Educational Journey

The University of Calgary team qualified for the 28th ACM
International Collegiate Programming Contest World Finals
in Prague, Czech Republic. Just by using the initials of team
members they got a very cunning team name: ACM (Alecs,
Celly and Monny). In order to prepare for the contest, they
have decided to travel to Edmonton to learn the tricks of trade
from Dilbert, Alberta-wide famous top-coder.
Due to a horrible miscommunication which is as welcome
as a plague among such teams, A, C and M drive from Calgary
to Edmonton in separate cars. To make things worse,
there was also a miscommunication with D, who being always
so helpful, decides to go to Calgary in order to save the team
a trip to the far, freezing North. All this happens on the same
day and each car travels at a constant (but not necessarily the
same) speed on the famous Alberta #2.
Then A passed C and M at time t1 and t2, respectively,
and met D at time t3. D met C and M at times t4 and t5,
respectively. The question is: at what time did C pass M?

Input
The input is a sequence of lines, each containing times t1, t2,
t3, t4 and t5, separated by white space. All times are distinct and given in increasing order. Each time
is given in the hh : mm : ss format on the 24-hour clock. A line containing ‘-1’ terminates the input.

Output
For each line of input produce one line of output giving the time when C passed M in the same format
as input, rounding the seconds in the standard way.

Sample Input
10:00:00 11:00:00 12:00:00 13:00:00 14:00:00
10:20:00 10:58:00 14:32:00 14:59:00 16:00:00
10:20:00 12:58:00 14:32:00 14:59:00 16:00:00
08:00:00 09:00:00 10:00:00 12:00:00 14:00:00
-1

Sample Output
12:00:00
11:16:54
13:37:32
10:40:00

题意：A,C,M在一条线上，v_a>v_c>v_m.

AC相遇在t1，AM相遇在t2，A到D的时间为t3，C到D为t4，M到d为t5

求CM相遇的时间

tip：注意时间是时刻，不是时间段，不是经过多少时间才到的；

　　数学运算得：t=(v_a-v_d)*(t2-t1)/(v_c-v_d)+0.5+t1;

　　0.5是为了误差，四舍五入；t1是原始要加的

 1 #include<iostream>//wrong
 2 #include<cmath>
 3 #include<cstring>
 4 #include<cstdio>
 5 
 6 using namespace std;
 7 
 8 char ss[100];
 9 int h,m,s;
10 
11 int change(char *ss)
12 {
13     sscanf(ss,"%d:%d:%d",&h,&m,&s);
14     return h*3600+m*60+s;
15 }
16 
17 int main()
18 {
19     while(scanf("%s",ss)==1&&strcmp(ss,"-1")!=0)
20     {
21         double t[10];
22         memset(t,0,sizeof(t));
23         t[1]=change(ss);
24         for(int i=2;i<=5;i++)
25         {
26             scanf("%s",ss);
27             t[i]=change(ss);
28         }
29 
30         double v_c=(t[3]-t[1])/(t[4]-t[1]);
31         double v_m=(t[3]-t[2])/(t[5]-t[2]);
32         int ans=(1.0-v_m)*(t[2]-t[1])/(v_c-v_m)+0.5+t[1];
33 
34         int ss=ans%60;
35         ans/=60;
36         int mm=ans%60;
37         ans/=60;
38         int hh=ans%24;
39         printf("%02d:%02d:%02d
",hh,mm,ss);
40 
41     }
42     return 0;
43 }