Codeforces Round #166 (Div. 2)暴力A. Beautiful Year

Posted 2022-02-10 SSL_ZZL

Beautiful Year

• 题目
• 题目大意
• 解题思路
• Code

Codeforces Round #166 (Div. 2) A. Beautiful Year
Codeforces Round #166 (Div. 2) B. Prime Matrix 题解
Codeforces Round #166 (Div. 2) D. Good Substrings 题解

---

题目

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.

Examples
input

1987

output

2013

input

2013

output

2014

---

题目大意

明显数的年份：没有出现重复的数字
找出一个严格大于给出年份的最小明显数年份
比如2022不是明显数年份，2031是2022后的第一个明显数年份

---

解题思路

开心开心开心开心，学长选的一套比赛好简单（哈哈虽然还是不会做）

因为年份4位数可以先提前预处理出答案
从后（9000）往前（1000），把当前最小的明显年份赋值给当前年份，如果当前年份是明显年份，那就更新最小明显年份
9000 * 4真的很小🙂

其实直接从 y+1 开始找明显年份也是可以的，但是我要写高级的程序(bushi，预处理在多组数据时比较吃香

---

Code

#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

int n, now, ans[11000], v[100];

int check(int i) 
	memset(v, 0, sizeof(v));
	while(i) 
		v[i % 10] ++;
		if(v[i % 10] > 1) return 0;
		i /= 10;
	
	return 1;


int main() 
	for(int i = 10000; i >= 1000; i --)   //比赛时天真的以为9000的答案是9123，然后被罚时了，暴躁开成10000
		ans[i] = now;
		if(check(i)) now = i;
	
	scanf("%d", &n);
	printf("%d", ans[n]);