UVa 10773 Back to Intermediate Math (数学&速度的分解&陷阱)

10773 - Back to Intermediate Math

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=467&page=show_problem&problem=1714

Umm! So, you claim yourself as an intelligent one? Let me check. As, computer science students always insist on optimization whenever possible, I give you an elementary problem of math to optimize.

You are trying to cross a river of width d meters. You are given that, the river flows at v ms-1 and you know that you can speed up the boat inu ms-1. There may be two goals how to cross the river: One goal (called fastest path) is to cross it in fastest time, and it does not matter how far the flow of the river takes the boat. The other goal (called shortest path) is to steer the boat in a direction so that the flow of the river doesn't take the boat away, and the boat passes the river in a line perpendicular to the boarder of the river. Is it always possible to have two different paths, one to pass at shortest time and the other at shortest path? If possible then, what is the difference (Let  P s) between the times needed to cross the river in the different ways?

Input

The first line in the input file is an integer representing the number of test cases. Each of the test cases follows below. Each case consists three real numbers (all are nonnegative, d is positive) denoting the value of d, v and u respectively.

Output

For each test case, first print the serial number of the case, a colon, an space and then print "can't determine" (without the quotes) if it is not possible to find different paths as stated above, else print the value of P corrected to three digits after decimal point. Check the sample input & output.

Sample Input

3

8 5 6

1 2 3

1 5 6

Sample Output

Case 1: 1.079

Case 2: 0.114

Case 3: 0.135

请回忆物理必修2。

完整代码：

/*0.015s*/

#include<cstdio>
#include<cmath>

int main(void)
{
	int test, i;
	int d, u, v;
	scanf("%d", &test);
	for (i = 1; i <= test; i++)
	{
		scanf("%d%d%d", &d, &v, &u);
		printf("Case %d: ", i);
		if (u > v && v)///陷阱：v不能为0，因为题目要求是"two different paths"
			printf("%.3f\n", d * (1.0 / sqrt(u * u - v * v) - 1.0 / u));
		else
			puts("can't determine");
	}
	return 0;
}