Problem:
Score: 100 points

Problem Statement

In this problem, a date is written as Y-M-D. For example, 2019-11-30 means November 30, 2019.

Integers M_1, D_1, M_2, and D_2 will be given as input.

It is known that the date 2019-M_2-D_2 follows 2019-M_1-D_1.

Determine whether the date 2019-M_1-D_1 is the last day of a month.

Constraints

Both 2019-M_1-D_1 and 2019-M_2-D_2 are valid dates in the Gregorian calendar.

The date 2019-M_2-D_2 follows 2019-M_1-D_1.

Input

Input is given from Standard Input in the following format:

M_1 D_1
M_2 D_2

Output

If the date 2019-M_1-D_1 is the last day of a month, print 1; otherwise, print 0.

Sample Input 1

11 16
11 17

Sample Output 1

0

November 16 is not the last day of a month.

Sample Input 2

11 30
12 1

Sample Output 2

1

November 30 is the last day of November.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define ll long long
#define ld long double
#define f first
#define s second
#define mp make_pair
#define pb push_back
#define vll vector<ll>
#define pll pair<ll,ll>

using namespace std;

const ll N = 3e5+100;
const ll MOD = 1e9+7;
bool poss[100010];
int main()
{

	ios_base::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);

	ll n;
	cin >> n;
	poss[0] = true;

	for(int i = 1;i<=n;i++)
	{
		for(auto x:{100,101,102,103,104,105})
		{
			if(i >= x)
			{
				if(poss[i-x]==true) poss[i] = true;
			}
		}
	}
	if(poss[n]) cout <<1  <<endl;else cout << 0 << endl;return 0;
	

}