Write a C++ program to calculate a to the power b without using power function.
#include<iostream.h> #include<conio.h> void main() { int a,b,i,temp=1; clrscr(); cout<<"Enter number for operation"; cin>>a>>b; for(i=1;i<=b;i++) { temp=temp*a; } cout<<endl<<"Result are:: "<<temp; getch(); } |
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Sir, this gives a^b+1
#include
#include
void main()
{
int n,i,c;
clrscr();
printf(“\nEnter a number: “);
scanf(“%d”,&n);
for(i=1;i<=n ;i++)
{
c=n*n;
}
printf("\npower of a number is: %d",c);
getch();
}
#include
#include
class power
{
int base,exp,i,temp;
public:
void get();
void cal();
void dis();
power()
{
temp=1;
}
};
void power::get()
{
cout<>base>>exp;
}
void power::cal()
{
for(i=1;i<=exp;i++)
{
temp=temp*base;
}
}
void power::dis()
{
cout<<"Calculated="<<temp;
}
void main()
{
clrscr();
power p;
p.get();
p.cal();
p.dis();
getch();
}
Above program gives wrong answers as well as they occupy high memory any more time to type the simple program is
# include
void main()
{
int a;
cout<>a;
cout<<"\n power of "<<a<<" is "<<a*a;
}
SORRY GUYS I MISTAKED IN ABOVE PROGRAM THE CORRECT CODE IS HERE
#include
void main()
{
int a;
cin>>a;
cout<<"\n power of "<<a<<" is
"<<a*a;
}
thanks
This problem can be solved by applying different methods.
1st Method:
#include
#include
void main()
{
int a,b,c,i;
clrscr;
printf(“Enter the base: “);
scanf(“%d”,&a);
printf(“Enter the power: “);
scanf(“%d”,&b);
c=a;
for(i=1;i<b;i++)
{
c=c*a;
}
printf("Result = %d",c);
getch();
}
2nd Method:
#include
#include
void power(int a, int b);
void main()
{
int a,b;
clrscr();
printf(“Enter the base: “);
scanf(“%d”,&a);
printf(“Enter the power: “);
scanf(“%d”,&b);
power(a,b);
getch();
}
void power(int a, int b)
{
int c;
c=a;
for(i=1;i<b;i++)
{
c=c*a;
}
printf("Result = %d",c);
}
There are so many different methods to solve this problem.