<e<<" "<<f<<" "<<g<<endl; //cout << ans2<<endl; a = 0, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0; if(mp[1][12] == 'X') a = 1; if(mp[0][14] == 'X') b = 1; if(mp[1][15] == 'X') c = 1; if(mp[3][14] == 'X') d = 1; if(mp[4][12] == 'X') e = 1; if(mp[6][14] == 'X') f = 1; if(mp[4][15] == 'X') g = 1; int ans3 = judge(a, b, c, d, e, f, g); a = 0, b = 0, c = 0, d = 0, e = 0, f = 0, g = 0; if(mp[1][17] == 'X') a = 1; if(mp[0][19] == 'X') b = 1; if(mp[1][20] == 'X') c = 1; if(mp[3][19] == 'X') d = 1; if(mp[4][17] == 'X') e = 1; if(mp[6][19] == 'X') f = 1; if(mp[4][20] == 'X') g = 1; int ans4 = judge(a, b, c, d, e, f, g); printf("%d%d:%d%d\n", ans1, ans2, ans3, ans4); } return 0; }