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Difficult C programs

Discussion in 'C++' started by aijaz, Apr 14, 2008.

  1. aijaz

    aijaz New Member

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    hello frends ... i need sme help in developing some programs in C

    1) i want the out like this 1
    1 1
    1 1 2
    1 1 2 3
    1 1 2 3 5
    1 1 2 3 5 8 ... & soon .. till the range i specify

    ... here you can see tht we are adding the Diagonal digits ... plz help

    2 ) program .. i want out like this .. in this program .....

    1
    1 2 1
    1 2 3 2 1
    1 2 3 4 3 2 1
     
  2. dipakdayanand

    dipakdayanand New Member

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    2)this is the answer:
    Code:
    #include<conio.h>
    #include<stdio.h>
    void main()
    {
    int i,j,k,before=0,array[10][10];
    clrscr();
    for(i=1;i<=5;i++)
    {
    for(j=1;j<=(2*i);j++)
    {
    if(j<=i)
    {
    array[i][j]=before+1;
    before++;
    }
    else
    {
    array[i][j]=before-1;
    before--;
    }
    }
    before=0;
    }
    for(i=1;i<=5;i++)
    {
    for(j=1;j<=((2*i)-1);j++)
    {
    printf("%d",array[i][j]);
    printf("\t");
    }
    printf("\n");
    }
    getch();
    }
     
    Last edited by a moderator: Jun 1, 2009
    abhix95 likes this.
  3. dipakdayanand

    dipakdayanand New Member

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    1)
    Code:
    #include<conio.h>
    #include<stdio.h>
    void main()
    {
    int a,b,c,i,j,array[10][10];
    clrscr();
    for(i=1;i<=5;i++)
    {
    a=0,b=1;
    for(j=1;j<=i;j++)
    {
    if(j==1)
    array[i][j]=1;
    else
    {
    c=a+b;
    array[i][j]=c;
    a=b;
    b=c;
    }
    }
    }
    for(i=1;i<=5;i++)
    {
    for(j=1;j<=i;j++)
    {
    printf("%d",array[i][j]);
    printf("\t");
    }
    printf("\n");
    }
    getch();
    }
     
    Last edited by a moderator: Jun 1, 2009
    abhix95 likes this.
  4. SaswatPadhi

    SaswatPadhi ~ Б0ЯИ Τ0 С0δЭ ~

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    Occupation:
    STUDENT !
    Location:
    Orissa, INDIA
    Home Page:
    Please post you code inside code-blocks.

    If you are not allowed to change your post, shabbir might change it for you.
     
  5. cvikas78

    cvikas78 New Member

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    Code:
    #include<stdio.h>
    int main(int argc , char *argv[])
    {
            int next = 1;
            int range = 1, rangeEnd = 0;
            int nextInRange = 0;
            int prev = 1;
            if(argv[1] == NULL) {
                    printf("Usage : ./a.out <range_number>\n");
            } else {
                    rangeEnd = atoi(argv[1]);
            }
            for(; range <= rangeEnd ; range++) {
                    next = 1;
                    prev = 1;
                    if(range >= nextInRange) {
                            printf("%d %d ", prev, next);
                            for(; next < range ;) {
                                    next = prev + next;
                                    printf("%d ", next);
                                    prev = next - prev;
                            }
                            nextInRange = next + prev;
                            putchar('\n');
                    }
            }
    }
     
    Last edited by a moderator: May 14, 2010
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  6. xpi0t0s

    xpi0t0s Mentor

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    OK, now where are the difficult C programs?
     
  7. sameerthapa2050

    sameerthapa2050 New Member

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    Last edited by a moderator: Feb 25, 2011
  8. sameerthapa2050

    sameerthapa2050 New Member

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    Code:
    public class Pattern {
    
    	public static void main(String[] args) {
    		
    		System.out.println("1");
    		
    		for(int i=1,j;i<5;i++){
    			
    			for(j=1;j<=i;j++){
    		
    				System.out.print(j+" ");
    			
    			}
    			
    			for(;j>=1;j--){
    			
    				System.out.print(j+" ");
    			
    			}
    			
    			System.out.println("");
    			
    		}
    	}
    }
     
    Last edited by a moderator: Feb 25, 2011
  9. sameerthapa2050

    sameerthapa2050 New Member

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  10. Sharad Sharma

    Sharad Sharma New Member

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    Code:
    #include<stdio.h>
    #include<conio.h>
    #include<math.h>
    void main()
    {
    long n,j,p=0,q;
    clrscr();
    printf("\n Enter the number of rows : ");
    scanf("%ld",&n);
    for(j=0;j<=n-1;j++)
    {
    p=1*pow(10,j)+p;
    q=p*p;
    printf("%ld",q);
    printf("\n");
    }
    getch();
    }
     
