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C Program for pyramid

What will be the code in c to get the following output?
A B C D E F G F E D C B A
A B C D E F F E D C B A
A B C D E E D C B A
A B C D D C B A
A B C C B A
A B B A
A A
Asked by: vrijesh28 | Member Since Dec-2006 | Asked on: Dec 20th, 2007

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Showing Answers 1 - 9 of 9 Answers
ravi_529

Answered On : Dec 24th, 2007

View all answers by ravi_529

Code
  1. int i, j, k;
  2. char a[] = { 'A', 'B', 'C', 'D', 'E', 'F', 'G' };
  3.  
  4. for (i = 0; i < 7; i++) {
  5.     for (k = 0; k < 7 - i; k++)
  6.         printf("%c ", a[k]);
  7.     for (j = 7 - (i + 1); j >= 0; j--)
  8.         printf("%c ", a[j]);
  9.     printf("n");
  10. }
  11.  

  
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jintojos

Answered On : May 23rd, 2008

View all answers by jintojos

Code
  1. #include<conio.h>
  2. #include<stdio.h>
  3. #include<string.h>
  4. void main()
  5. {
  6. char arr[]="ABCDEFG";
  7. int i,j,length=strlen(arr);
  8. clrscr();
  9. while(length>0)
  10. {
  11. for(i=0;i<length;i++) printf(" %c",arr[i]);
  12. for(j=length-2;j>=0;j--) printf(" %c",arr[j]);
  13. printf("n");
  14. length--;
  15. }
  16. getch();
  17. }
  18.  

  
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Code
  1. #include<stdio.h>
  2. int main()
  3. {
  4.     /*This code is for any input upto 'Z' from 'A' */
  5.     char i, j, k;
  6.     int a;
  7.     scanf(" %c", &j);
  8.     for (a = j; a <= j; a++) {
  9.         for (i = 'A'; i <= a; i++)
  10.             printf("%c", i);
  11.         for (k = a - 1; k >= 'A'; k--)
  12.             printf("%c", k);
  13.     }
  14.     printf("n");
  15.     for (a = j - 1; a >= 65; a--) {
  16.         for (i = 'A'; i <= a; i++)
  17.             printf("%c", i);
  18.         for (k = a; k >= 'A'; k--)
  19.             printf("%c", k);
  20.         printf("n");
  21.     }
  22.     return 0;
  23. }
  24.  

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GNANAPRAKASH.S

Answered On : Aug 9th, 2011

Code
  1. #include<stdio.h>
  2. #include<conio.h>
  3. void main()
  4. {
  5.     int i, j, k, n;
  6.     char a[30];
  7.     clrscr();
  8.     scanf("%d", &n, printf("Enter the number of lines upto alpahbets:"));
  9.     for (i = n; i >= 1; i--) {
  10.         for (j = 1; j <= i; j++) {
  11.             a[j] = 64 + j;
  12.             printf("%c", a[j]);
  13.         }
  14.         for (k = j - 1; k >= 1; k--) {
  15.             printf("%c", a[k]);
  16.         }
  17.         printf("
  18.    ");
  19.     }
  20.     getch();
  21. }
  22.  

  
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Code
  1. #include
  2. #include
  3. using namespace std;
  4. int main()
  5. {
  6.     char a[] = "abcdefg";
  7.     char b[] = "fedcba";
  8.     cout <<< int m, n;
  9.     m = strlen(a);
  10.     n = strlen(b);
  11.     for (int i = 0; i cout < cout << int k = 1; while (m > 1 && k
  12.   {                                            
  13.     int l = 0; for (int i = 0; i cout < for (int i = k; i cout < cout < m--; k++;
  14.   }
  15.     cin >> z; return 0;
  16. }
  17. <<<
  18.  

  
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This one works

Code
  1. #include<iostream>
  2.  
  3. #include<cstring>
  4.  
  5. using namespace std;
  6.  
  7. int main()
  8. {
  9.  
  10.     char a[] = "abcdefg";
  11.  
  12.     char b[] = "fedcba";
  13.  
  14.     cout << a << b << endl;
  15.  
  16.     int m, n;
  17.  
  18.     m = strlen(a);
  19.  
  20.     n = strlen(b);
  21.  
  22.     for (int i = 0; i < m - 1; i++)
  23.  
  24.         cout << a[i];
  25.  
  26.     cout << b << endl;
  27.  
  28.     int k = 1;
  29.  
  30.     while (m > 1 && k < n)
  31.     {
  32.  
  33.         int l = 0;
  34.  
  35.         for (int i = 0; i < m - 2; i++)
  36.  
  37.             cout << a[i];
  38.  
  39.         for (int i = k; i < n; i++)
  40.  
  41.             cout << b[i];
  42.  
  43.         cout << endl;
  44.  
  45.         m--;
  46.  
  47.         k++;
  48.  
  49.     }
  50.  
  51.     return 0;
  52.  
  53. }
  54.  

  
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Anil

Answered On : Oct 7th, 2011

Code
  1. #include<stdio.h>
  2. #include<conoio.h>
  3. void main()
  4. {
  5.     int len, i, j, k;
  6.     char a[] = { 'A', 'B', 'C', 'D', 'E', 'F', 'G' };
  7.     clrscr();
  8.     len = a.count();
  9.     for (i = 0; i < len; i++) {
  10.         for (j = 0; j < len - i; j++) {
  11.             printf(a[j]);
  12.         }
  13.         for (k = ((len - 2) - i); k >= 0; k--) {
  14.             printf(a[k]);
  15.         }
  16.         printf("
  17.    ");
  18.     }
  19.     getch();
  20. }
  21.  

  
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Gaurav Bhadauria

Answered On : Oct 13th, 2011

Code
  1. int i, j, k;
  2. char a[] = { 'A', 'B', 'C', 'D', 'E', 'F', 'G' };
  3.  
  4. for (i = 0; i < 7; i++) {
  5.     for (k = 0; k < 7 - i; k++)
  6.         printf("%c ", a[k]);
  7.     for (j = 7 - (i + 1); j >= 0; j--)
  8.         printf("%c ", a[j]);
  9.     printf("n");
  10. }
  11.  

  
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Jimmy Jack

Answered On : Apr 25th, 2012

Code
  1. #include <iostream>
  2. #include <fstream>
  3.  
  4. void display_sequence2(int depth)
  5. {
  6.     int output_count = 0;
  7.     for (int count = depth - 1; count >= 0; --count) {
  8.         for (int index = 0; index <= count; ++index)
  9.             std::cout << (char) ((((output_count++ == 0) || (index == 2)) ? A : a) + index) << " ";
  10.         for (int index = (count == (depth - 1)) ? count - 1 : count; index >= 0; --index) {
  11.             std::cout << (char) ((((output_count++ == 0) || (index == 2)) ? A : a) + index);
  12.             if (index > 0)
  13.                 std::cout << " ";
  14.         }
  15.     }
  16. }
  17.  
  18. int main()
  19. {
  20.     display_sequence2(7);
  21. }
  22.  

  
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