Codeforces Intel Code Challenge Final Round (Div. 1 + Div. 2, Combined) A. Checking the Calendar（水题）

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Description

You are given names of two days of the week.

Please, determine whether it is possible that during some non-leap year the first day of some month was equal to the first day of the week you are given, while the first day of the next month was equal to the second day of the week you are given. Both months should belong to one year.

In this problem, we consider the Gregorian calendar to be used. The number of months in this calendar is equal to 12. The number of days in months during any non-leap year is: 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.

Names of the days of the week are given with lowercase English letters: "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".

Input

The input consists of two lines, each of them containing the name of exactly one day of the week. It's guaranteed that each string in the input is from the set "monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".

Output

Print "YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).

Sample Input

mondaytuesday

sundaysunday

saturdaytuesday

Sample Output

NO

YES

YES

思路

题意：给出一星期中的两天，问他们有没有可能是平年中连续的两个月的头一天。

注意tuesday thursday和thursday tuesday不是同种情况就可以了。

 

#include<bits/stdc++.h>
using namespace std;

int main()
{
	int x,y;
	int a[5] = {29,31,32};
	string str1,str2;
	cin >> str1 >> str2;
	if (str1 == "monday")	x = 1;		if (str2 == "monday")	y = 1;
	if (str1 == "tuesday")	x = 2;		if (str2 == "tuesday")	y = 2;
	if (str1 == "wednesday")	x = 3;	if (str2 == "wednesday")	y = 3;
	if (str1 == "thursday")	x = 4;		if (str2 == "thursday")	y = 4;
	if (str1 == "friday")	x = 5;		if (str2 == "friday")	y = 5;
	if (str1 == "saturday")	x = 6;		if (str2 == "saturday")	y = 6;
	if (str1 == "sunday")	x = 7;		if (str2 == "sunday")	y = 7;
	bool flag = false;
	if (y >= x)
	{
		for (int i = 0;i < 3;i++)
		{
			if (a[i] % 7 == y - x + 1)
			{
				flag = true;
				break;
			}
		}
		if (flag)	printf("YES\n");
		else	printf("NO\n");
	}
	else
	{
		for (int i = 0;i < 3;i++)
		{
			if (a[i] % 7 == 7 - x + 1 + y)
			{
				flag = true;
				break;
			}
		}
		if (flag)	printf("YES\n");
		else	printf("NO\n");
	}
	return 0;
}