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<iostream>
using namespace std;
int main(){
  long long x;
  cin>>x;
  long long n=1000000;
  long long ok=0;
  for(long long a=0;a<=n;a++){
    long long sum=100*a;
    if(sum==x){
      ok=1;
    }
    for(long long b=0;b<=n;b++){
       sum+=101*b;
    if(sum==x){
      ok=1;
      break;
    }
      else if(sum>x){
          break;
        }
      for(long long c=0;c<=n;c++){
      sum=sum+102*c;
    if(sum==x){
      ok=1;
      break;
    }
        else if(sum>x){
          break;
        }
      for(long long d=0;d<=n;d++){
       sum+=103*d;
    if(sum==x){
      ok=1;
      break;
    }
          else if(sum>x){
          break;
        }
      for(long long e=0;e<=n;e++){
      sum+=104*b;
    if(sum==x){
      ok=1;
      break;
    }
          else if(sum>x){
          break;
        }
      for(long long f=0;f<=n;f++){
       sum+=105*f;
    if(sum==x){
      ok=1;
      break;
    }
          else if(sum>x){
          break;
        }
      
    }
    }
    }
    }
    }
    if(ok==1) break;
  }
         
  cout<<ok<<endl;
 return 0; 
}