UVA-10995 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 andMonny). 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.

A passed C and M at time t1 and t2, respectively, and met D at time t3. D met Cand M at times t4 and t5, respectively. The question is: at what time time did Cpass M?

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.

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

Output for sample input

12:00:00

11:16:54

13:37:32

10:40:00

题目大意：A、C、M三人去拜访D，他们处在同一条直线上，位置分布为A、C、M、D。A、C、M均以匀速（并不相等）往D的方向前进，D以匀速往A、C、M的方向前进。

t1时刻，A超过C，t2时刻A超过M，t3时刻A遇到D,t4时刻C遇到D，t5时刻M遇到D。求C超过M的时刻。其中，t1、t2、t3、t4、t5是单调递增的。

题目解析：不妨将Vd视为0。t1时刻时，设|AD|=L，则|CD|=L，则可计算出Vc，Va。由Va及时刻数据，可算出Vm和t2时刻的|CM|。进而算出C超过M的时刻。

最终推导出的公式为：tx=t2+(t5-t2)*(t4-t3)*(t2-t1)/((t3-t1)*(t5-t2)-(t4-t1)*(t3-t2))。

代码如下：

# include<iostream>

# include<cstdio>

# include<cstring>

# include<algorithm>

using namespace std;

int h[6],m[6],s[6],t[8];

int main()

{

int i,hx,mx,sx;

char p[10];

while(scanf("%s",p))

{

if(p[0]=='-')

break;

h[1]=(p[0]-'0')*10+p[1]-'0';

m[1]=(p[3]-'0')*10+p[4]-'0';

s[1]=(p[6]-'0')*10+p[7]-'0';

for(i=2;i<=5;++i){

scanf("%d",&h[i]);

getchar();

scanf("%d",&m[i]);

getchar();

scanf("%d",&s[i]);

}

t[0]=0;

for(i=1;i<=5;++i){

t[i]=h[i]*3600+m[i]*60+s[i];

}

double tx=1.0*(t[5]-t[2])*(t[4]-t[3])*(t[2]-t[1])/(double)((t[3]-t[1])*(t[5]-t[2])-(t[4]-t[1])*(t[3]-t[2]));

tx+=t[2];

//cout<<tx<<endl;

int tt=tx+0.5;

sx=tt;

mx=sx/60;

sx%=60;

hx=mx/60;

mx%=60;

printf("%02d:%02d:%02d\n",hx,mx,sx);

}

return 0;

}