A. Checking the Calendar
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A. Checking the Calendar
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output
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).
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.
这是平年31%7=3;30%7=2;28%7=0;这是一个月中下一个月与前一个月第一天差的天数。
所以这要确定每个月第一天和下一个月差的几天就可以确实是不是一年的。
#include<stdio.h>#include<string.h>#include<math.h>int judge(char s[]){ if(strcmp(s,"monday")==0) return 1; if(strcmp(s,"tuesday")==0) return 2; if(strcmp(s,"wednesday")==0) return 3; if(strcmp(s,"thursday")==0) return 4; if(strcmp(s,"friday")==0) return 5; if(strcmp(s,"saturday")==0) return 6; if(strcmp(s,"sunday")==0) return 7;}///把英文转化为数字int main(){ char s1[20],s2[20],n,m; int i; while(~scanf("%s%s",s1,s2)) { n=judge(s1); m=judge(s2); int a=(m-n+7)%7;///这是最值钱的一个地 if(a==3||a==2||a==0) printf("YES\n"); else printf("NO\n"); } return 0;}
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