HackerEarth Counting Frog Paths problem solution

In this HackerEarth Counting Frog Paths problem solution There is a frog initially placed at the origin of the coordinate plane. In exactly 1 second, the frog can either move up 1 unit, move right 1 unit, or stay still. In other words, from position (x,y), the frog can spend 1 second to move to:

1. (x+1, y)
2. (x, y+1)
3. (x, y)

After T seconds, a villager who sees the frog reports that the frog lies on or inside a square of side-length s with coordinates (X, Y), (X+s, Y), (X, Y+s), (X+s, Y+s. Calculate how many points with integer coordinates on or inside this square could be the frog's position after exactly T seconds.

HackerEarth Counting Frog Paths problem solution.

#include <bits/stdc++.h>

using namespace std;


string twodig(int i){
    string res = "";
    while(i>0){
        res = char('0' + i%10) + res;
        i/=10;
    }
    while(res.length() < 2) res = "0"+res;
    return res;
}
int solve(int X, int Y, int s, int T){
    int counter = 0;
    for(int positionX = X; positionX <= X+s; positionX++){
        for(int positionY = Y; positionY <= Y+s; positionY++){
            if(positionY + positionX <= T){
                counter++;
            }
        }
    }
    return counter;
}
void gen(){
    int iters = 50;
    for(int i = 6; i <= iters; i++){
        ofstream fout("in"+twodig(i)+".txt");
        int x = rand()%101, y = rand()%101, s = (1 + rand()%100), t = rand()%401;
        fout << x << " " << y << " "<< s << " " << t << endl;
        fout.close();
    }
}
void solveall(){
    int iters = 50;
    for(int i = 1; i <= iters; i++){
        ifstream fin("in"+twodig(i)+".txt");
        int x,y,s,t;
        fin >> x >> y >> s >> t;
        int ans = solve(x,y,s,t);
        ofstream fout("out"+twodig(i)+".txt");
        fout << ans << endl;
        fout.close();
        fin.close();
    }
}
void validateall(){
    int iters = 50;
    for(int i = 1; i <= iters; i++){
        ifstream fin("in"+twodig(i)+".txt");
        int x,y,s,t;
        fin >> x >> y >> s >> t;
        assert(0<=x);
        assert(x<=100);
        assert(0<=y);
        assert(y<=100);
        assert(1<=s);
        assert(s<=100);
        assert(0<=t);
        assert(t<=400);
    }
}
int main()
{
    int x,y,s,t;
    cin >> x >> y >> s >> t;
    cout << solve(x,y,s,t) << endl;
}

Second solution

#include<bits/stdc++.h>

using namespace std;

typedef complex<double> base;
typedef long double ld;
typedef long long ll;

#define pb push_back
#define pii pair<int,int>
#define pll pair< ll , ll >
#define vi vector<int>

const int maxn=(int)(1e5+5);
const ll mod=(ll)(998244353);
int a[maxn];

int main()
{
    ios_base::sync_with_stdio(0);

    int x,y,s,t,res=0;

    cin>>x>>y>>s>>t;

    assert(min(x,y)>=0 && max(x,y)<=100);

    assert(s>=1 && s<=100);

    assert(t>=0 && t<=400);

    for(int i=x;i<=x+s;i++)
    {
        for(int j=y;j<=y+s;j++)
        {
            int dis=i+j;

            if(dis<=t)
            {
                res++;
            }
        }
    }

    cout<<res<<endl;

}