New Year and Days

小熊Limak的糖果计划

最新推荐文章于 2017-07-28 21:37:23 发布

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小熊Limak决定在2016年定期存糖果，根据不同的存糖果计划（每周固定某天或每月固定某天），计算一年内能存多少糖果。示例包括周四存糖果或每月30号存糖果的情况。

Description

Today is Wednesday, the third day of the week. What's more interesting is that tomorrow is the last day of the year 2015.

Limak is a little polar bear. He enjoyed this year a lot. Now, he is so eager to the coming year 2016.

Limak wants to prove how responsible a bear he is. He is going to regularly save candies for the entire year 2016! He considers various saving plans. He can save one candy either on some fixed day of the week or on some fixed day of the month.

Limak chose one particular plan. He isn't sure how many candies he will save in the 2016 with his plan. Please, calculate it and tell him.

Input

The only line of the input is in one of the following two formats:

"x of week" where x (1 ≤ x ≤ 7) denotes the day of the week. The 1-st day is Monday and the 7-th one is Sunday.
"x of month" where x (1 ≤ x ≤ 31) denotes the day of the month.

Output

Print one integer — the number of candies Limak will save in the year 2016.

Sample Input

Input

4 of week

Output

Input

30 of month

Output

Hint

Polar bears use the Gregorian calendar. It is the most common calendar and you likely use it too. You can read about it on Wikipedia if you want to – https://en.wikipedia.org/wiki/Gregorian_calendar. The week starts with Monday.

In the first sample Limak wants to save one candy on each Thursday (the 4-th day of the week). There are 52 Thursdays in the 2016. Thus, he will save 52 candies in total.

In the second sample Limak wants to save one candy on the 30-th day of each month. There is the 30-th day in exactly 11 months in the 2016 — all months but February. It means that Limak will save 11 candies in total.

2016年是闰年，有52周，各多出一天周五和周六，每周第一天是周日。

败给了常识...

#include<stdio.h>
#include<string.h>
int main(){
	int n,sum,len;
	char str[100];
	while(~scanf("%d",&n)){
		gets(str);
		len=strlen(str);
		if(str[len-1]=='k'){
			if(n<=4){
				sum=52;
			}
			else if(n==6){
				sum=53;
			}
			else if(n==5){
				sum=53;
			}
			else if(n==7){
				sum=52;
			}
		}
		else{
			if(n<=29){
				sum=12;
			}
			else if(n==30){
				sum=11;
			}
			else if(n==31){
				sum=7;
			}
		}
		printf("%d\n",sum);
		memset(str,0,sizeof(str));
	}
	return 0;
}