Morris' Blog 我在摸索我存在的意義、生命的意義、內涵及價值。 沒有真正神手般的厲害，套用模式是大部分的情況。 沒那麼厲害的我存在的意義為何，襯托出神的存在？ 有時候不能怪別人不給我機會，我已經開始省思， 為什麼我沒有給別人機會的原因，而是全要靠別人。 有時候我認為我與眾不同，那是因為我做不到很多人能辦到的事情。 我聽不到、玩不起來、說得不夠明瞭、理解得不夠多， 我缺少的語文能力，就是一切一切無能的判定。 為此，我不曉得能做些對別人有幫助的事情。 我知道我自己是個不服輸的性格，但現在我不得不認輸。 我沒有能力，因此我必須認輸。 我到底來這世界作什麼？處處充滿不確定性， 而我總是無法出類拔萃，想當個一般人似乎也回不了頭了。 越強大的挑戰需要越強力的支持，而我的支持是什麼？ 最近嘗試與人談話，但都是失敗的結局，沒人能接受， 不需要言語，不需要溝通，一切所想都表達不出來。

49愛的鼓勵 6訂閱站台

Return of the Aztecs

Problem D:	How Far?
Time Limit: 4 seconds Memory Limit: 32 MB

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As you probably know, the earth moves round the sun, and the moon moves round the earth. Both the earth and the moon follow elliptical paths. But for this problem, we will consider their paths to be circular. So the earth moves round the sun in a circular path with the sun in the center, and likewise the moon moves round the earth in a circular path with the earth in the center. This same kind of planetary system can be observed elsewhere in the galaxy. So, for a general case, let there are n such bodies b1, b2, b3, ... bn, where b1 moves round the sun, which is stationary, from a distance of r1. Body b2 moves round b1 from a distance of r2, and so on. Body bi completes a cycle in nonzero-time ti. Given the r�s and the t�s, you have to find out the distances d�s of the bodies from the sun at a given time T. At T = 0, all the bodies lie in their farthest positions from the sun.

Input

Input consists of multiple test cases. Each case starts with n and T in a line. The following n lines each contains first ri and then ti for i = 1, 2, ... n. Input is terminated by EOF. All the inputs are positive integers. There won�t be more than 50 bodies in a single solar system.

Output

For each case, there should a new line. Print all the d�s in that line separated by spaces. Each d should have 4-digits after the decimal point.

Sample Input

3 5
20 5
30 5
40 5

Sample Output

20.0000 50.0000 90.0000

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[ The image is from http://www.earthsky.com ]

Problem setter: Sadrul Habib Chowdhury (Adil)
Solution author: Monirul Hasan (Tomal)

Man has his will, - but woman has her way.
-- Oliver Wendell Holmes

題目是講很多星球, 誰繞太陽, 誰繞誰, 誰繞誰 ... 分別有公轉週期與公轉半徑。
將太陽座標化(0, 0) 然後對於每個點進行公轉周期的計算。

#include <stdio.h>
#include <math.h>

int main() {
    int n, i;
    double T;
    const double pi = acos(-1);
    while(scanf("%d %lf", &n, &T) == 2) {
        double theta = 0, r, t;
        double x = 0, y = 0;
        for(i = 0; i < n; i++) {
            scanf("%lf %lf", &r, &t);
            theta = 2*pi*T/t;
            x += r*sin(theta);
            y += r*cos(theta);
            if(i)   putchar(' ');
            printf("%.4lf", hypot(x, y));
        }
        puts("");
    }
    return 0;
}