Codeforces Round #386 (Div. 2) C. Tram 数学、讨论

2609次C. Tramtime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputThe。

2609次

C. Tram

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

The tram in Berland goes along a straight line from the point0to the pointsand
back, passing1meter pert1seconds
in both directions. It means that the tram is always in the state of uniform rectilinear motion, instantly turning around at pointsx = 0andx = s.

Igor is at the pointx1.
He should reach the pointx2.
Igor passes1meter pert2seconds.

Your task is to determine the minimum time Igor needs to get from the pointx1to
the pointx2,
if it is known where the tram is and in what direction it goes at the moment Igor comes to the pointx1.

Igor can enter the tram unlimited number of times at any moment when his and the tram's positions coincide. It isnot obligatorythat 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 than1meter pert2seconds).
He can also stand at some point for some time.

Input

The first line contains three integerss,x1andx2(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 integerst1andt2(1 ≤ t1, t2 ≤ 1000)—
the time in seconds in which the tram passes1meter and the time in seconds in which Igor passes1meter.

The third line contains two integerspandd(1 ≤ p ≤ s - 1,dis
either1or)—
the position of the tram in the moment Igor came to the pointx1and
the direction of the tram at this moment. If,
the tram goes in the direction from the pointsto the point0.
Ifd = 1, the tram goes in the direction from the point0to
the points.

Output

Print the minimum time in seconds which Igor needs to get from the pointx1to
the pointx2.

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 pass2meters and it takes8seconds
in total, because he passes1meter per4seconds.

In the second example Igor can, for example, go towards the pointx2and
get to the point1in6seconds
(because he has to pass3meters, but he passes1meters
per2seconds). At that moment the tram will be at the point1,
so Igor can enter the tram and pass1meter in1second.
Thus, Igor will reach the pointx2in7seconds
in total.

Source

Codeforces Round #386 (Div. 2)

My Solution

题意:从x1 动身到 x2,走路速度是t2 s/ 单位长度 ,坐车是t1s/ 单位长度,车子在0~s间不断来回,此时车在p位置,且方向是d,(正向 d == 1,反向 d == ⑴) 问从x1到x2的最短时间

数学、讨论

另外题可以分成2大类,到达的时候是走路的则只斟酌全走路,如果到达的时候是坐车则只斟酌坐车,

即x1到达x2所花的时间为此时p位置到最后以x1指向x2的方向经过x2时的总时间。

具体分成以下6类讨论

当 x1 < x2 时
ans = (x2 - x1) * t2;
if(d == 1){
if(p <= x1){ //p 在 x1、 x2 左侧边
ans = min(ans, (x2 - p) * t1);
}
else{
ans = min(ans, (s + x2 + s - p) * t1);
}
}
else{
ans = min(ans, (p + x2) * t1);
}

当 x1 > x2 时
ans = (x1 - x2) * t2;
if(d == 1){
ans = min(ans, (s - p + s - x2) * t1); //
}
else{
if(p >= x1){ //p 在 x1、 x2 右侧
ans = min(ans, (p - x2) * t1);
}
else{
ans = min(ans, (p + s + s - x2) * t1);
}
}

#include <iostream>
#include <cstdio>

using namespace std;
typedef long long LL;
const int maxn = 1e6 + 8;


int main()
{
    #ifdef LOCAL
    freopen("c.txt", "r", stdin);
    //freopen("c.out", "w", stdout);
    int T = 4;
    while(T--){
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    LL s, x1, x2, t1, t2, p, d, ans = 9e18;
    cin >> s >> x1 >> x2;
    cin >> t1 >> t2;
    cin >> p >> d;
    if(x1 < x2){
        ans = (x2 - x1) * t2;
        if(d == 1){
            if(p <= x1){
                ans = min(ans, (x2 - p) * t1);
            }
            else{
                ans = min(ans, (s + x2 + s - p) * t1);
            }
        }
        else{
            ans = min(ans, (p + x2) * t1);
        }
    }
    else{
        ans = (x1 - x2) * t2;
        if(d == 1){
            ans = min(ans, (s - p + s - x2) * t1); //(©Ð£ß©Ð)
        }
        else{
            if(p >= x1){
                ans = min(ans, (p - x2) * t1);
            }
            else{
                ans = min(ans, (p + s + s - x2) * t1);
            }
        }
    }

    cout << ans << endl;

    #ifdef LOCAL
    cout << endl;
    }
    #endif ** LOCAL
    return 0;
}

Thank you!

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