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CodeForces Round #503(Div.2)

August 15, 2018 • Read: 293 • CodeForces

A. New Building for SIS

题解

签到题,无非就是在同一座塔内和不在同一座塔内分情况考虑

代码
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int h = cin.nextInt();
        int a = cin.nextInt();
        int b = cin.nextInt();
        int k = cin.nextInt();
        for(int i = 0;i < k;i++) {
            int ta = cin.nextInt();
            int fa = cin.nextInt();
            int tb = cin.nextInt();
            int fb = cin.nextInt();
            if(ta == tb)//在一座塔内
                System.out.println(Math.abs(fa - fb));
            else {//不在同一座塔内
                long tmp = Math.abs(ta - tb);
                if(fa <= a && fb <= a) {
                    long tmp1 = Math.abs(fa - a);
                    long tmp2 = Math.abs(fb - a);
                    System.out.println(tmp + tmp1 + tmp2);
                }
                else if(fa >= b && fb >= b) {
                    long tmp1 = Math.abs(fa - b);
                    long tmp2 = Math.abs(fb - b);
                    System.out.println(tmp + tmp1 + tmp2);
                }
                else {
                    long tmp1 = Math.abs(fa - fb);
                    System.out.println(tmp + tmp1);
                }
            }
        }
    }
}

B. Badge

题解

题意是有n个学生干了一些不可描述的事,然后老师要抓人,当抓到一个学生的时候就给这个学生标记1,然后这个学生会说是另一个学生让他干的,然后老师就会找另一个学生,直到老师找到已经被标记过了两次的学生为止,输出这个学生的编号

也是水题,模拟一下就行了,从1号开始遍历,找到标记两次的,然后清空map,再从2号开始遍历,找到标记两次的,然后清空map,直到到n为止

代码
import java.util.*;

public class Main {

    public static void main(String[] args) {
        Scanner cin = new Scanner(System.in);
        int n = cin.nextInt();
        int[] stu = new int[n + 1];
        for(int i = 1;i <= n;i++)
            stu[i] = cin.nextInt();
        HashMap<Integer,Integer> map = new HashMap<>();
        for(int i = 1;i <= n;i++) {
            map.clear();
            int curStu = i;
            while(true) {
                if(!map.containsKey(curStu))
                    map.put(curStu,1);
                else {
                    int times = map.get(curStu);
                    map.put(curStu,++times);
                    if(times == 2) {
                        System.out.print(curStu + " ");
                        break;
                    }
                }
                curStu = stu[curStu];
            }
        }
    }
}

C. Elections

最后编辑于: October 7, 2018
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