杭电OJ题 1491 Octorber 21st 解题报告

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本文介绍了一个简单的程序设计问题，即计算任意给定日期（限定于2006年）到2006年10月21日之间的天数差。如果所给日期在10月21日之后，则输出提示信息；如果正好是10月21日，则输出特定消息。

Octorber 21st

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1934 Accepted Submission(s): 1175

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

Sample Output

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

——————————————————————————————————————————————————————————————————————————————————————————
/****************************
 *Name:Octorber 21st.c
 *Tags:ACM water
 ****************************/

#include <stdio.h>

int main()
{
      int t, m, d, days, i;
      char str[10];
      int month[12] = {31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
      scanf("%d", &t);
      /* getchar(); */
      while(t--) {
	    scanf("%d%d", &m, &d);
	    /*gets(str);
	    if(str[1] == ' ') {
		  m = str[0] - '0';
	    } else {
		  m = (str[0] - '0') * 10 + str[1] - '0';
	    }
	    if(str[3] == ' ') {
		  d = str[4] - '0';
	    } else {
		  d = (str[3] - '0') * 10 + str[4] - '0';
	    }*/
	    /* printf("m=%d d=%d\n", m, d);*/
	    if(m > 10 || (m == 10 && d > 21)) {
		  printf("What a pity, it has passed!\n");
	    }
	    else if(m == 10 && d == 21) {
		  printf("It's today!!\n");
	    } else {
		  days = 0;
		  while(m < 10) {
			days += month[m-1];
			m++;
		  }
		  days += 21;
		  days -= d;
		  printf("%d\n", days);
	    }
      }
      return 0;
}