Question

我正在制作一个神奇的数字程序。它几乎正确显示输出，但它不适用于像38＆amp; 46。

代码：

#include<stdio.h>
#include<conio.h>

int answer=0;
char resp;
void printanswer()
{
    printf("\n\n\n\n\tMAGIC GAME");
    printf("\n\n\n\n\n\n\n\t\tThe number in you mind is %d",answer);
    getch();
}

void screen6()
{
    printf("\n 32 37 42 47 52 57");
    printf("\n 33 38 43 48 53 58");
    printf("\n 34 39 44 49 54 59");
    printf(" \n 35 40 45 50 55 60");
    printf("\n 36 41 46 51 56 ");

    printf("\n\nPress 'Y' if number is present");
    printf("\n the list else press any key\n");
    resp=getche();
    if(resp=='y'||resp=='Y')
    {
        answer+=32;
        resp='n';
    }


    printanswer();
}

void screen5()
{
    printf(" \n16 21 26 31 52 57");
    printf("\n 17 22 27 48 53 58");
    printf(" \n 18 23 28 49 54 59");
    printf("\n 19 24 29 50 55 60");
    printf("\n 20 25 30 51 56 ");

    printf("\n\nPress 'Y' if number is present");
    printf("\n\n the list else press any key\n");

    resp=getche();
    if(resp=='y'||resp=='Y')
    {
        answer+=16;
        resp='n';
    }

    screen6();
}

void screen4()
{
    printf( "\n 8 13 26 31 44 57");
    printf( "\n 9 14 27 40 45 58");
    printf( "\n 10 15 28 41 46 59");
    printf("\n 11 24 29 42 47 60");
    printf("\n 12 25 30 43 56 ");

    printf("\n\nPress 'Y' if number is present");
    printf("\n\n the list else press any key\n");

    resp=getche();
    if(resp=='y'||resp=='Y')
    {
        answer+=8;
        resp='n';
    }

    screen5();
}

void screen3()
{
    printf( "\n  4 13 22 31 44 53");
    printf( " \n 5 14 23 36 45 54");
    printf( " \n 6 15 28 37 46 55");
    printf( " \n 7 20 29 38 47 60");
    printf( "\n 12 21 30 39 52 ");
    printf("\n\nPress 'Y' if number is present");
    printf("\n\n the list else press any key\n");
    resp=getche();
    if(resp=='y'||resp=='Y')
    {
        answer+=4;
        resp='n';
    }

    screen4();
}

void screen2()
{
    printf( " \n 2 11 22 31 42 51");
    printf( " \n 3 14 23 34 43 54");
    printf( " \n 6 15 26 35 46 55");
    printf( " \n 7 18 27 38 47 58");
    printf( "\n 10 19 30 39 50 59");

    printf("\n\nPress 'Y' if number is present");
    printf("\n \nthe list else press any key\n");

    resp=getche();
    if(resp=='y'||resp=='Y')
    {
        answer+=2;
        resp='n';
    }

    screen3();
}

void screen1()
{
    printf("\n 1 11 21 31 41 51");
    printf(" \n 3 13 23 33 46 53");
    printf("\n 5 15 25 35 45 55");
    printf("\n 7 17 27 37 47 57");
    printf("\n 9 19 29 39 49 59");

    printf("\n\nPress 'Y' if number is present");
    printf("\n \nthe list else press any key\n");

    resp=getche();
    if(resp=='y'||resp=='Y')
    {
        answer+=1;
        resp='n';
    }

    screen2();
}


void main()
{   clrscr();
    answer=0;
    printf("\nSelect any number");
    printf("\n between 1 to 60");
    printf("  \n in your mind...");
    printf("\n Press Enter to Continue...");
    getch();

    screen1();
}

如果您能帮我解决问题，我将非常感谢您。谢谢。

Answer 1

您的screen1函数应该只有奇数。所以它看起来像这样：

void screen1()
{
    printf("\n 1 11 21 31 41 51");
    printf(" \n 3 13 23 33 43 53"); // 43 instead of 46
    printf("\n 5 15 25 35 45 55");
    printf("\n 7 17 27 37 47 57");
    printf("\n 9 19 29 39 49 59");

    printf("\n\nPress 'Y' if number is present");
    printf("\n \nthe list else press any key\n");

    resp=getche();
    if(resp=='y'||resp=='Y')
    {
        answer+=1;
        resp='n';
    }

    screen2();
}

魔数编号程序

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