Codeforces 746C Tram(水题)

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我是靠谱客的博主称心铃铛，最近开发中收集的这篇文章主要介绍Codeforces 746C Tram(水题)，觉得挺不错的，现在分享给大家，希望可以做个参考。

概述

The tram in Berland goes along a straight line from the point 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.

Igor is at the point x1. He should reach the point x2. Igor passes 1 meter per t2 seconds.

Your task is to determine the minimum time Igor needs to get from the point 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.

Igor can enter the tram unlimited number of times at any moment when his and the tram's positions coincide. It is 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 seconds). He can also stand at some point for some time.

Input

The first line contains three integers 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.

The second line contains two integers t1 and t2 (1 ≤ t1, t2 ≤ 1000) — the time in seconds in which the tram passes 1 meter and the time in seconds in which Igor passes 1 meter.

The third line contains two integers p and d (1 ≤ p ≤ s - 1, d is either 1 or $\text{[math]}$ ) — the position of the tram in the moment Igor came to the point x1 and the direction of the tram at this moment. If $\text{[math]}$ , 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

Print the minimum time in seconds which Igor needs to get from the point x1 to the point x2.

Examples

input
4 2 4
3 4
1 1

output
8

input
5 4 0
1 2
3 1

output
7

Note

In the first example it is profitable for Igor to go by foot and not to wait the tram. Thus, he has to pass 2 meters and it takes 8 seconds in total, because he passes 1 meter per 4 seconds.

In the second example Igor can, for example, go towards the point 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 1 meter in 1 second. Thus, Igor will reach the point x2 in 7 seconds in total.

题意：在一条直线上，坐标从0到S。人要从X1位置向X2位置移动，前进一个单位要t2时间。当人在X1位置，车子在p位置，前进一个单位要t1秒。方向为d，d==1表示从正方向前进， d == -1表示负方向前进。人可以在与车子相遇时，坐上车子。

题解：亏了，跳过C题去肝D题。貌似这个C更省时间。。。只要判断一下在满足车子和人至少一次同方向时，谁先到X2位置就好了。

代码如下：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
int main()
{
int s,t1,t2,x1,x2,p,d;
while(scanf("%d%d%d",&s,&x1,&x2)!=EOF)
{
scanf("%d%d",&t1,&t2);
scanf("%d%d",&p,&d);
int dix=abs(x2-x1);
int time1=dix*t2;
int time2;
if(d==1)
{
if(p<=x1&&x1<x2)
time2=(x2-p)*t1;
else if(x1<x2)
time2=(s-p+s+x2)*t1;
else if(x2<x1)
time2=(s-p+s-x2)*t1;
}
else if(d==-1)
{
if(p>=x1&&x2<x1)
time2=(p-x2)*t1;
else if(x2<x1)
time2=(p+s+s-x2)*t1;
else if(x1<x2)
time2=(p+x2)*t1;
}
if(time2>time1)
printf("%dn",time1);
else
printf("%dn",time2);
}
return 0;
}