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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Newbie Member
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| 22Aug2011,18:04 | #11 |
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Newbie Member
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| 22Aug2011,18:08 | #12 |
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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(); }
abhix95
like this
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Newbie Member
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| 18Jan2012,23:14 | #13 |
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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(); } |
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Newbie Member
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| 29Jul2012,21:44 | #14 |
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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();
}
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Newbie Member
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| 3Aug2012,16:26 | #15 |
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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");
}
}
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Newbie Member
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| 3Aug2012,16:27 | #16 |
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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");
}
}
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Mentor
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![]() |
| 3Aug2012,17:30 | #17 |
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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!
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Newbie Member
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| 17Aug2012,16:25 | #18 |
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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");
}
}
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Newbie Member
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| 10Jan2013,10:40 | #19 |
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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");
}
}
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Mentor
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![]() |
| 10Jan2013,13:51 | #20 |
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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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