HDU 1491 Octorber 21st【计算日期】

/*
中文题目 10月21号
中文翻译-大意  求解距离10月21号还有多少天（都在2006年里面取日期）
关键点：计算日期
解题人：lingnichong
解题时间：2014-08-30 07:54:49
解题体会：简单题
*/

Octorber 21st

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2934    Accepted Submission(s): 1806

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

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!

 

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

 

#include<stdio.h> int a[]={0,31,28,31,30,31,30,31,31,30,31,30,31};int main(){int n,m,d,i;int sum1,sum2;scanf("%d",&n);while(n--){sum1=294;sum2=0;scanf("%d%d",&m,&d);if(m==10&&d==21){printf("It's today!!\n");continue;}else if(m==10&&d>21||m>10){printf("What a pity, it has passed!\n");continue;}else{for(i=0;i<m;i++)sum2+=a[i];sum2+=d;printf("%d\n",sum1-sum2);}}return 0;}