【63.63%】【codeforces 724A】Checking the Calendar

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

【题解】

31%7=3,30%7=2,28%7=0;
所以后一天是前一天加上3，2，0的星期。
我是想说每个月的第一天在不同的年星期一到星期天都存在吧。
所以年份什么的就不用考虑了。

#include <cstdio>
#include <iostream>
#include <string>

using namespace std;

string s[10],s1,s2;

int main()
{
    //freopen("F:\rush.txt", "r", stdin);
    s[1] = "monday";
    s[2] = "tuesday";
    s[3] = "wednesday";
    s[4] = "thursday";
    s[5] = "friday";
    s[6] = "saturday";
    s[7] = "sunday";
    cin >> s1;
    cin >> s2;
    int num1, num2;
    for (int i = 1; i <= 7; i++)
        if (s1 == s[i])
            num1 = i;
    for (int i = 1; i <= 7; i++)
        if (s2 == s[i])
            num2 = i;
    if (num2 < num1)
        num2 += 7;
    int temp = num2 - num1;
    if (temp == 0 || temp == 2 || temp == 3)
        puts("YES");
    else
        puts("NO");
    return 0;
}