    Last edited by a moderator: Aug 22, 2011
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  11. Sharad Sharma

    Sharad Sharma New Member

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    1
    121
    12321

    #include<stdio.h>
    #include<conio.h>
    #include<math.h>
    void main()
    {
    long n,j,p=0,q;
    clrscr();
    printf("\n Enter the number of rows : ");
    scanf("%ld",&n);
    for(j=0;j<=n-1;j++)
    {
    p=1*pow(10,j)+p;
    q=p*p;
    printf("%ld",q);
    printf("\n");
    }
    getch();
    }
     
  12. Sharad Sharma

    Sharad Sharma New Member

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    (2)
    1
    121
    12321

    #include<stdio.h>
    #include<conio.h>
    #include<math.h>
    void main()
    {
    long n,j,p=0,q;
    clrscr();
    printf("\n Enter the number of rows : ");
    scanf("%ld",&n);
    for(j=0;j<=n-1;j++)
    {
    p=1*pow(10,j)+p;
    q=p*p;
    printf("%ld",q);
    printf("\n");
    }
    getch();
    }
     
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  13. abhix95

    abhix95 New Member

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    1
    11
    112
    11235
    112358

    #include<iostream.h>
    #include<conio.h>
    void main()
    {
    int a, b,c,r,p=0,n=1;
    clrscr();
    cout<<"Enter number of line";
    cin>>a;
    for(b=1;b<=a;b++)
    {
    cout<<1<<"\t";
    for(c=1;c<b;c++)
    {
    r=p+n;
    p=n;
    n=r;
    cout<<r<<"\t";
    }
    cout<<endl;
    n=1;
    p=r=0;
    }
    getch();
    }
     
  14. cwizard

    cwizard New Member

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    Code:
    #include<iostream.h>
    #include<conio.h>
    void main()
    {
    clrscr();
    int a[20],n;
    a[0]=1;
    a[1]=1;
    cout<<"Enter the limit"<<endl;
    cin>>n;
    for(int j=1;j<=n;j++)
     {
      for(int i=2;i<=j;i++)
      {
       a[i]=a[i-1]+a[(i-1)-1];
      }
     }
    for(j=0;j<=n;j++)
     {
      for(int i=0;i<=j;i++)
      {
      cout<<a[i];
      }
     cout<<endl;
     }
    getch();
    }
     
    Last edited by a moderator: Jul 30, 2012
  15. vijayuit

    vijayuit New Member

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    Code:
    #include<stdio.h>
    void main(){
    int n,i,j,f1=0,f2=1,f3=0,k,a[25];
    printf("ENTER THE NUMBER\n");
    scanf("%d",&n);
    for(i=1;i<n;i++)
    {
    f1=f2;
    f2=f3;
    f3=f2+f1;
    a[i]=f3;
    for(j=1;j<=i;j++)
    {
    printf("%d\t",a[j]);
    }
    printf("\n");
    }
    }
     
    Last edited by a moderator: Aug 3, 2012
  16. vijayuit

    vijayuit New Member

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    Code:
    #include<stdio.h>
    void main(){
    int n,i,j,k;
    printf("ENTER THE NUMBER\n");
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
    for(j=1;j<=i;j++)
    printf("%d",j);
    for(k=j-2;k>0;k--)
    printf("%d",k);
    printf("\n");
    }
    }
     
    Last edited by a moderator: Aug 3, 2012
  17. xpi0t0s

    xpi0t0s Mentor

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    But where are the *difficult* C programs? These 10-15 line tiddlers might be challenging for beginners for about 5 minutes, but that doesn't qualify them to be called difficult. IOCCC - now THERE are some difficult C programs!
     
