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  • codeforces

    C. View Angle
    time limit per test
    2 seconds
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Flatland has recently introduced a new type of an eye check for the driver's licence. The check goes like that: there is a plane with mannequins standing on it. You should tell the value of the minimum angle with the vertex at the origin of coordinates and with all mannequins standing inside or on the boarder of this angle.

    As you spend lots of time "glued to the screen", your vision is impaired. So you have to write a program that will pass the check for you.

    Input

    The first line contains a single integer n (1 ≤ n ≤ 105) — the number of mannequins.

    Next n lines contain two space-separated integers each: xi, yi (|xi|, |yi| ≤ 1000) — the coordinates of the i-th mannequin. It is guaranteed that the origin of the coordinates has no mannequin. It is guaranteed that no two mannequins are located in the same point on the plane.

    Output

    Print a single real number — the value of the sought angle in degrees. The answer will be considered valid if the relative or absolute error doesn't exceed 10 - 6.

    Examples
    input
    2
    2 0
    0 2
    
    output
    90.0000000000
    
    input
    3
    2 0
    0 2
    -2 2
    
    output
    135.0000000000
    
    input
    4
    2 0
    0 2
    -2 0
    0 -2
    
    output
    270.0000000000
    
    input
    2
    2 1
    1 2
    
    output
    36.8698976458
    
    Note

    Solution for the first sample test is shown below:

    Solution for the second sample test is shown below:

    Solution for the third sample test is shown below:

    Solution for the fourth sample test is shown below:


    思路:

    要想使最小的角度包含所有的点,必定会有两个相邻点所夹得角没有被覆盖,而且只有一个。现在,把这个角找出来就可以了。

    #include <bits/stdc++.h>
    using namespace std;
    const double PI = acos(-1.0);
    const int MAXN = 1e5 + 10;
    double ang[MAXN];
    bool cmp(double x, double y) {return x > y;}
    int main () {
        int n; double a, b;
        while (scanf("%d", &n) != EOF) {
            for (int i = 0; i < n; i++) {
                scanf("%lf%lf", &a, &b);
                ang[i] = atan(b/a)/PI*180;
                if (ang[i] < 0) ang[i] += 180; 
                if (b < 0 || a < 0 && b == 0) ang[i] += 180;
            }
            sort(ang, ang + n, cmp);
            double ans = 360 - ang[0] + ang[n - 1]; //用最后一个减去第一个,从而形成一个环。 
            for (int i = 0; i < n - 1; i++) {
                ans = max(ans, ang[i] - ang[i + 1]);
            }
            printf("%.8lf
    ", 360 - ans);
        }
        return 0;
    }



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  • 原文地址:https://www.cnblogs.com/cniwoq/p/6770775.html
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