Finding nth power of integer m through C program w

问题:

C program to find nth power of integer m without pow().

Input:

m=3 n=2
output:
9.000

Tests to validate the program works as expected!

1. For negative M

Input : -2  3
output : -8.000

2. For negative N

Input : 2  -3  
output : 0.125000

3. For negative M and N

Input : -2  -3
output : -0.125000

However I am not getting the desired output

void main() 
{

    signed int m, n;
    int i;
    float p;
    clrscr();
    printf("Enter the number and its power (exponent)\n");
    scanf("%d%d",&m,&n);
    p=1;
    if (n==0)
    {
        printf("%d raised to %d is: %f",m,n,p);
    }

    if (n>0)
    {
        for( i = 0 ; i < n ; i++ )
            p*=m;
        if(m>0)
            printf("%d raised to %d is: %f",m,n,p); 
        if(m<0) 
            printf("%d raised to %d is: %f",m,n,-p); 
    }

    if (n<0)

    {
        n=-n;
        for( i = 0 ; i < n ; i++ )
            p*=m;
        if(m>0)
            printf("%d raised to %d is: %f",m,-n,1/p);
        if(m<0)
            printf("%d raised to %d is: %f",m,-n,-(1/p)); 
    }
    getch(); 
}

Can u kindly provide the correct program for the test cases?

I can't declare signed float as it is giving an error.

回答1:

The code for negatives is incorrect. You cannot just blindly negate the result when the base m is negative. , but . Also, you're not printing anything if m is zero!

And ints are signed by default so signed int is noise. floats are signed too; but here you could as well use a double for more precision. The return value of main should be int.

Therefore the fixed code would be (add nonstandard clrscrs and getchs to your taste ;):

#include <stdio.h>
#include <stdlib.h>

int main() 
{
    int m, n, i;
    double p = 1.0;
    printf("Enter the number and its power (exponent)\n");
    scanf("%d%d",&m,&n);

    if (n==0) {
        printf("%d raised to %d is: %f",m,n,p);
    }

    else if (n > 0) {
        for(i = 0; i < n; i++)
            p*=m;
        printf("%d raised to %d is: %f",m,n,p); 
    }

    else { // n < 0
        n = -n;
        for (i = 0 ; i < n ; i++)
            p*=m;
        printf("%d raised to %d is: %f", m, -n, 1 / p);
    }
}

回答2:

Try changing:

void main() 
{

    signed int m, n;
    int i;
    float p;
    printf("Enter the number and its power (exponent)\n");
    scanf("%d%d",&m,&n);
    p=1;
    if (n==0)
    {
        printf("%d raised to %d is: %f",m,n,p);
    }

    if (n>0)
    {
        for( i = 0 ; i < n ; i++ )
            p*=m;
        printf("%d raised to %d is: %f",m,n,p); 
    }

    if (n<0)

    {
        n=-n;
        for( i = 0 ; i < n ; i++ ){
            p*=m;
        }
         printf("%d raised to %d is: %f",m,-n,1/p);

    }

}

回答3:

@Antti Haapala well identified the errors in OP's code.

---

As an interview question relating to pow() substitute, certainly a solution that takes log2 N steps (as below) is preferable to one that takes N steps (OP's approach) where N is the exponent.

The key is to square the base on each iteration.
Using log2 N steps rather than N results in less error in the least significant digits.

With a negative exponent, simple invert the base.

Use double to provide to provide at least 10, typically 17, decimal digits of precision.

double powdii(int x, int y) {
  double z = 1.0;
  double base = x;
  if (y < 0) {
    base = 1.0 / base;
  }
  while (y != 0.0) {
    if (y % 2) {
      z *= base;
    }
    y /= 2;
    base *= base;
  }
  return z;
}