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	cff_0102 & aqua_qaq | 团子群 185800038 | 如果我死了说明我 AFO 了

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P8248 简单数列 题解

这道题，很多人都觉得 $500$ 个数，这得试到猴年马月才能过啊，还是直接枚举/搜索吧。

诶，但我就打表过了，而且没有靠程序枚举输出。

题意补充

这个题意即使这么简便，还是有人看不懂。所以我补充一下。

• 数列里面只能有 $1,2,3,4$ 这 $4$ 个数。所以 1 2 3 4 5 6 7 是不允许的。
• 类似 1 2 3 1 2 3，1 2 3 4 1 3 1 2 3 4 1 3，1 1 这些情况都不能出现，而 1 2 3 4 1 2 3 就能出现，即没有任何相邻的的子串相同。

思路

打表。首先构造一组比较可以的数列，至少能拿 $20$ 分。这个时候把整个数列复制几遍，先把这个初始的数列称为一段。每段之间加上几个数字，使得数字总数超过 $500$，并且如果把一段看做 5 的话，这个数列要满足没有任何相邻的的子串相同。

接着，自己写SPJ或者直接交上来查看哪里有误，接着进行微调。微调也要有技巧，我调了半个多小时才 AC。

代码

这里提供 C++ 和 Python 代码。

C++：

#include<bits/stdc++.h>
using namespace std;
int s[500]={1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,1,2,3,4,1,3,1,4,3,4,2,3,4,1,3,1,2,3,4,3,1,2,3,1,4,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,1,2,3,4,1,3,1,4,1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,1,4,1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,4,3,4,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,3,2,1,2,3,4,1,3,1,2,3,4,2,3}//呜呜呜累死了
int main(){
	ios::sync_with_stdio(false);
	int a;cin>>a;//输入
	for(int i=0;i<a;i++){
		cout<<s[i]<<" ";//这个数组满足所有的，因为输入500是可以的，所以更小的数字也行
	}
	return 0;
}

Python：

#这段代码如果自动识别语言会识别成Pascal
a=int(input())
s=[1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,1,2,3,4,1,3,1,4,3,4,2,3,4,1,3,1,2,3,4,3,1,2,3,1,4,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,1,2,3,4,1,3,1,4,1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,1,4,1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,3,2,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,4,3,4,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,1,2,3,4,3,1,2,3,1,3,2,3,4,3,1,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,2,1,2,3,4,1,3,4,2,3,4,1,3,1,2,3,4,2,4,1,2,3,4,1,3,1,2,3,4,3,2,1,2,3,4,1,3,1,2,3,4,2,3#这里真的很难调
for i in range(a):
    print(s[i],end=" ")

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