有n个人围成一圈,顺序排号。从第一个人开始报数(从1到3报数),凡报到3的人退出圈子,问最后留下的是原来第几号的那位。

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答案:
import java.util.Scanner;
public class Ex37 {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        boolean[] arr = new boolean[n];
        for(int i=0; i<arr.length; i++) {
            arr[i] = true;//下标为TRUE时说明还在圈里
        }
        int leftCount = n;
        int countNum = 0;
        int index = 0;
        while(leftCount > 1) {
            if(arr[index] == true) {//当在圈里时
                countNum ++; //报数递加
                if(countNum == 3) {//报道3时
                    countNum =0;//从零开始继续报数
                    arr[index] = false;//此人退出圈子
                    leftCount --;//剩余人数减一
                }
            }
            index ++;//每报一次数,下标加一
            if(index == n) {//是循环数数,当下标大于n时,说明已经数了一圈,
                index = 0;//将下标设为零重新开始。
            }
        }
        for(int i=0; i<n; i++) {
            if(arr[i] == true) {
                System.out.println(i);
            }
        }
    }
}