wap to print the following pattern.

wap to print the following pattern in c----

  


 #include<stdio.h>
void main()
{
int i,j,k,f=1;
for(i=65;i<=71;i++)
printf("%c",i);
for(i=70;i>=65;i--)
printf("%c",i);
printf("\n");
for(i=1;i<=6;i++)
{
  {
   for(j=65;j<=71-i;j++)
   printf("%c",j);
   }
     {
      for(k=1;k<=f;k++)
       printf(" ");
       f=f+2;
      for(k=71-i;k>=65;k--)
      printf("%c",k);
     }
 printf("\n");
}
}




NOTE-  i have compiled this program in codeblocks if u do it in turbo c++  use  conio.h  header file in the beginning  and getch()  a the  last.

a simple code...see carefully...

include<stdio.h>
include<conio.h>
main()
{
int a,b;
for(a=0;a,=149;a++);
for(b=0;b=149;b++);
b=a+b;
printf("the sum of the number is%d ",b);
}


output--
300

wap to input a five digit number and add 1 to each digit such that the output is as follows ..e.g 12569...so the output should be..23670..... this is to be done without using any loop....

#include<stdio.h>
void main()
{
long int num,a,b,c,d,e,f,g,h,eureka;
printf("enter the five digit number");
scanf("%d",&num);
a=num/10000;
b=num%10000;
c=b/1000;
d=b%1000;
e=d/100;
f=d%100;
g=f/10;
h=f%10;
a=((a+1)%10)*10000;
c=((c+1)%10)*1000;
e=((e+1)%10)*100;
g=((g+1)%10)*10;
h=((h+1)%10)*1;
eureka=a+c+e+g+h;
printf("the increased no. is %d",eureka);
}

write a function factorial to find the factorial of a number and use this function to solve the following series:::: y=(1!)+(1!+2!)+(1!+2!+3!)+............n terms

the following program  came in my sessional. and i wrote it   and is in front of u ,  check it out.:::::::::::







#include<stdio.h>
#include<conio.h>
int fact(int i);
int main()
{
int sum=0 ,i,n,e;
printf("enter the no. til u want the series");
scanf("%d",&n);
for(i=1;i<=n;i++)
{
e=fact(i);
sum=e+sum;
}
printf("the sum of the series is  %d",sum);
}
int fact(int i)
{
int t,j,a=1,d=0;
for(t=1;t<=i;t++)
{
for(j=1;j<=t;j++)
{
a=a*j;

}
d=d+a;
a=1;
printf("%d \n",d);
}
return(d);
}