Monday, May 30, 2016

2016 campus recruitment of mushroom street programming problem resolution technology


http://codepub.cn/2015/09/23/2016-campus-recruitment-of-mushroom-street-programming-problem-resolution-technology/

解题思路:既然通过添加一个字母可以变为回文串,那么通过删除与添加的字母相对位置的字符,应该亦为回文串。
  • ‘abcb’在末尾添加’a’ —> ‘abcba’为回文串
    ‘abcb’删除与想要添加的字符’a’对应位置的字符 —> ‘bcb’亦为回文串
  • ‘aabbaab’在头部添加’b’ —> ‘baabbaab’为回文串
    ‘aabbaab’删除与想要添加的字符’b’对应位置的字符 —> ‘aabbaa’亦为回文串
    final String Y = "YES";
    final String N = "NO";

    public String isPalindrome(String input) {
        if (input == null || "".equals(input)) {
            return Y;
        }
        int length = input.length();
        //题目说明不超过10个字符,那么超过的话,直接返回NO
        if (length > 10) {
            return N;
        }
        StringBuilder sb = new StringBuilder(input);
        for (int i = 0; i < length; i++) {
            sb.deleteCharAt(i);
            String temp = sb.toString();
            if (sb.reverse().toString().equals(temp)) {
                return Y;
            } else {
                sb = new StringBuilder(input);
                continue;
            }
        }
        return N;
    }

  1. 小蘑的时间假设为[a,b],小菇的时间假设是[c+t,d+t],小菇起床的时间是t∈[l,r]
  2. 那么当"a < b < (c+t) < (d+t)"或者"(c+t) < (d+t) < a < b"的情况时,小蘑和小菇无法聊天,由题目条件已知"a < b""c < d",那么推出"(c+t) < (d+t)"
  3. 所以仅仅当"b < (c+t)"或者"(d+t) < a"时无法聊天,其余情况都是可以聊天的
    private static void isLegal(int num, boolean flag) {
//        flag为true标识判断[1,50],flag为false标识判断[0,1000]
        if (flag) {
            if (!(1 <= num && num <= 50)) {
//                System.out.println("数据非法");
                System.exit(0);
            }
        } else {
            if (!(0 <= num && num <= 1000)) {
//                System.out.println("数据非法");
                System.exit(0);
            }
        }
    }


    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        while (cin.hasNextInt()) {
            int p = 0, q = 0, l = 0, r = 0;
            p = cin.nextInt();
            isLegal(p, true);
            q = cin.nextInt();
            isLegal(q, true);
            l = cin.nextInt();
            isLegal(l, false);
            r = cin.nextInt();
            isLegal(r, false);
            int[] time_A_B = new int[p * 2];//标识小蘑的时间
            int[] time_C_D = new int[q * 2];//标识小菇的时间
            for (int i = 0; i < time_A_B.length; i++) {
//            接收p行的数据,每一行数据是一个时间对
                int temp = cin.nextInt();
                isLegal(temp, false);
                time_A_B[i] = temp;
            }

            for (int i = 0; i < time_C_D.length; i++) {
                int temp = cin.nextInt();
                isLegal(temp, false);
                time_C_D[i] = temp;
            }
            int count = 0;//标识小菇能有多少个合适的起床时间
            begin:
            for (int t = l; t <= r; t++) {
                for (int i = 0; i < time_A_B.length; i += 2) {
                    for (int j = 0; j < time_C_D.length; j += 2) {
                        if (!(time_C_D[j] + t > time_A_B[i + 1] || time_C_D[j + 1] + t < time_A_B[i])) {
                            count++;
                            continue begin;
                        }
                    }
                }
            }
            System.out.println(count);
        }
    }

  1. length = sqrt((x1-x2)^2+(y1-y2)^2)先计算两个圆心点之间的距离
  2. 参考上图,以(2 0 0 0 4)作为输入数据进行说明,当两个圆心点之间的距离lenght<(2*r+1)的时候,我们是不能够沿着两个圆心之间的连线进行移动的,而由两点之间直线最短,可知,沿两圆心的连线进行移动是最短的距离,换句话说,旋转一次移动的最长距离就是2*r,而在旋转之前需要先移动一步,所以阈值设为2*r+1
  3. 那么当lenght<(2*r+1)的时候,我们该如何进行旋转呢?正确的做法是以(x1,y1)为圆心r为半径作圆,与以(x,y)为圆心的圆的交叉点就是支点,固定此点旋转即可
  4. 根据以上的分析,再以(2 0 0 0 5)作为输入数据进行说明,当length>=(2*r+1),那么length中有几个(2*r+1),我们就需要走几步,如果相除不为整数,那么加1即可
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        while (cin.hasNextInt()) {
            int r = cin.nextInt();
            if (r < 1 || r > 100000) {
                System.exit(0);
            }
            int x = cin.nextInt();
            int y = cin.nextInt();
            int x1 = cin.nextInt();
            int y1 = cin.nextInt();
            if (x < -100000 || x > 100000) {
                System.exit(0);
            }
            if (y < -100000 || y > 100000) {
                System.exit(0);
            }
            if (x1 < -100000 || x1 > 100000) {
                System.exit(0);
            }
            if (y1 < -100000 || y1 > 100000) {
                System.exit(0);
            }
            double length = Math.sqrt(Math.pow(x - x1, 2) + Math.pow(y - y1, 2));
            int count;
            //向上取整之后强转为int型即可
            count = (int) Math.ceil(length / (2 * r + 1));
            System.out.println(count);
        }
    }


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