Marathon（精度 + 分析模拟）

 

B. Marathon

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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 d meters 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

 

      题意：

      给出 a（1 ~ 105） ，d（1 ~ 105）代表一个边长为a，停水处是每 d 距离就有一处。后给出 n（1 ~ 105），要求输出 n 次 x 和 y 的坐标，代表这 n 次停下来的位置是什么，以逆时针绕行。

 

      思路：

      精度 + 分析。总距离 % （4 * a ），但是模运算只对整型，故对一个不是整型的数要 k - floor（k / （4 * a）） * （4 * a）这样子求得余数。求得余数后，再分析它在哪一段上面，分析每段坐标输出结果就行。除此之外比较大小的时候也要考虑到精度问题，因为题目要求后四位小数相同才算相等。还有对于浮点型的数运算，有乘法的话若直接随循环累加会导致精度损失，根据累加运算次数增加，误差也就增大了，那么最好还是直接用乘法。

 

      AC：

#include <stdio.h>
#include <math.h>
#define eps 1e-6

int main() {
    double a,d;
    int n;
    scanf("%lf%lf",&a,&d);
    scanf("%d",&n);
    for(int i = 1;i <= n;++i) {
        double ans;
        ans = (i * d) - floor((i * d) / (4 * a)) * (4 * a);
        if (ans - a <= eps)            printf("%f %f\n",ans,0.0);
        else if (ans - 2 * a <= eps)   printf("%f %f\n",a,ans - a);
        else if (ans - 3 * a <= eps)   printf("%f %f\n",3 * a - ans,a);
        else                           printf("%f %f\n",0.0,4 * a - ans);
    }
    return 0;
}