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 <cstring>
#include <map>
#include <unordered_map>
#include <string>
#include <list>
#include <vector>
#include <algorithm>
#include <queue>
#include <stack>
#include <cstdio>
#include <iostream>
#include <set>
#include <unordered_set>
using namespace std;

#define MAX(a,b) (a>b?a:b)
#define MIN(a,b) (a<b?a:b)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int ui;
typedef pair<int, int> pii;

int gi() {
    int a;
    scanf("%d", &a);
    return a;
}

ll gli() {
    ll a;
    scanf("%lld", &a);
    return a;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    int m = gi();
    int d = gi();
    int y = 0;
    if (m == 1 || m == 3 || m == 5 || m == 7 || m == 8 || m == 10 || m == 12)
        y = (d == 31);
    else if (m == 2)
        y = (d == 28);
    else
        y = (d == 30);

    printf(y ? "1\n" : "0\n");

    return 0;
}