World Racing 2(世界赛车2）怎么操作？

Posted 2023-04-08

进入游戏后不知道方向键，谁可以告诉我怎么操作！（俺不懂英文）
尽量解说详细！！

游戏中：
键盘上下左右为前进、后退、向左、向右转
空格为手刹
shift为氧化亚氮加速器
a提档、z降档（自动挡不用）
c切换驾驶视觉
b向后看
v截图
h喇叭
r还原（车子陷入困境时用）
q左转灯、e右转灯、w双跳灯
此外按Esc可以详细调整其他设置

游戏外：
上下左右为调整选项
insert为改变车辆外观（q和w调整角度）

游戏结束时：
insert回放、delete重来
进游戏先定账号，进career可以和电脑比赛
freeride自己比赛，其他不常用

ps：比赛时那些要求
position是名次要求
drift是漂移点数
avrage speed是平均速度
high speed是到达最高速度次数
lead time是领先第二名时间
计速表右上角的%是绿色的话是伤害限制
达不到这些要求比赛还是失败 参考技术A 那个老爷车那关怎么也过不了

I - Long Distance Racing（第二季水）

Description

Bessie is training for her next race by running on a path that includes hills so that she will be prepared for any terrain. She has planned a straight path and wants to run as far as she can -- but she must be back to the farm within M seconds (1 ≤ M ≤ 10,000,000).

The entire path she has chosen is T units (1 ≤ T ≤ 100,000) in length and consists of equal-length portions that are uphill, flat, or downhill. The input data describes path segment i with a single character Si that is u, f, or d, indicating respectively uphill, flat, or downhill.

Bessie takes U seconds (1 ≤ U ≤ 100) to run one unit of uphill path, F (1 ≤ F ≤ 100) seconds for a unit of flat path, and D (1 ≤ D ≤ 100) seconds for a unit of downhill path. Note that, when returning home, uphill paths become downhill paths and downhill paths become uphill paths.

Find the farthest distance Bessie can get from the farm and still make it back in time.

Input

* Line 1: Five space-separated integers: M, T,U, F, and D * Lines 2..T+1: Line i+1 describes path segment i with a single letter: S_i

Output

* Line 1: A single integer that is the farthest distance (number of units) that Bessie can get from the farm and make it back in time.

Sample Input

13 5 3 2 1
u
f
u
d
f

Sample Output

3

太水，不解释

#include<iostream>
using namespace std;
char s[100000];
int main()
{
    int m,t,u,f,d,k=0,a;
    cin>>m>>t>>u>>f>>d;
    for(int i=0;i<t;i++)cin>>s[i];
    int i;
    for(i=0;i<t;i++){
        if(s[i]==‘u‘||s[i]==‘d‘)k+=u+d;
        else k+=2*f;
        if(k>m)break;
        a=k;
    }
    cout<<i<<endl;
    //system("pause");
    return 0;
}