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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

class Program
{
    static void Main(string[] args)
    {
        var waaaa = new waaaa();
        waaaa.Solve();
    }
}

public class waaaa
{
    const int MOD = 1000000007;
    const int INF = 1 << 30;
    const long INFL = 1 << 62;
    const int MOD2 = 998244353;

    myinout IO = new myinout();
    Math2 mmath = new Math2();

    public void Solve()
    {
        //ここから
        var o = IO.IntM;
        int N = o[0];
        int M = o[1];        
        var o0 = IO.IntM;
        int N1 = o[0];
        int M1 = o[1];

        if(M1==1)IO.wri(1);
        else IO.wri(0);

    }



}

//b-aで降順

public class Math2
{
    public long[] Fact;  //Nまでの階乗
    public long[] invFact;  //階乗の逆元

    //Nまでの階乗をセットしておく
    public void FactoricalModSet(long N, long MOD = 1000000007)
    {
        Fact = new long[N + 1];
        invFact = new long[N + 1];

        Fact[0] = 1;
        for (long i = 1; i <= N; ++i)
        {
            Fact[i] = Fact[i - 1] * i % MOD;
        }
        invFact[N] = pow(Fact[N], MOD - 2);

        for (long i = N; i > 0; --i)
        {
            invFact[i - 1] = invFact[i] * i % MOD;
        }

    }

    //nCr 引数:n r (mod)
    public long nCr(long n, long r, long MOD = 1000000007)
    {
        return (Fact[n] * invFact[r] % MOD * invFact[n - r] % MOD);
    }

    //階乗(O(logN)) 引数:底x n乗　(MOD)
    public long pow(long x, long n, long MOD = 1000000007)
    {
        long res = 1;
        while (n > 0)
        {
            if ((n & 1) != 0) res = res * x % MOD;
            x = x * x % MOD;
            n >>= 1;
        }
        return res;
    }

    //最大公約数 引数:整数x,y
    public long GCD(long x, long y)
    {
        while (y != 0)
        {
            var r = x % y;
            x = y;
            y = r;
        }
        return x;
    }

    //最小公倍数 引数:整数x,y
    public long LCM(long x, long y)
    {
        return x * y / GCD(x, y);
    }

    //素因数分解 引数:整数N
    public Dictionary<long, int> PrimeFactorization(long N)
    {
        var ret = new Dictionary<long, int>();

        for (long i = 2; i * i <= N; ++i)
        {
            int cnt = 0;
            while (N % i == 0)
            {
                cnt++;
                N /= i;
            }
            if (cnt != 0) ret[i] = cnt;

        }
        if (N >= 2) ret[N] = 1;

        return ret;

    }

    //約数列挙
    public List<long> divisorenumrate(long N)
    {
        var ret = new List<long>();

        for (long i = 1; i * i <= N; ++i)
        {

            if (N % i == 0)
            {
                ret.Add(i);
                ret.Add(N / i);
            }

        }

        return ret;

    }

}

class myinout
{
    //in
    public string Str => Console.ReadLine();

    public string[] StrM => Str.Split(' ');

    public long[] LongM => StrM.Select(long.Parse).ToArray();
    public int[] IntM => StrM.Select(int.Parse).ToArray();
    public char[] CharM => Str.ToArray();
    public long Long1 => LongM[0];
    public int Int1 => IntM[0];

    //out
    public void wri<T>(T x)
    {
        Console.WriteLine(x);
    }

    public void yn(bool x)
    {
        if (x) wri("Yes");
        else wri("No");
    }

}