Evaluating polynomials

You are given a polynomial of degree n. The polynomial is of the form P(x) = anxn + an-1xn-1 + … + a0. For given values k and m, You are required to find P(k) at the end of the mth iteration of Horner’s rule. The steps involved in the Horner’s rule are given below,

Pn (x) = an

Pn-1 (x) = an-1 + x * Pn (x)                              1st iteration.

Pn-2 (x) = an-2 + x * Pn-1 (x)                           2nd iteration.

.

.

P0 (x) = a0 + x * P1 (x)                                  nth iteration.

In general, Pi (x) = ai + x * Pi + 1 (x) and P0(x) is the final result.

Input

n m k an an-1 …. a0

(Space separated)

Output: P(k) value at the end of the mth iteration.

Sample Input:

2 2 5

3 2 1

Sample Output:

86

Constraints:

1 <= n, k, m <= 10

0 <= ai <=10


Solution -:


#include <stdio.h> int main() { int n, m, k, a[30],j; int ans,i; scanf("%d %d %d", &n, &m, &k); ans = 0; for( i = 0; i <= n; ++i) scanf("%d", &a[i]); ans=a[0]; for(j=1;j<=m;j++) { ans=ans*k+a[j]; } printf("%d",ans); return 0; }