NBUT - 1219 C - Time

题目连接:
https://vjudge.net/problem/NBUT-1219

Time
时间限制: 1000 ms　内存限制: 131072 K
问题描述
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.
输入
There are several test cases.
Each case contains 4 integers in a line, separated by space.
Proceed to the end of file.
输出
For each test case, output the time expressed by the digital clock such as Sample Output.
样例输入
1 2 5 6
2 3 4 2
样例输出

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

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

这道题的意思就是输入数字然后模拟时钟，就是一个简单的给字符串赋值的过程
可以定义一个二维的数组来模拟输出的这几个数字，后来看到有大佬直接暴力定义三个一维数组来模拟，学习一波。

方法一
#include<iostream>
#include<algorithm>
using namespace std;
char s1[35]="     _  _     _  _  _  _  _  _ ";
char s2[35]="   | _| _||_||_ |_   ||_||_|| |";
char s3[35]="   ||_  _|  | _||_|  ||_| _||_|";
int main()
{
    int a,b,c,d;
    while(cin>>a>>b>>c>>d)
    {
        getchar();
        if(a==0) a=10;
        if(b==0) b=10;
        if(c==0) c=10;
        if(d==0) d=10;
        cout<<s1[a*3-2]<<s1[a*3-1]<<s1[a*3]<<s1[b*3-2]<<s1[b*3-1]<<s1[b*3]<<s1[c*3-2]<<s1[c*3-1]<<s1[c*3]<<s1[d*3-2]<<s1[d*3-1]<<s1[d*3]<<endl;
        cout<<s2[a*3-2]<<s2[a*3-1]<<s2[a*3]<<s2[b*3-2]<<s2[b*3-1]<<s2[b*3]<<s2[c*3-2]<<s2[c*3-1]<<s2[c*3]<<s2[d*3-2]<<s2[d*3-1]<<s2[d*3]<<endl;
        cout<<s3[a*3-2]<<s3[a*3-1]<<s3[a*3]<<s3[b*3-2]<<s3[b*3-1]<<s3[b*3]<<s3[c*3-2]<<s3[c*3-1]<<s3[c*3]<<s3[d*3-2]<<s3[d*3-1]<<s3[d*3]<<endl;
    }
    return 0;
}


方法二
#include<iostream>
#include<algorithm>
using namespace std;
char c[5][35]=
{
    " _     _  _     _  _  _  _  _ ",
    "| |  | _| _||_||_ |_   ||_||_|",
    "|_|  ||_  _|  | _||_|  ||_| _|",
};
int main()
{
    int a[5];
    while(cin>>a[0]>>a[1]>>a[2]>>a[3])
        for(int k=0; k<3; k++)
        {
            for(int i=0; i<4; i++)
                for(int j=a[i]*3; j<a[i]*3+3; j++)
                    cout<<c[k][j];
            cout<<endl;
        }
    return 0;
}