CodeForces-271A-Beautiful Year

A. Beautiful Year

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

It seems like the year of 2013 came only yesterday. Do you know a curious fact? The year of 2013 is the first year after the old 1987 with only distinct digits.

Now you are suggested to solve the following problem: given a year number, find the minimum year number which is strictly larger than the given one and has only distinct digits.

Input

The single line contains integer y (1000 ≤ y ≤ 9000) — the year number.

Output

Print a single integer — the minimum year number that is strictly larger than y and all it's digits are distinct. It is guaranteed that the answer exists.

Sample test(s)

input

1987

output

2013

input

2013

output

2014

最初做法：运用java字符串与数字间的切换，再将数字导入集合，判断基本元素个数多少来判断是否每个数字都不一样

import java.util.*;
public class BeautifulYear {
	public static void main(String[]args){
		Scanner in=new Scanner(System.in);
		String xString=in.next();
		StringBuffer xBuffer=new StringBuffer(xString);
		Set set=new HashSet();
		int temp;
		while(true){
			temp=Integer.parseInt(xBuffer.toString());
			temp++;
			xBuffer=new StringBuffer(String.valueOf(temp));
			for(int i=0;i<4;i++){
				set.add(xBuffer.charAt(i));
			}
			if(set.size()==4)
				break;
			set.clear();
		}
		System.out.println(temp);
	}
}

改进，运用单纯的‘%’以及‘/‘操作符进行取余以及求整操作，得到每个数字的各位的数值，%的数值，/筛选出下个数量级的数据继续重复步骤，代码改用c++操作

#include<iostream>
#include<string>

using namespace std;

int main()
{
	int q, b, s, g, x;
	int y, z;
	cin >> x;
	x++;
	
	while (1)
	{
		q = x / 1000;
		y = x % 1000;
		b = y / 100;
		z = y % 100;
		s = z / 10;
		g = z % 10;
		if (q != b&&q != s&&q != g&&b != s&&b != g&&g != s)
			break;
		x++;
	}
	cout << x;

	system("pause");
	return 0;
}