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本文探讨了计数绵羊问题的算法实现,通过具体的示例讲解了如何帮助虚拟绵羊Bleatrix Trotter更快入睡的方法。介绍了输入输出格式、限制条件及样例解释,并提供了一段Java代码实现。

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2016年没有参赛,在师兄的介绍下,试了一下简单的一题,需要注意的是读写数据的形式还有具体代码。

2016资格赛 A题

Problem

Bleatrix Trotter the sheep has devised a strategy that helps her fall asleep faster. First, she picks a number N. Then she starts naming N, 2 × N, 3 × N, and so on. Whenever she names a number, she thinks about all of the digits in that number. She keeps track of which digits (0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) she has seen at least once so far as part of any number she has named. Once she has seen each of the ten digits at least once, she will fall asleep.

Bleatrix must start with N and must always name (i + 1) × N directly after i × N. For example, suppose that Bleatrix picks N = 1692. She would count as follows:

  • N = 1692. Now she has seen the digits 1, 2, 6, and 9.
  • 2N = 3384. Now she has seen the digits 1, 2, 3, 4, 6, 8, and 9.
  • 3N = 5076. Now she has seen all ten digits, and falls asleep.

What is the last number that she will name before falling asleep? If she will count forever, print INSOMNIA instead.

Input

The first line of the input gives the number of test cases, TT test cases follow. Each consists of one line with a single integer N, the number Bleatrix has chosen.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the last number that Bleatrix will name before falling asleep, according to the rules described in the statement.

Limits

1 ≤ T ≤ 100.

Small dataset

0 ≤ N ≤ 200.

Large dataset

0 ≤ N ≤ 106.

Sample


Input 
 

Output 
 
5
0
1
2
11
1692

Case #1: INSOMNIA
Case #2: 10
Case #3: 90
Case #4: 110
Case #5: 5076


In Case #1, since 2 × 0 = 0, 3 × 0 = 0, and so on, Bleatrix will never see any digit other than 0, and so she will count forever and never fall asleep. Poor sheep!

In Case #2, Bleatrix will name 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. The 0 will be the last digit needed, and so she will fall asleep after 10.

In Case #3, Bleatrix will name 2, 4, 6... and so on. She will not see the digit 9 in any number until 90, at which point she will fall asleep. By that point, she will have already seen the digits 0, 1, 2, 3, 4, 5, 6, 7, and 8, which will have appeared for the first time in the numbers 10, 10, 2, 30, 4, 50, 6, 70, and 8, respectively.

In Case #4, Bleatrix will name 11, 22, 33, 44, 55, 66, 77, 88, 99, 110 and then fall asleep.

Case #5 is the one described in the problem statement. Note that it would only show up in the Large dataset, and not in the Small dataset.

 

 

特殊情况就是输入为0的时候,永远也不会结束,其他情况就不会出现。本以为可能会有某个数导致答案非常大,发生int越界,但是也没有。文件名好像无所谓。

import java.io.BufferedReader;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.Scanner;

public class CountingSheep {
    public static void main(String[] args) throws FileNotFoundException {
        FileInputStream fis = new FileInputStream("A-large.in");
        PrintStream out = new PrintStream(new FileOutputStream("A-large.out"));
        System.setIn(fis);
        System.setOut(out);
        Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        while (in.hasNext()) {
            int t = in.nextInt();
            for (int i = 1;i < t+1;i++) {
                int n = in.nextInt();
                if (n == 0) {
                    System.out.println("Case #"+i+": "+"INSOMNIA");
                } else {
                    int[] ds = new int[10];
                    int count = 0;
                    for (int j = 1;j != 0;j++) {
                        int temp = n*j;
                        while (temp != 0) {
                            int digit = temp%10;
                            if (ds[digit] == 0) {
                                ds[digit] = 1;
                                count++;
                            }
                            temp = temp/10;
                            if (count == 10) {
                                System.out.println("Case #"+i+": "+n*j);    
                                j = -1;
                                break;
                            }
                        }
                    }            
                }
                    
            }
        }
    }
}

 

转载于:https://www.cnblogs.com/fisherinbox/p/5470594.html

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