题目链接：点击打开链接
A. Checking the Calendar
time limit per test
memory limit per test
input
output
You are given names of two days of the week.
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.
31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31.
monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
Input
monday", "tuesday", "wednesday", "thursday", "friday", "saturday", "sunday".
Output
YES" (without quotes) if such situation is possible during some non-leap year. Otherwise, print "NO" (without quotes).
Examples
input
monday tuesday
output
NO
input
sunday sunday
output
YES
input
saturday tuesday
output
YES
Note
In the second sample, one can consider February 1 and March 1 of year 2015. Both these days were Sundays.
In the third sample, one can consider July 1 and August 1 of year 2017. First of these two days is Saturday, while the second one is Tuesday.
大意：判断是否存在相邻月份的第一天满足给出的例子
题解：题目已经确定是平年，星期的周期是 7，而月份相差天数只有三种情况 28，30，31；其对应的周期余数为 0，2，3。
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
char a[10],b[10];
// 0 2 3
int main()
{
while(~scanf("%s%s",a,b))
{
if(a[0]=='m')
{
if(b[0]=='m'||b[0]=='w'||(b[0]=='t'&&b[1]=='h'))
puts("YES");
else puts("NO");
}
if(a[0]=='t'&&a[1]=='u')
{
if(b[0]=='t'||b[0]=='f')
puts("YES");
else puts("NO");
}
if(a[0]=='w')
{
if(b[0]=='w'||b[0]=='f'||(b[0]=='s'&&b[1]=='a'))
puts("YES");
else puts("NO");
}
if(a[0]=='t'&&a[1]=='h')
{
if((b[0]=='t'&&b[1]=='h')||b[0]=='s')
puts("YES");
else puts("NO");
}
if(a[0]=='f')
{
if(b[0]=='f'||(b[0]=='s'&&b[1]=='u')||b[0]=='m')
puts("YES");
else puts("NO");
}
if(a[0]=='s'&&a[1]=='a')
{
if((b[0]=='s'&&b[1]=='a')||b[0]=='m'||(b[0]=='t'&&b[1]=='u'))
puts("YES");
else puts("NO");
}
if(a[0]=='s'&&a[1]=='u')
{
if((b[0]=='s'&&b[1]=='u')||(b[0]=='t'&&b[1]=='u')||b[0]=='w')
puts("YES");
else puts("NO");
}
}
return 0;
}