  18. manojgdhv

    manojgdhv New Member

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    1)
    Code:
    void main()
    {
    int f1=1,f2=1,f3=0,n=5,i,j;
    for(i=1;i<=5;i++)
    {
       printf("\n%d %d ",f1,f2);
       for(j=2;j<=i;j++)
       {
        
        f3=f1+f2;
        printf("%d ",f3);
        f1=f2;
        f2=f3;
       }
    }
    }
     2)
    void main()
    {
    int i=0,j=0;
    for(i=1;i<=4;i++)
    {
        for(j=1;j<=i;j++)
      printf("%d",j);
      for(j=i-1;j>=1;j--)
      printf("%d",j);
    printf("\n");
    }
    }
     
    Last edited by a moderator: Aug 17, 2012
  19. Kush Pandey

    Kush Pandey New Member

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    here is the output of program to print
    11
    112
    1123
    11235
    112358


    Code:
    #include<stdio.h>
    #include<conio.h>
     main(){
        int i,j,a=1,b=1,c;
    
        for(i=1;i<=5;i++)
        {
            printf("%d\t%d\t",a,b);
        for(j=3;j<=i+1;j++)
        {
            c=a+b;
            printf("%d\t",c);
            a=b;
            b=c;
        }
        a=1;b=1;
        printf("\n");
        }
    }
     
    Last edited by a moderator: Jan 10, 2013
  20. xpi0t0s

    xpi0t0s Mentor

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    Just thought I'd borrow something from IOCCC to show you people what a difficult C program really looks like. These piddly examples to draw stars in triangles or whatever are nothing compared to these:
    Code:
    #include<stdio.h> /******** SpigotQuine -- usage: ./spigot [pi or e] ********/
    char*s="G1%%xJ{;Q7wunmuGuu%%uu#include<stdio.h>/*Spigot_Quine*/#include<stdli"
    "b.h>/*_IOCCC2012_*/int*e,"    "i,j,k,n"     ";char*q"    ",*a,*d,*z,*p=%s%c;"
    "int" "%cmain(){a=calloc("                                 "1,1e4+n*2);;for(*"
    "a=\0@3,z=d=a+n+1,j=n*8-7;"    "k=0,j-1"     ";j-=2){"    "for(a[1]+=2;--z-a;"
    "*z=k%%10,k/=10)k+=j/2**z;;for(;k=k%%j*"     "10+*++z,z<d;)*z=k/" "j;;\0@2,z="
    "d=a+n*2,*z=1,j=0;++j<n;){for(;k=k%%"           "j*10+*z,a-z;*z"   "--=k/j)a+"
    "+;for(k=0;z-d;*a--=k%%10,k/=10)k+"               "=*++z+*a\0"     "@;}d+=spr"
    "intf(q=d-20,p,p,34,32,n+1)+2;;;;"                 "for(n=n*2"     "0-400;k<n"
    ";++k%%n?j=!puts("                                                 "d):(d[j]="
    "47,d++,d[j-2"                                                     "]=42),k%%"
    "20<1?puts(d"                                                      "-1),a++:0"
    "){for(i=-1"                                                       ";i++<32;!"
    "*z?q[662]"          "=0,z=q+207:"                 "*z+z[1]<6"     "5?z+=11:*"
    "z==34?p=0"         ":0)d[i]=((k/2"               "0-1?275*q["     "*a+10]-8*"
    "q[*a+0]-8"         ":128)>>(i/11+k/"           "4%%5*3))&1?k"     "/3*!j&&p?"
    "j=34:(j="           "i+1,*z++):32;k/3*"     "j--&&p?d[z--,j]=3"   "4:0;}}int"
    "*y,n=%d;/*..~",*f="nnLa5~z23~|22t$q(s82r&q(s82q'q(s8;q(s8;q(s8:" "r(s8:r(s8:"
    "q)s89r)sLr#t+" "sLx,uJw-yGu/wnnnU",*g="nnLa<z::t$u88t(u67t*u57s,t56t,t56~v56"
    "tF6tF6tF8t1p"   "Nu/qOv+rS}Xxnng";int main(int m,char**v){char a[2012],b[2012
    ],*p=a,*r=m>1     &&*v[1]=='e'?g:f,*q=b,*t=r;;sprintf(a,"%s%s%s",s,r==g?s+281:
    s+168,s+386);     sprintf(b,a+22,a,34,32,24);for(sprintf(a,"%.33s/*%.28s*/%.3"
    "3s/*%.28s*/%"   ".33s\"%s*/",b,b+66,b+33,b+76,b+66,b+99);*r;r++){;for(m=0;m++
    <(*r-34)%77;*q++=*r>111?32:*p++)(q-b)%66<1?*q++=10:0;*r-110&&*r-126&&r-t<(t-g?
    62:45)?*q++=34,((q-b)%66<1?*q++=10,*q++=34:0):0;}*q=0;puts(b+1);}/*IOCCC2012*/
    
     

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