Codeforces Round #237 (Div. 2) B题模拟题

最新推荐文章于 2022-11-19 18:52:08 发布

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最新推荐文章于 2022-11-19 18:52:08 发布

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原文链接：http://www.cnblogs.com/ccccnzb/p/3915030.html

本文介绍了一个马拉松比赛中运动员在特定体育场内跑步路径的计算方法。通过输入体育场边长和每段距离，算法可以计算出运动员在每个补给点的位置坐标。

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链接：http://codeforces.com/contest/404/problem/B

Valera takes part in the Berland Marathon. The marathon race starts at the stadium that can be represented on the plane as a square whose lower left corner is located at point with coordinates (0, 0) and the length of the side equals a meters. The sides of the square are parallel to coordinate axes.

As the length of the marathon race is very long, Valera needs to have extra drink during the race. The coach gives Valera a bottle of drink each d meters of the path. We know that Valera starts at the point with coordinates (0, 0) and runs counter-clockwise. That is, when Valera covers a meters, he reaches the point with coordinates (a, 0). We also know that the length of the marathon race equalsnd + 0.5 meters.

Help Valera's coach determine where he should be located to help Valera. Specifically, determine the coordinates of Valera's positions when he covers d, 2·d, ..., n·d meters.

Input

The first line contains two space-separated real numbers a and d (1 ≤ a, d ≤ 105), given with precision till 4 decimal digits after the decimal point. Number a denotes the length of the square's side that describes the stadium. Number d shows that after each dmeters Valera gets an extra drink.

The second line contains integer n (1 ≤ n ≤ 105) showing that Valera needs an extra drink n times.

Output

Print n lines, each line should contain two real numbers xi and yi, separated by a space. Numbers xi and yi in the i-th line mean that Valera is at point with coordinates (xi, yi) after he covers i·d meters. Your solution will be considered correct if the absolute or relative error doesn't exceed 10 - 4.

Note, that this problem have huge amount of output data. Please, do not use cout stream for output in this problem.

Sample test(s)

input

2 5
2

output

1.0000000000 2.0000000000
2.0000000000 0.0000000000

input

4.147 2.8819
6

output

2.8819000000 0.0000000000
4.1470000000 1.6168000000
3.7953000000 4.1470000000
0.9134000000 4.1470000000
0.0000000000 2.1785000000
0.7034000000 0.0000000000

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记录下，没事的时候拿来看看！！！

 1 #include <cstdio>
 2 #include <cmath>
 3 
 4 using namespace std;
 5 
 6 double a, d, n;
 7 
 8 void init() {
 9 
10     scanf("%lf%lf%lf", &a, &d, &n);
11 }
12 
13 void solve() {
14 
15     double sum = 0;
16 
17     for (int i = 1; i <= n; i++) {
18 
19         sum += d;
20 
21         long long m = floor(sum/a);
22         double r = sum-m*a;
23         m %= 4;
24 
25         long long tmp = floor(sum/(4*a));
26         sum -= tmp*4*a;
27 
28 
29         switch (m) {
30 
31         case 0 :  printf("%.10lf %.10lf\n", r, 0.0);
32             break;
33         case 1 :  printf("%.10lf %.10lf\n", a, r);
34             break;
35         case 2 :  printf("%.10lf %.10lf\n", a-r, a);
36             break;
37         case 3 :  printf("%.10lf %.10lf\n", 0.0, a-r);
38             break;
39         }
40     }
41 }
42 
43 int main()
44 {
45 
46     init();
47     solve();
48 }