DPOLY-Editorial

PROBLEM LINK:

Contest Division 1
Contest Division 2
Contest Division 3
Contest Division 4

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

DIFFICULTY:

793

PREREQUISITES:

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];
vector <int> adj[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 readString(int l,int r,char endd){
    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){
    return readInt(l,r,' ');
}
long long readIntLn(long long l,long long r){
    return readInt(l,r,'\n');
}
string readStringLn(int l,int r){
    return readString(l,r,'\n');
}
string readStringSp(int l,int r){
    return readString(l,r,' ');
}
void solve()
{
    int n;
    n=readInt(1,1000,'\n');
    
    bool oneNonZero = false;
    int a[n];
    for(int i = 0; i<n-1; i++){
        int x = readInt(-1000,1000,' ');
        a[i] = x;
        if(x!=0) oneNonZero = true;
    }
    int x = readInt(-1000,1000,'\n');
    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);
    int T=readInt(1,100,'\n');
    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); }
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
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();
}