196 条题解

  • 5
    @ 2018-01-15 23:54:32

    运动轨迹

    对于每一个时刻,每个人的运动方向都沿着两人间的连线方向,大小为v,这样,在指向等边三角形中心方向上有一个分速度v’,大小为v在指向中心方向上的正投影大小,v’=v*cos30°。同理每个人在指向中心方向上的位移l’=l*(√3)/3,两者相除得t=(2/3)*(l/v)。由于运动的等时性,总路程为
    s=t*v=(2/3)*l。

    #include <cstdio>
    double l,v;
    int main() {
        scanf("%lf%lf",&l,&v);
        printf("%.1f\n%.1f",l*2/3,l*2/3/v);
        return 0;
    }
    
  • 2
    @ 2014-08-04 14:35:30

    #include<cstdio>
    using namespace std;
    main()
    {
    double l,v,s,t;
    scanf("%lf%lf",&l,&v);
    s=2*l/3;
    t=s/v;
    printf("%.1lf\n%.1lf",s,t);
    }

  • 1
    @ 2015-04-25 13:20:43

    ###pascal code
    program P1072;
    var v,s,l,t,k:real;
    begin
    read(l,v);
    s:=l/(sqrt(3));
    k:=v*(sqrt(3)/2);
    t:=s/k;
    s:=v*t;
    writeln(s:0:1); writeln(t:0:1);
    end.

  • 1
    @ 2010-03-28 10:43:46

    每时每刻每个人指向三角形中心的分速度大小均为V1=V*cos30

    而每个人到中心的距离为X=(1/2L)/cos30,所以T=X/V1=(2/3)*(L/V)

    所以路程S=V*T=(2/3)*L

  • 1
    @ 2009-11-01 22:06:12

    纯物理题...

    program x1072;

    var l,v:extended;

    begin

    readln(l);

    readln(v);

    writeln(l*2/3:0:1);

    writeln(l*2/3/v:0:1);

    end.

  • 1
    @ 2009-10-23 21:40:57

    program Project1(inpot,output);

    var

    l,v,t,s:real;

    begin

    readln(l);

    readln(v);

    t:=(2*l)/(3*v);

    s:=t*v;

    writeln(s:0:1);

    writeln(t:0:1);

    end.

  • 0
    @ 2009-10-08 09:21:49

    program p1072(input,output);

    var l,v,t,s:real;

    begin

    readln(l);

    readln(v);

    t:=(2*l)/(3*v);

    s:=2*l/3;

    writeln(s:0:1);

    writeln(t:0:1);

    end.

  • 0
    @ 2009-10-05 22:21:19

    我连它们是怎么跑的都不懂.............靠三个白*,瞎跑什么啊...........

  • 0
    @ 2009-10-03 18:30:01

    C++的保留小数很猥琐 - -!

    #include

    #include

    using namespace std;

    int main()

    {

    double a,b;

    cin>>a>>b;

    cout

  • 0
    @ 2009-09-27 12:53:37

    hjjqq

    var

    l,v,s,t:real;

    begin

    read(l); read(v);

    s:=((2*l)/3);

    t:=((2*l)/(3*v));

    writeln(s,t);

    end.

  • 0
    @ 2009-09-17 17:00:10

    var s,t,l,v:real;

    begin

    readln(l,v);

    t:=(2*l)/(3*v);

    s:=(2*l)/3;

    writeln(s:0:1);

    writeln(t:0:1);

    end.

  • 0
    @ 2009-08-18 13:06:55

    我物理很烂!!!

    t:=(2*l)/(3*v);

    s:=2*l/3;

    啥意思呃......

  • 0
    @ 2009-08-10 11:18:31

    program p1072(input,output);

    var l,v,t,s:real;

    begin

    readln(l);

    readln(v);

    t:=(2*l)/(3*v);

    s:=2*l/3;

    writeln(s:0:1);

    writeln(t:0:1);

    end.

  • 0
    @ 2009-08-06 10:17:38

    var

    l,v,s,t:real;

    begin

    read(l); read(v);

    s:=((2*l)/3);

    t:=((2*l)/(3*v));

    writeln(s,t);

    end.

  • 0
    @ 2009-08-05 15:21:44

    var

    l,v,s,t:real;

    begin

    readln(l);

    readln(v);

    s:=l/3*2;

    t:=s/(3*v);

    writeln(s:0:1);

    wrietln(t:0:1);

    end.

  • 0
    @ 2009-08-03 23:05:42

    推理过程:

    设经t'(t'很小)三人构成一较小L三角形边长为L1

    则L1=L-v*t'-v*t'*cos60。

    同理 Ln=L-n*3/2*t'*v

    所以 令Ln=0 解得

    t=L*2/3/v;

  • 0
    @ 2009-08-03 16:50:24

    var l,v,s,t:real;

    begin

    readln(l);

    readln(v);

    s:=l/3*2;

    t:=s/(3*v);

    writeln(s:0:1);

    wrietln(t:0:1);

    end.

  • 0
    @ 2009-07-27 13:55:03

    var l,v,s,t:real;

    begin

    readln(l);

    readln(v);

    s:=2*l/3;

    t:=(2*l)/(3*v);

    writeln(s:0:1);

    writeln(t:0:1);

    end.

    黄河之水题中来!!!

  • 0
    @ 2009-07-19 17:24:53

    #include

    int main(){

    double l,v,s,t;

    scanf("%lf",&l);

    scanf("%lf",&v);

    s=l*2.0/3.0;

    t=l/(1.5*v);

    printf("%.1lf\n%.1lf",s,t);

    return 0;

    }

  • 0
    @ 2009-07-17 15:18:09

    program dsa;

    var l,v:real;

    begin

    read(l,v);

    writeln(l/3*2:0:1);

    write(((l/3*2)/v):0:1);

    end.

    农夫山泉

    这代码不是人敲的

信息

ID
1072
难度
1
分类
其他 | 数学 点击显示
标签
递交数
3374
已通过
2248
通过率
67%
被复制
12
上传者