A - Calandar

On a planet far away from Earth, one year is composed of 12 months, and each month always consists of 30 days.

Also on that planet, there are 5 days in a week, which are Monday, Tuesday, Wednesday, Thursday and Friday. That is to say, if today is Monday, then tomorrow will be Tuesday, the day after tomorrow will be Wednesday. After 3 days it will be Thursday, after 4 days it will be Friday, and after 5 days it will again be Monday.

Today is the d_1d1-th day in the m_1m1-th month of year y_1y1. Given the day of today on that planet, what day will it be (or was it) on the d_2d2-th day in the m_2m2-th month of year y_2y2 on that planet?

Input

There are multiple test cases. The first line of the input contains an integer TT (about 100), indicating the number of test cases. For each test case:

The first line contains three integers y_1y1m_1m1d_1d1 (2000 \le y_1 \le 10^92000y1109, 1 \le m_1 \le 121m112, 1 \le d_1 \le 301d130) and a string ss, indicating the date and day of today on that planet. It's guaranteed that ss is either "Monday", "Tuesday", "Wednesday", "Thursday" or "Friday".

The second line contains three integers y_2y2m_2m2 and d_2d2 (2000 \le y_2 \le 10^92000y2109, 1 \le m_2 \le 121m212, 1 \le d_2 \le 301d230), indicating the date whose day we want to know.

Output

For each test case output one line containing one string, indicating the day of the d_2d2-th day in the m_2m2-th month of year y_2y2 on that planet.

Sample Input

4
2019 5 12 Monday
2019 5 14
2019 5 12 Tuesday
2019 12 30
2019 5 12 Friday
1000000000 1 1
1000000000 1 1 Wednesday
2019 5 12

Sample Output

Wednesday
Friday
Thursday
Thursday

 

代码     ps:只需要看day,与year和month无关

 1 #include<iostream>
 2 using namespace std;
 3 int main(){
 4     int t;
 5     int y,m,d;
 6     int y1,m1,d1;
 7     char s[10];
 8     scanf("%d",&t);
 9     while(t--){
10         scanf("%d %d %d %s",&y,&m,&y,s);
11         scanf("%d %d %d",&y1,&m1,&y1);
12         int n=(y1-y)%5;
13         int k;
14         if(s[0]=='M') k=0;
15         else if(s[0]=='T'&&s[1]=='u') k=1;
16         else if(s[0]=='W') k=2;
17         else if(s[0]=='T'&&s[1]=='h') k=3;
18         else if(s[0]=='F') k=4;
19         if(n<=0) k=(k+5+n)%5;
20         else k=(k+n)%5;
21         if(k==0) printf("Monday\n");
22         else if(k==1) printf("Tuesday\n");
23         else if(k==2) printf("Wednesday\n");
24         else if(k==3) printf("Thursday\n");
25         else if(k==4) printf("Friday\n");
26     }
27 }
posted @ 2020-03-18 15:50  Caecae_with_island  阅读(287)  评论(0编辑  收藏  举报