Kooky Clock

The first line contains one integer $T$ ( $0\le T\le 1000$ ), which denotes the times the three hands have rotated.

The second line contains three integers $l_1,l_2$ and $l_3$ ( $1\le l_1,l_2,l_3 \le 1000$ ), denotes the length of each hand.

The third line contains three integers $t_1,t_2$ and $t_3$ ( $1\le t_1,t_2,t_3 \le 1000$ ), denotes the time of each hand required to rotate in a 360-degree.

Print two numbers to denote the position of the third hand is $(x,y)$ .

Your answer is considered correct if its absolute or relative error does not exceed $10^{-6}$ .

Formally, let your answer be $a$ , and the jury's answer be $b$ . Your answer is accepted if and only if $\frac{|a - b|}{\max{(1,|b|)}} \le 10^{-6}$ .

C++(clang++ 11.0.1) 解法, 执行用时: 4ms, 内存消耗: 468K, 提交时间: 2022-10-08 19:37:34

#include<bits/stdc++.h>
using namespace std;
const double pai=2*3.1415926535;
double T,l1,l2,l3,t1,t2,t3;
int main(){
	cin>>T>>l1>>l2>>l3>>t1>>t2>>t3;
	printf("%.15lf ",l1*sin(pai*T/t1)+l2*sin(pai*T/t2)+l3*sin(pai*T/t3));
	printf("%.15lf",l1*cos(pai*T/t1)+l2*cos(pai*T/t2)+l3*cos(pai*T/t3));
	
}

pypy3 解法, 执行用时: 82ms, 内存消耗: 53788K, 提交时间: 2022-10-16 15:05:46

import math
t = int(input())
l1,l2,l3 = map(int,input().split())
t1,t2,t3 = map(int,input().split())
d1,d2,d3 = map(lambda x:math.pi*2*(t/x),(t1,t2,t3))
x = l1*math.cos(d1)+l2*math.cos(d2)+l3*math.cos(d3)
y = l1*math.sin(d1)+l2*math.sin(d2)+l3*math.sin(d3)
print("%.10f %.10f" % (y,x))

Python3 解法, 执行用时: 45ms, 内存消耗: 4868K, 提交时间: 2022-10-07 11:38:35

import math
a=int(input())
c=list(map(int,input().split()))
b=list(map(int,input().split()))
x=0
y=0
for i in range(3):
    x=x+c[i]*math.sin((a%b[i])/b[i]*(2*math.pi))
    y=y+c[i]*math.cos((a%b[i])/b[i]*(2*math.pi))
print('%.10f' % x +" "+"%.10f" % y)

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