模拟时钟

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1003: Time

Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 113 Solved: 67
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Description

Digital clock use 4 digits to express time, each digit is described by 3*3 characters (including”|”,”_”and” “).now given the current time, please tell us how can it be expressed by the digital clock.

Input

There are several test cases.

Each case contains 4 integers in a line, separated by space.

Proceed to the end of file.

Output

For each test case, output the time expressed by the digital clock such as Sample Output.

Sample Input

1 2 5 62 3 4 2

Sample Output

    _  _  _   | _||_ |_   ||_  _||_| _  _     _  _| _||_| _||_  _|  ||_

HINT

The digits showed by the digital clock are as follows:   _  _     _  _  _  _  _  _  | _| _||_||_ |_   ||_||_|| | ||_  _|  | _||_|  ||_| _||_|

Source

辽宁省赛2010

模拟题：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <climits>#include <cassert>#include <complex>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>using namespace std;#pragma comment(linker, "/STACK:102400000,102400000")typedef long long LL;typedef double DB;typedef unsigned uint;typedef unsigned long long uLL;const int MOD = int(1e9)+7;const int INF = 0x3f3f3f3f;const LL INFF = 0x3f3f3f3f3f3f3f3fLL;const DB EPS = 1e-9;const DB OO = 1e20;const DB PI = acos(-1.0); //M_PI;const int maxn = 1001;int a[4];string s1[maxn] = { " _ " , "   " , " _ " , " _ " , "   " , " _ " , " _ " , " _ " , " _ " , " _ " };string s2[maxn] = { "| |" , "  |" , " _|" , " _|" , "|_|" , "|_ " , "|_ " , "  |" , "|_|" , "|_|" };string s3[maxn] = { "|_|" , "  |" , "|_ " , " _|" , "  |" , " _|" , "|_|" , "  |" , "|_|" , " _|" };int main(){    #ifdef xxz    freopen("in.txt","r",stdin);    #endif    while(cin >> a[0] >>a[1] >> a[2] >> a[3])    {        for(int i = 0 ; i < 4 ; i ++)        {            cout << s1[a[i]];        }        cout << endl;        for(int i = 0 ; i < 4 ; i ++)        {            cout << s2[a[i]];        }        cout << endl;        for(int i = 0 ; i < 4 ; i ++)        {            cout << s3[a[i]];        }        cout << endl;    }}

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