C. Tram
time limit per test
memory limit per test
input
output
0 to the point s and back, passing 1 meter per t1 seconds in both directions. It means that the tram is always in the state of uniform rectilinear motion, instantly turning around at points x = 0 and x = s.
x1. He should reach the point x2. Igor passes 1 meter per t2
x1 to the point x2, if it is known where the tram is and in what direction it goes at the moment Igor comes to the point x1.
not obligatory that points in which Igor enter and exit the tram are integers. Assume that any boarding and unboarding happens instantly. Igor can move arbitrary along the line (but not faster than 1 meter per t2
Input
s, x1 and x2 (2 ≤ s ≤ 1000, 0 ≤ x1, x2 ≤ s, x1 ≠ x2) — the maximum coordinate of the point to which the tram goes, the point Igor is at, and the point he should come to.
t1 and t2 (1 ≤ t1, t2) — the time in seconds in which the tram passes 1 meter and the time in seconds in which Igor passes 1
p and d (1 ≤ p ≤ s - 1, d is either 1 or
) — the position of the tram in the moment Igor came to the point x1 and the direction of the tram at this moment. If
, the tram goes in the direction from the point s to the point 0. If d = 1, the tram goes in the direction from the point 0 to the point s.
Output
x1 to the point x2.
Examples
input
4 2 43 4 1 1
output
8
input
5 4 01 2 3 1
output
7
Note
2 meters and it takes 8 seconds in total, because he passes 1 meter per 4
x2 and get to the point 1 in 6 seconds (because he has to pass 3meters, but he passes 1 meters per 2 seconds). At that moment the tram will be at the point 1, so Igor can enter the tram and pass 1meter in 1 second. Thus, Igor will reach the point x2 in 7
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这个题第一次看想多了，看了别人的博客才知道这题的思路。
问人从x1处，如何最快到达x2 处。t1,t2分别表示车和人每走一米的时间。
先考虑一种情况：如果车速比人慢，就没有上车的必要了，以下讨论车速快于人的速度
一开始我猜想，人可能走一段后上车，那我们假设我这种猜想成立，那么人最后一定是在终点x2处下车，那么无论人从哪里上车，都是同样的时间下车（人上下车的时间是忽略不计的），所以我们干脆从一开始就不要步行，直接等车。
所以，我们只需要比较，人步行到达终点的时间，和车到达终点的时间，取其小的。
用暴力模拟一下这个过程即可。
/*****************
代码如下：
/**
#include<stdio.h>
int s,x1,x2,t1,t2,p,d;
int main()
{
while(scanf("%d%d%d%d%d%d%d",&s,&x1,&x2,&t1,&t2,&p,&d)!=EOF)
{
int time=(x2-x1)*t2;//步行所用的时间
if(time<0)
time=-time;
int timecar=0;
int pc=p; //pc表示此时此刻车的位置
int f1=0; //f1标记x1是否经过
while(1)
{
if(pc==s||pc==0)//如果车到了两端，那就换方向
d=-d;//转向
if(pc==x1)//经过起点，只有经过起点，才能让人上车
f1=1;
if(pc==x2 && f1==1)//经过终点，人下车
break;
pc+=d; //向当前方向走一米
timecar+=t1;//加上车，这一秒的用时
}
printf("%d\n",time<timecar ? time : timecar);//取时间短的输出
}
return 0;
}