uva 10773 - Back to Intermediate Math(数论)

题目链接:uva 10773 - Back to Intermediate Math

题目大意:有一天河,宽d,水流速度v,船速u,问说垂直过河和最快过河的时间差,如果不能过河输出“can‘t determine”。

解题思路:将u的速度分解成水平方向和竖直方向的两个速度,使水平方向速度恰好为v,船即可垂直过河,速度为竖直方向速度。

#include <cstdio>
#include <cstring>
#include <cmath>

const double eps = 1e-6;

int main () {
	int cas;
	scanf("%d", &cas);
	for (int i = 1; i <= cas; i++) {
		double d, u, v;
		scanf("%lf%lf%lf", &d, &v, &u);

		printf("Case %d: ", i);
		if (u-v < eps || u < eps || v < eps)
			printf("can‘t determine\n");
		else {
			double x = sqrt(u*u - v*v);
			printf("%.3lf\n", d/x - d/u);
		}
	}
	return 0;
}

uva 10773 - Back to Intermediate Math(数论),码迷,mamicode.com

时间: 2024-10-25 04:18:42

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