HLJUOJ1003(预处理)
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1003: Time
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 27 Solved: 13
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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: _ _ _ _ _ _ _ _ | _| _||_||_ |_ ||_||_|| | ||_ _| | _||_| ||_| _||_|
解题思路:
题意要求把相应的数字转换成时钟对应的图形输出。
首先预处理出0~9所有数字的图形,这里很多细节,比如每个数字占三个位置,如果是1的话,前面两个补空格什么的····具体详见代码···此处坑无限多
完整代码:
#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;/** Constant List .. **/ //{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 DoubleQ 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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