# DPOLY-Editorial

Setter: Kanhaiya Mohan
Tester: Manan Grover, Lavish Gupta
Editorialist: Devendra Singh

793

None

# 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.

# EXPLANATION:

The answer to the above problem is the largest index i (0-based) in the array A which is non-zero.Let ans be the largest index with non-zero value. Keep on iterating on the array A from the beginning and if A_i!=0, set ans =i.
In the end output ans.

# TIME COMPLEXITY:

O(N) for each test case.

# SOLUTION:

Setter's solution
//Utkarsh.25dec
#include <bits/stdc++.h>
#define ll long long int
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define vl vector <ll>
#define all(c) (c).begin(),(c).end()
using namespace std;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
ll modInverse(ll a){return power(a,mod-2);}
const int N=500023;
bool vis[N];
long long readInt(long long l,long long r,char endd){
long long x=0;
int cnt=0;
int fi=-1;
bool is_neg=false;
while(true){
char g=getchar();
if(g=='-'){
assert(fi==-1);
is_neg=true;
continue;
}
if('0'<=g && g<='9'){
x*=10;
x+=g-'0';
if(cnt==0){
fi=g-'0';
}
cnt++;
assert(fi!=0 || cnt==1);
assert(fi!=0 || is_neg==false);

assert(!(cnt>19 || ( cnt==19 && fi>1) ));
} else if(g==endd){
if(is_neg){
x= -x;
}

if(!(l <= x && x <= r))
{
cerr << l << ' ' << r << ' ' << x << '\n';
assert(1 == 0);
}

return x;
} else {
assert(false);
}
}
}
string ret="";
int cnt=0;
while(true){
char g=getchar();
assert(g!=-1);
if(g==endd){
break;
}
cnt++;
ret+=g;
}
assert(l<=cnt && cnt<=r);
return ret;
}
long long readIntSp(long long l,long long r){
}
long long readIntLn(long long l,long long r){
}
}
}
void solve()
{
int n;

bool oneNonZero = false;
int a[n];
for(int i = 0; i<n-1; i++){
a[i] = x;
if(x!=0) oneNonZero = true;
}
a[n-1] = x;
if(x!=0) oneNonZero = true;

assert(oneNonZero==true);

for(int i = n-1;i>=0; i--){
if(a[i]!=0){
cout<<i<<endl;
return;
}
}
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
#endif
ios_base::sync_with_stdio(false);
cin.tie(NULL),cout.tie(NULL);
while(T--)
solve();
assert(getchar()==-1);
cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
}

Editorialist's Solution
#include "bits/stdc++.h"
using namespace std;
#define ll long long
#define pb push_back
#define all(_obj) _obj.begin(), _obj.end()
#define F first
#define S second
#define pll pair<ll, ll>
#define vll vector<ll>
ll INF = 1e18;
const int N = 1e5 + 11, mod = 1e9 + 7;
ll max(ll a, ll b) { return ((a > b) ? a : b); }
ll min(ll a, ll b) { return ((a > b) ? b : a); }
void sol(void)
{
int n;
cin >> n;
vll v(n);
for (int i = 0; i < n; i++)
cin >> v[i];
int ans = 0;
for (int i = 0; i < n; i++)
if (v[i])
ans = i;
cout << ans << '\n';
return;
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL), cout.tie(NULL);
int test = 1;
cin >> test;
while (test--)
sol();
}