【杭电oj】2180-时钟（打表，水）

时钟

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1795    Accepted Submission(s): 459

Problem Description

从a点b分到s点t分时针和分针重合多少次? 

 

Input

有多组数据，每组1行4个数 a,b,s,t. 1<=a,s <=12, 0<=b,t<60. 0 0 0 0结束.

 

Output

参看Sample output 

 

Sample Input

   
   
   
   

    
    
    
    12 50 1 2 
3 8 3 20 
2 45 11 0 
11 0 3 20 
1 2 12 50 
3 20 3 8
0 0 0 0
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    0
1
8
4
11
10
   
   
   
   

 

Author

zhousc

 

Source

ECJTU 2008 Summer Contest

事实证明，想一些乱七八糟的计算，不如直接把结果列出来方便。

代码如下：

#include <stdio.h>
int main()
{
	double h[25]={0,65.45,130.91,196.36,261.82,327.27,392.73,458.18,523.64,589.09, 
                	654.55,720.00,785.45,850.91,916.36,981.82,1047.27,1112.73,1178.18, 
                	1243.64,1309.09,1374.55,1440.00};
	int a,b,s,t;
	int t1,t2;
	int c;
	while (~scanf ("%d %d %d %d",&a,&b,&s,&t) && (a||b||s||t))
	{
		c=0;
		t1=a*60+b;
		t2=s*60+t;
		if (t1>t2)
			t2+=720;
		for (int i=0;i<=22;i++)
		{
			if (t1<=h[i] && t2>=h[i])
				c++;
		}
		printf ("%d\n",c);
	}
}