HDU 3257 Hello World!（位运算&模拟）

神头鬼脸的位运算
(●ˇ∀ˇ●)o(^▽^)o
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Problem Description
Your task is to print … er … “Hello World” … in a fantastic way – using a beautiful font.

I’ve sent you a nice font for you to use, but I’m too busy to tell you how. Can you help yourself?

Input
The first line contains a single integer T (T <= 20), the number of test cases.
Each case begins with an integer C (1 <= C <= 80) in a single line, then each of the following C lines contains five two-digit numbers in hex (letters will be in uppercase). Don’t ask me what they mean, I’m too busy…

Output
For each test case, print the case number in the first line, then followed by a blank line.
After that, print all T characters. Use a single blank column of spaces between two consecutive characters. Each line should have exactly 6C-1 character (again, don’t ask me why).
Don’t forget to print another blank line after the output of each test case.

Sample Input
2
11
7F 08 08 08 7F
38 54 54 54 18
00 41 7F 40 00
00 41 7F 40 00
38 44 44 44 38
00 00 00 00 00
3F 40 38 40 3F
38 44 44 44 38
7C 08 04 04 08
00 41 7F 40 00
38 44 44 48 7F
5
14 08 3E 08 14
04 02 01 02 04
40 40 40 40 40
04 02 01 02 04
14 08 3E 08 14

Sample Input
2
11
7F 08 08 08 7F
38 54 54 54 18
00 41 7F 40 00
00 41 7F 40 00
38 44 44 44 38
00 00 00 00 00
3F 40 38 40 3F
38 44 44 44 38
7C 08 04 04 08
00 41 7F 40 00
38 44 44 48 7F
5
14 08 3E 08 14
04 02 01 02 04
40 40 40 40 40
04 02 01 02 04
14 08 3E 08 14


Sample Output
Case 1:

#   #        ##    ##               #   #              ##       #
#   #         #     #               #   #               #       #
#   #  ###    #     #    ###        #   #  ###  # ##    #    ## #
##### #   #   #     #   #   #       # # # #   # ##  #   #   #  ##
#   # #####   #     #   #   #       # # # #   # #       #   #   #
#   # #       #     #   #   #       # # # #   # #       #   #   #
#   #  ###   ###   ###   ###         # #   ###  #      ###   ####

Case 2:

        #           #        
  #    # #         # #    #  
# # # #   #       #   # # # #
 ###                     ### 
# # #                   # # #
  #                       #  
            #####  

要点：
- scanf(“%x”,&balabala…)读入十六进制
- 输出先列后行，两个十六进制字符为一列，输入的每五组十六进制对应输出的一个字符
- 格式：输出的每个字符间有空格，最后一个没有(6*C-1的原因)，依题意最后还要回车
- 位运算判断每个二进制的值
- eg.0X7F = 0111 1111从上往下，为零的是空格，为一的是#
- 判断第k位是否为零：a[j]&1<<k
- 出题人真的很皮欸 (～￣▽￣)→))￣▽￣)o

#include <stdio.h>
int main()
{
    int a[5],c,t,cnt=0;
    char ans[7][512];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&c);
        for(int i=0; i<c; i++)
            for(int j=0; j<5; j++)
            {
                scanf("%X",&a[j]);
                for(int k=6; k>=0; k--)
                    ans[k][i*5+j] = (a[j]&1<<k)?'#':' ';
            }
        printf("Case %d:\n\n",++cnt);
        for(int j=0; j<7; j++)
        {
            for(int k=0; k<5*c; k++)
                (k%5==0&&k!=0)?printf(" %c",ans[j][k]):printf("%c",ans[j][k]);
            printf("\n");
        }
        printf("\n");
    }
    return 0;
}

Emmmm….
后来想了想其实可以直接输出的…
之前做复杂了
,,ԾㅂԾ,,

#include <stdio.h>
int main()
{
    int a[400],c,t,cnt = 0;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&c);
        for(int i=0; i<c; i++)
            for(int j=1; j<=5; j++)
                scanf("%X",&a[i*5+j]);
        printf("Case %d:\n\n",++cnt);
        for(int i=0; i<7; i++)
        {
            for(int j=1,k=1; j<=5*c; j++,k++)
            {
                if(k%6==0)
                    printf(" "),k=1;
                printf("%c",(a[j]&1)?'#':' ');
                a[j] >>= 1;
            }
            printf("\n");
        }
        printf("\n");
    }
    return 0;
}