行舟问题
输入样例
48.6
6:32:24.2 9:37:18.5
12:46:15.3 17:4:19.8
输出样例
13.53
2.24
#inclue<stdio.h>
int main()
{
double d,t1,t2,t3,t4;
//
//
}
输入样例
48.6
6:32:24.2 9:37:18.5
12:46:15.3 17:4:19.8
输出样例
13.53
2.24
#inclue<stdio.h>
int main()
{
double d,t1,t2,t3,t4;
//
//
}
#include<stdio.h>
int main() {
double d;
int h[4], m[4];
double s[4];
double t1, t2;
double v1, v2;
scanf("%lf", &d);
for (int i = 0; i < 4; ++i) {
scanf("%d:%d:%lf", &h[i], &m[i], &s[i]);
}
t1 = (h[1] * 3600 + m[1] * 60 + s[1] - h[0] * 3600 - m[0] * 60 - s[0]) / 3600;
t2 = (h[3] * 3600 + m[3] * 60 + s[3] - h[2] * 3600 - m[2] * 60 - s[2]) / 3600;
if (t1 < 0)
t1 += 24;
if (t2 < 0)
t2 += 24;
v1 = (d / t1 + d / t2) / 2;
v2 = (d / t1 - d / t2) / 2;
printf("%.2lf\n%.2lf", v1, v2);
}