杭电1491Octorber 21st（数学）

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#杭电 #1491 #Octorber 21st #数学

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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，特别适用于纪念特殊日子或事件倒计时。文章提供了完整的C++代码实现，并通过样例输入输出进行了验证。

Octorber 21st

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

Total Submission(s): 3688 Accepted Submission(s): 2268

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!

简单数学题！

#include<cstring>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
	int m,n,i,j,M,a[12]={0,31,28,31,30,31,30,31,31,30,31};
	scanf("%d",&M);
	int sum=0;
	while(M--)
	{
		scanf("%d%d",&m,&n);
		if(m>10||(m==10&&n>21))
		printf("What a pity, it has passed!\n");
		else
		{
			if(m==10)
			{
				if(n==21)
				{
					printf("It's today!!\n");
					continue;
				}
				else{
					printf("%d\n",21-n);
					continue;
				}
			}
			sum=a[m]-n;
			for(i=m+1;i<10;i++)
			sum+=a[i];
			sum+=21;
			printf("%d\n",sum);
		}
	}
	return 0;
 }