hdu 1491 Octorber 21st(数学题)

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1491

Problem Description

HDU‘s 50th birthday, on Octorber 21st, is coming. What an exciting day!! As a student of HDU, I always want to know how many days are there between today and Octorber 21st.So, write a problem and tell me the answer.Of course, the date I give you is always in
2006.

Input

The input consists of T test cases. The number of T is given on the first line of the input file.Following T lines, which represent dates, one date per line. The format for a date is "month day" where month is a number between 1 (which indicates January) and
12 (which indicates December), day is a number between 1 and 31.All the date in the input are in 2006, you can assume that all the dates in the input are legal(合法).

Output

For each case, if the date is before Octorber 21st, you should print a number that between the date and Octorber 21st.If the day is beyond Octorber 21st, just print "What a pity, it has passed!".If the date is just Octorber 21st, print"It‘s today!!".

Sample Input

7
10 20
10 19
10  1
10 21
9   1
11 11
12 12

Sample Output

1
2
20
It‘s today!!
50
What a pity, it has passed!
What a pity, it has passed!

Author

8600

Source

“2006校园文化活动月”之“校庆杯”大学生程序设计竞赛暨杭州电子科技大学第四届大学生程序设计竞赛

代码如下：

#include <cstdio>
int mon(int x)
{
    if(x == 1)
        return 31;
    else if(x == 2)
        return 28;
    else if(x == 3)
        return 31;
    else if(x == 4)
        return 30;
    else if(x == 5)
        return 31;
    else if(x == 6)
        return 30;
    else if(x == 7)
        return 31;
    else if(x == 8)
        return 31;
    else if(x == 9)
        return 30;
}
int main()
{
    int t;
    int m, d;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d %d",&m,&d);
        if(m == 10 && d == 21)
        {
            printf("It's today!!\n");
            continue;
        }
        if(m>10 ||(m==10&&d>21))
        {
            printf("What a pity, it has passed!\n");
            continue;
        }
        int sum = 0;
        for(int i = m; i < 10; i++)
        {
            sum+=mon(i);
        }
        printf("%d\n",sum-d+21);
    }
    return 0;
}