196 条题解
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5吴哥——传奇 LV 8 @ 2018-01-15 23:54:32
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; }
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22014-08-04 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);
} -
12015-04-25 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. -
12010-03-28 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 -
12009-11-01 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. -
12009-10-23 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. -
02009-10-08 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. -
02009-10-05 22:21:19@
我连它们是怎么跑的都不懂.............靠三个白*,瞎跑什么啊...........
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02009-10-03 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 -
02009-09-27 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.
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02009-09-17 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. -
02009-08-18 13:06:55@
我物理很烂!!!
t:=(2*l)/(3*v);
s:=2*l/3;
啥意思呃...... -
02009-08-10 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. -
02009-08-06 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.
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02009-08-05 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. -
02009-08-03 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; -
02009-08-03 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. -
02009-07-27 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!!!
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02009-07-19 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;
} -
02009-07-17 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