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: Accepted)

#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <cmath>
#include <string>
#include <cstring>

#define roop(i, n) for(int i=0;i<n;i++)

using namespace std;
using ll = long long int;
using ull = unsigned long long;

constexpr ull mod = 1e9+7;
constexpr int N = 13;
ull dp[100010][N];

float myPower(int a, int n) { // aのn乗の計算
	float x = 1;
	while (n > 0) { // 全てのbitが捨てられるまで
		if ( n & 1) { // 一番右のbitが１のとき
			x *= a;
		}
		a *= a;
		n >>= 1; // bit全体を右に1つシフトして一番右を捨てる
	}
	return x;
}

int main() {
	int m1, m2, d1, d2;
	cin >> m1 >> d1 >> m2 >> d2;
	if (m1 != m2) {
		cout << "1";
	} else {
		cout << "0";
	}
	return 0;
}