Question

我在制作这个节目时遇到了一些问题。我需要做的是为我班上的一个项目。我需要做的是这样的：http://patorjk.com/software/taag/

我可以一次打印一个字母，但问题是我无法打印像“ABC”那样的东西，它将它们全部打印在同一个地方。所以，如果你们中的任何一个人能够帮助我，并告诉我应该如何做的印刷将是如此美好。

#include <iostream>
#include <string.h>
#include<windows.h>
#include <stdio.h>
using namespace std;

char A[4][12]={
"    //   ",
"  // //  ",
" /////// ",
"//     //"};

char B[5][12]={
" /////   ",
" //  //  ",
" /////   ",
" //  //  ",
" /////   "};

char C[5][12]={
"  /////  ",
" //      ",
" //      ",
" //      ",
"  /////  "};

char D[5][12]={
" /////   ",
" //  //  ",
" //   // ",
" //  //  ",
" /////   "};

char E[5][12]={
" /////   ",
" //      ",
" ////    ",
" //      ",
" /////   "};

char F[5][12]={
" /////   ",
" //      ",
" ////    ",
" //      ",
" //      "};

char G[5][12]={
"  /////  ",
" //      ",
"//  ///  ",
" //   // ",
"  /////  "};

char H[5][12]={
" //   // ",
" //   // ",
" /////// ",
" //   // ",
" //   // "};

char I[5][12]={
"    **   ",
"    //   ",
"    //   ",
"    //   ",
"    //   "};

char J[5][12]={
"      // ",
"      // ",
"      // ",
"  //  // ",
"   ////  "};

char K[5][12]={
" //  //  ",
" // //   ",
" ///     ",
" // //   ",
" //  //  "};



int main()
{
    int x, k = 0;
    char a[999];
    cin.get(a, 999);
    x = strlen(a);
    while (k < 6)
    {
        for (int i = 0; i <= x; i++)
        {
            if (a[i] == 'A')
                cout << A[k] << endl;
            if (a[i] == 'B')
                cout << B[k] << endl;
            if (a[i] == 'C')
                cout << C[k] << endl;
            if (a[i] == 'D')
                cout << D[k] << endl;
            if (a[i] == 'E')
                cout << E[k] << endl;
            if (a[i] == 'F')
                cout << F[k] << endl;
            if (a[i] == 'G')
                cout << G[k] << endl;
            if (a[i] == 'H')
                cout << H[k] << endl;
            if (a[i] == 'I')
                cout << I[k] << endl;
            if (a[i] == 'J')
                cout << J[k] << endl;
            if (a[i] == 'K')
                cout << K[k] << endl;
        }
        k = k + 1;
   }

    return 0;
}

Answer 1

我写了一个非常相似的程序，通过打印散列来打印大尺寸的整数。这是程序：

/**
Author:- Mayukh Datta
**/
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#define H 7
#define W 8 //one extra room in the char array is required for storing '\0'
void main()
{
    char num[11];  //here too one extra room is needed for the '\0'
    char c;  //for option
    int i, j, k;
    //declaring char 2D arrays and initializing with hash-printed digits
    char zero[H][W]={" ##### ", //H=0
                     " #   # ", //H=1
                     " #   # ", //H=2
                     " #   # ", //H=3
                     " #   # ", //H=4
                     " #   # ", //H=5
                     " ##### "},//H=6

         one[H][W]={"   #   ",
                    "  ##   ",
                    "   #   ",
                    "   #   ",
                    "   #   ",
                    "   #   ",
                    " ##### "},

         two[H][W]={" ##### ",
                    "     # ",
                    "     # ",
                    " ##### ",
                    " #     ",
                    " #     ",
                    " ##### "},

         three[H][W]={" ##### ",
                      "     # ",
                      "     # ",
                      " ##### ",
                      "     # ",
                      "     # ",
                      " ##### "},

         four[H][W]={" #     ",
                     " #   # ",
                     " #   # ",
                     " ##### ",
                     "     # ",
                     "     # ",
                     "     # "},

         five[H][W]={" ##### ",
                     " #     ",
                     " #     ",
                     " ##### ",
                     "     # ",
                     "     # ",
                     " ##### "},

         six[H][W]={" ##### ",
                    " #     ",
                    " #     ",
                    " ##### ",
                    " #   # ",
                    " #   # ",
                    " ##### "},

         seven[H][W]={" ##### ",
                      "     # ",
                      "     # ",
                      "  #### ",
                      "     # ",
                      "     # ",
                      "     # "},

         eight[H][W]={" ##### ",
                      " #   # ",
                      " #   # ",
                      " ##### ",
                      " #   # ",
                      " #   # ",
                      " ##### "},

         nine[H][W]={" ##### ",
                     " #   # ",
                     " #   # ",
                     " ##### ",
                     "     # ",
                     "     # ",
                     "     # "};

    do
    {
        printf("Enter a number upto 10 digits:- ");
        fflush(stdin);
        gets(num);
        if(strlen(num)>10)
           printf("\nYou must enter a number upto 10 digits.\nTry again!\n");
        else
        {
            printf("\n");

            k=1;
            j=0;  //controls H of each digit
            while(k<=7)  //controls height
            {
                for(i=0;i<strlen(num);i++)  //reads each digit
                {
                    if(num[i]=='0')
                        printf("%s", zero[j]);
                    else if(num[i]=='1')
                        printf("%s", one[j]);
                    else if(num[i]=='2')
                        printf("%s", two[j]);
                    else if(num[i]=='3')
                        printf("%s", three[j]);
                    else if(num[i]=='4')
                        printf("%s", four[j]);
                    else if(num[i]=='5')
                        printf("%s", five[j]);
                    else if(num[i]=='6')
                        printf("%s", six[j]);
                    else if(num[i]=='7')
                        printf("%s", seven[j]);
                    else if(num[i]=='8')
                        printf("%s", eight[j]);
                    else if(num[i]=='9')
                        printf("%s", nine[j]);
                }
                printf("\n");
                k++;
                j++;
            }
        }
        printf("\nEnter Y to continue:- ");
        fflush(stdin);
        scanf("%c", &c);
    }while(c=='Y'||c=='y');
}

了解循环在我的程序中的工作方式。链接：http://www.thecoducer.com/2017/08/printing-the-number-in-large-size.html

在控制台中打印Ascii Art字母

1 个答案: