hdu 1491 Octorber 21st(數學題)

標籤：hdu   數學   

題目連結：http://acm.hdu.edu.cn/showproblem.php?pid=1491

Problem DescriptionHDU‘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.


 
InputThe 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). 
OutputFor 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

710 2010 1910  110 219   111 1112 12

 
Sample Output

1220It‘s today!!50What a pity, it has passed!What a pity, it has passed!

 
Author8600 
Source“2006校園文化活動月”之“校慶杯”大學生程式設計競賽暨杭州電子科技大學第四屆大學生程式設計競賽

代碼如下：

#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;}

hdu 1491 Octorber 21st(數學題)