Date comparison

D105 – Time Limit: 1.000 s – Memory Limit: 256 MB

Write a program to compare two dates: $y_{1} - m_{1} - d_{1}$ and $y_{2} - m_{2} - d_{2}$ .

Input and Output

The input consists of two lines. The first line contains $y_{1}$ , $m_{1}$ and $d_{1}$ separated by spaces. The second line contains $y_{1}$ , $m_{1}$ and $d_{1}$ separated by spaces.

$1000 \leq y_{1}, y_{2} \leq 9999$ , and the dates are guaranteed to be valid.

If $y_{1} - m_{1} - d_{1}$ is earlier, output Before.

If $y_{2} - m_{2} - d_{2}$ is earlier, output After.

If the two dates are the same, output Same.

Sample Tests

	Input	Output
1	2015 3 1 2015 7 5	Before

2	2014 6 20 2012 11 2	After

3	2046 5 13 2046 5 13	Same

Solutions

Note: There are many other algorithms/implementations that can solve this task correctly.

C++


#include <iostream>
using namespace std;
int main() {
    int y1, m1, d1, y2, m2, d2;
    cin >> y1 >> m1 >> d1;
    cin >> y2 >> m2 >> d2;
    int a = 10000 * y1 + 100 * m1 + d1;
    int b = 10000 * y2 + 100 * m2 + d2;
    if (a < b) {
        cout << "Before" << endl;
    } else if (a == b) {
        cout << "Same" << endl;
    } else {
        cout << "After" << endl;
    }
    return 0;
}

Python


y1, m1, d1 = map(int, input().split())
y2, m2, d2 = map(int, input().split())
a = 10000 * y1 + 100 * m1 + d1
b = 10000 * y2 + 100 * m2 + d2
if a < b:
    print('Before')
elif a == b:
    print('Same')
else:
    print('After')

Hidden Tests

	Input	Output
4	2015 11 30 2015 12 1	Before

5	2000 9 2 2000 9 2	Same

6	2000 9 2 2000 9 3	Before

7	2000 9 3 2000 9 2	After

8	2000 10 1 2000 9 5	After

9	2000 8 21 2000 9 3	Before

10	2004 2 29 2046 3 01	Before

11	2006 1 1 2005 12 31	After

12	1000 1 1 9999 12 31	Before

13	9876 12 31 9877 1 1	Before