Degree of Polynomial : DPOLY | CodeChef Solution

Problem

In mathematics, the degree of polynomials in one variable is the highest power of the variable in the algebraic expression with non-zero coefficient.

Chef has a polynomial in one variable $x$ with $N$ terms. The polynomial looks like $A_0\cdot x^0 + A_1\cdot x^1 + \ldots + A_{N-2}\cdot x^{N-2} + A_{N-1}\cdot x^{N-1}$ where $A_{i-1}$ denotes the coefficient of the $i^{th}$ term $x^{i-1}$ for all $(1\le i\le N)$ .

Find the degree of the polynomial.

Note: It is guaranteed that there exists at least one term with non-zero coefficient.

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Input Format

First line will contain $T$ , number of test cases. Then the test cases follow.
First line of each test case contains of a single integer $N$ - the number of terms in the polynomial.
Second line of each test case contains of $N$ space-separated integers - the $i^{th}$ integer $A_{i-1}$ corresponds to the coefficient of $x^{i-1}$ .

Output Format

For each test case, output in a single line, the degree of the polynomial.

Constraints

$1 \leq T \leq 100$
$1 \leq N \leq 1000$
$-1000 \le A_i \le 1000$
$A_i \ne 0$ for at least one $(0\le i \lt N)$ .
$1 \leq T \leq 1000$

Example / Sample:

Input:

Output:

Explanation:

Test case $1$ : There is only one term $x^0$ with coefficient $5$ . Thus, we are given a constant polynomial and the degree is $0$ .

Test case $2$ : The polynomial is $-3\cdot x^0 + 3\cdot x^1 = -3 + 3\cdot x$ . Thus, the highest power of $x$ with non-zero coefficient is $1$ .

Test case $3$ : The polynomial is $0\cdot x^0 + 0\cdot x^1 + 5\cdot x^2= 0+0 + 5\cdot x^2$ . Thus, the highest power of $x$ with non-zero coefficient is $2$ .

Test case $4$ : The polynomial is $1\cdot x^0 + 2\cdot x^1+ 4\cdot x^2 + 0\cdot x^3= 1 + 2\cdot x + 4\cdot x^2$ . Thus, the highest power of $x$ with non-zero coefficient is $2$ .

Solution

#include <iostream>
using namespace std;

// Solution from : Code Radius [ https://radiuscode.blogspot.com/ ]

int main() {
 int a,t;
    cin>>t;
    for(a=0;a<t;a++)
    {
        int n,a[1000],i;
        cin>>n;
        for(i=0;i<n;i++)
        {
            cin>>a[i];
        }
        for(i=n-1;i>=0;i--)
        {
            if(a[i]!=0)
            {
                break;
            }
        }
        cout<<i<<endl;
        
    }
 return 0;

}

Please First Try to Solve Problem by Yourself.

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