[E] Attempt the following:
(a)Write a program to print all prime numbers from 1 to 300. (Hint: Use nested loops, break and continue)
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Answer:
void main()
{
int no,i,rem,j,flag=0;
for(no=2;no<=300;no++) { flag=0; j=no/2; for(i=j;i>=2;i--)
{
rem=no%i;
if(rem==0)
{
flag=1;
break;
}
}
if(flag!=1)
printf("%d\n",no);
}
}
(b)Write a program to fill the entire screen with a smiling face. The smiling face has an ASCII value 1.
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Answer:
void main()
{
int smiling=1,i;
for(i=1;i<=2000;i++)
{
printf("%c",smiling); }
}
(c)Write a program to add first seven terms of the following series using a for loop: 1/1! + 2/2! + 3/3! + .........
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Answer:
void main()
{
int factorial,fact,i;
float sum=0.0;
i=0;
while(i<=7) { fact=i;//for finding factorial see note: 1. factorial=1; while(fact>=1)
{
factorial=factorial*fact;
fact--;
}
sum=sum+(float)i/factorial;//int/int gives integer value so
// we converted into float value.
i++;
}
printf("sum upto 7th term is %f",sum);
}
(d)Write a program to generate all combinations of 1, 2 and 3 using for loop.
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Answer:
void main()
{
int i,j,k,l,count=0;
for(i=1;i<=4;i++)
{
for(j=1;j<=4;j++)
{
for(k=1;k<=4;k++)
{
for(l=1;l<=4;l++)
{
if(i!=j&&j!=k&&i!=k&&i!=l&&j!=l&&k!=l)
{
printf("\n%d %d %d %d",i,j,k,l); count++; }
}
}
}
}
printf("\ncount = %d",count);
}
(e) According to a study, the approximate level of intelligence of a person can be calculated using the following formula: i = 2 + ( y + 0.5 x ) Write a program, which will produce a table of values of i, y and x, where y varies from 1 to 6, and, for each value of y, x varies from 5.5 to 12.5 in steps of 0.5.
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Answer:
void main()
{
int y;
float i,x;
for(y=1;y<=6;y++)
{
for(x=5.5;x<=12.5;)//x varies from5.5 to 12.5 so
{
i=2+(y+0.5*x); //formula
printf("i=%f y=%d x=%f\n",i,y,x);//displaying
x=x+0.5; //x increase in step of 0.5 by question
}
}
}
(f)Write a program to produce 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
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Answer:
void main()
{
int i,j,k,count1,count2=0;
for(i=104;i>=97;i--)
{
for(j=97;j<=104;j++)
{
if(i>=j)
printf("%c ",j);
}
for(count1=0;count1=97;k--)
{
if(k<=i&&k!=104)
printf("%c ",k);
}
if(i==103)
count2=count2+1;
else
count2=count2+2;
printf("\n"); //new line for each i.
}
}
[7]int i=10;
(g)Write a program to fill the entire screen with diamond and heart alternatively. The ASCII value for heart is 3 and that of diamond is 4.
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Answer:
void main()
{
int i;
for(i=0;i<=600;i++)
{
printf(" %c %c",3,4);
}
}
(h)Write a program to print the multiplication table of the number entered by the user. The table should get displayed in the following form. 29 * 1 = 29 29 * 2 =58 ....
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Answer:
void main()
{
int no,i,multiply;
printf("enter number for multiplication table:");
scanf("%d",&no);
for(i=1;i<=9;i++)//we can make upto any number.
{
multiply=no*i;
printf("\n%d * %d = %d",no,i,multiply);
}
}
(i) Write a program to produce the following output:
1
2 3
4 5 6
7 8 9 10
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Answer:
void main()
{
int i,j,k,cnt=1,icpy,icpy1,space,count=10;
for(i=1;i<=5;i++) { for(space=count;space>=1;space--)
/*this is for managing spaces. atfirst there will be 10 spaces
which gradually decreases by 1.In output 2 is at 1
space left of 1.so, we decrease by 1*/
{
printf(" ");
}
count=count-1;
icpy=i;
for(j=1;j<=i;j++)
{
printf("%d ",icpy);
icpy++;
}
printf("\n");
}
}
(j)Write a program to produce the following output:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
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Answer:
void main()
{
int a,b,c,line,space,i,j,n,r,ncr,fact_r,fact_n,factn_r;
printf("please enter the number of lines of the pascal triangle: \n");
scanf("%d",&line);
printf("\n");
for(i=0;i<=line;i++)
{
fact_n=1;
fact_r=1;
factn_r=1;
for(space=line;space>=i;space--)
{
printf(" ");
}
for(j=0;j<=i;j++) { n=i; r=j; fact_n=1; fact_r=1; factn_r=1; if(n==0) fact_n=1; else { for(a=n;a>=1;a--)
fact_n=fact_n*a;
}
//factorial of fact_r.
if(r==0)
fact_r=1;
else
{
for(b=r;b>=1;b--)
fact_r=fact_r*b;
}
if((n-r)==0)
factn_r=1;
else
{
for(c=(n-r);c>=1;c--)
factn_r=factn_r*c;
}
ncr=fact_n/(fact_r*factn_r);
c(n,r)
printf("%2d",ncr); //2 space for each ncr
}
printf("\n"); //new line after each row.
}
}
(k)A machine is purchased which will produce earning of Rs. 1000 per year while it lasts. The machine costs Rs. 6000 and will have a salvage of Rs. 2000 when it is condemned. If 12 percent per annum can be earned on alternate investments what would be the minimum life of the machine to make it a more attractive investment compared to alternative investment?
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Answer:
void main()
{
int year=0,inv,altn;
while(altn>inv)
{
year++;
altn=120*year; // 12%of 1000 = 120
inv=(1000*year)-4000;
}
printf("The minimum year is %d",year);
}
(l) When interest compounds q times per year at an annual rate of r % for n years, the principle p compounds to an amount a as per the following formula
a = p ( 1 + r / q ) ^ nq
Write a program to read 10 sets of p, r, n & q and calculate the corresponding as.
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Answer:
void main()
{
int p,r,n,q,a,i,j,product;
float amt=1;
for(i=0;i<10;i++)
{
amt=1;
printf("enter the values of p,r,n and q: "); scanf("%d%d%d%d",&p,&r,&n,&q);
product=n*q;
for(j=1;j<=product;j++)
{
amt = amt * ( 1 +(float) r / q );
}
amt= p *amt;
printf("%f",amt);
}
}
(m)The natural logarithm can be approximated by the following series.
(x-1)/x + 1/2((x-1)/2)^2 + 1/2((x-1)/2)^3 + ...
If x is input through the keyboard, write a program to calculate the sum of first seven terms of this series.
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Answer:
void main()
{
int x,i,j;
float sum=0,power=1;
printf("enter x for sum upto 7th term: ");
scanf("%d",&x);
for(i=1;i<=6;i++)
{
power=1;
for(j=0;j<=i;j++)
{
power = power * ((x-1.0)/2.0);
}
sum = (1.0/2) * power + sum;
}
sum=sum + (float)(x-1.0)/x; printf("%f",sum);
}
