杭电1491Octorber 21st（数学）

忘是亡心i 2022-08-17 14:25 12阅读 0赞

# 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.  
  
![c48-1001-1.jpg][]

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!

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;
     }

[c48-1001-1.jpg]: /images/20220731/43920a47c52f4304914a5d6697aa1a02.png