HDU 2180 时钟

时钟
Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2186    Accepted Submission(s): 626


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

题意：》》》》

思路：直接暴力找出重合时间点，遍历一遍即可；

下面附上代码：

#include<bits/stdc++.h>
using namespace std;
int main()
{
	double h[25];
	int a,b,s,t;
	for(int i=0;i<23;i++)
		h[i]=1.0*i*30*2/11+i*60;
	while(~scanf("%d %d %d %d",&a,&b,&s,&t)&&(a||b||s||t))
	{
		int c=0;
		int t1,t2;
		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);
	} 
	return 0;
}