patterncMinor

Evaluating a polynomial

Submitted by: @import:stackexchange-codereview·Mar 10, 2026·

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polynomialevaluatingstackoverflow

Problem

You are given a polynomial of degree n. The polynomial is of the form

P(x) = an · xn + an-1 · xn-1 + … + a0

where the ai‘s are the coefficients. Given an integer
x, write a program that will evaluate P(x).

You are provided with a function named power( ) that takes two
positive integers x and y and returns xy. If y is
0, the function returns 1.

The prototype of this function is

int power(int x, int y);

INPUT:

Line 1 contains the integers n and x separated by whitespace.

Line 2 contains the coefficients an, an-1, …, a0 separated by whitespace.

OUTPUT:

A single integer which is P(x).

CONSTRAINTS:

The inputs will satisfy the following properties. It is not necessary
to validate the inputs.

1 <= n <= 10
1 <= x <= 10
0 <= ai <=10

My code

#include

int power(int x, int y);

int main()
{

int n,x,a[11],i; unsigned long long int sum=1;
scanf("%d%d",&n,&x);
if(n10) return 0;
if(x10) return 0;

for(i=0;i10) return 0;
}
for(i=0;i<n;i++)
{

sum+=power(x,n-i)*a[i];
}
printf("%llu",sum);

    return 0;
}

int power(int x, int y)
 {
   int result = x,i;

   if(y == 0) return 1;
   if(x < 0 || y < 0) return 0;

   for(i = 1; i < y; ++i)
   result *= x;

   return result;
}

Solution

-
Recommend a uniform and more indented style. E.g.

for(i=0;i<n;i++)
{
    sum+=power(x,n-i)*a[i];
}

-
Both a[11] and if(... n>10) rely on the same value. Also to avoid undocumented magic numbers, consider the following (also the same for x and a[]):

#define n_MAX (10)
#define n_MIN (0)
int a[n_MAX + 1];
if(nn_MAX) return 0;

-
unsigned long long int sum serves little value in protecting against numbers larger than INT_MAX as power() is the most likely candidate to overflow. (Strictly speaking, it is not even known that LLONG_MAX is greater than INT_MAX.) So if you want a larger range, re-write power() as unsigned long long power(int x, int y), otherwise leaving sum as int is OK.

Else suggest for marginal range improvement:

sum += (unsigned long long) power(x,n-i) * a[i];

-
BTW: should not it be:

unsigned long long int sum = 0;  //( 0 not 1)

-
Note: Stated objective says "It is not necessary to validate the inputs.". Still I like the testing you have done. Possible that the final reviewer (teacher?) may object to unneeded code. Could wrap in an #if DEBUG macro...

-
Faster way to do power()

-
See no reason for

x

- 
Do

power() with unsigned rather than int

 as " function named power( ) that takes two positive integers x and y".

- 
@Constructor comment leads to an excellent suggestion.  Calculate

sum with a loop that does the below. No need for a power function. Running sum can be unsigned long long

 and code gets great range.  Its faster & simple.

// sum = (((a[n]*x + a[n-1])*x + a[n-2])*x + .... )*x + a[0];

// Something like
unsigned long long sum = 0;
for (int i=n; i>=0; i--) {
  sum *= x;
  sum += a[i];
}


- 
Lastly, even the array

a` is not needed:

#include
int main() {
  int n;
  unsigned x, a;
  unsigned long long sum = 0;
  scanf("%d%u", &n, &x);
  while (n-- >= 0) {
    scanf("%u", &a);
    sum *= x;
    sum += a;
  }
  printf("%llu", sum);
  return 0;
}

Code Snippets

for(i=0;i<n;i++)
{
    sum+=power(x,n-i)*a[i];
}

#define n_MAX (10)
#define n_MIN (0)
int a[n_MAX + 1];
if(n<n_MIN || n>n_MAX) return 0;

sum += (unsigned long long) power(x,n-i) * a[i];

unsigned long long int sum = 0;  //( 0 not 1)

if (y < 0) return 0;

Context

StackExchange Code Review Q#45488, answer score: 4

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