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 $d_1$-th day in the $m_1$-th month of year $y_1$. Given the day of today on that planet, what day will it be (or was it) on the $d_2$-th day in the $m_2$-th month of year $y_2$ on that planet?

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

The first line contains three integers $y_1$, $m_1$, $d_1$ ($2000 \le y_1 \le 10^9$, $1 \le m_1 \le 12$, $1 \le d_1 \le 30$) and a string $s$, indicating the date and day of today on that planet. It’s guaranteed that $s$ is either “Monday”, “Tuesday”, “Wednesday”, “Thursday” or “Friday”.

The second line contains three integers $y_2$, $m_2$ and $d_2$ ($2000 \le y_2 \le 10^9$, $1 \le m_2 \le 12$, $1 \le d_2 \le 30$), 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 $d_2$-th day in the $m_2$-th month of year $y_2$ 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

#include<bits/stdc++.h>
using namespace std;

char n[5][10]= {"Monday","Tuesday","Wednesday","Thursday","Friday"};
int main()
{
    std::ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--)
    {
        long long int y,m,d;
        char s[10];
        cin>>y>>m>>d>>s;
        long long int y2,m2,d2;
        cin>>y2>>m2>>d2;
        long long int sum1=0;
        sum1=-(y*360+m*30+d)+(y2*360+m2*30+d2);
        int week=sum1%5;
        if(week<0)//注意是負數的情況
            week+=5;
        int j;
        for(j=0; j<5; j++)
        {
            if(strcmp(n[j],s)==0)
                break;
        }
        cout<<n[(j+week)%5]<<endl;
    }
    return 0;
}