Ural 2037. Richness of binary words (打表+构造)

最新推荐文章于 2017-07-28 00:44:54 发布

原创最新推荐文章于 2017-07-28 00:44:54 发布 · 409 阅读

0 ·

CC 4.0 BY-SA版权

ACM 同时被 3 个专栏收录

360 篇文章

订阅专栏

好题

185 篇文章

订阅专栏

易错

50 篇文章

订阅专栏

本文介绍了一种构造特定数量不同回文子串的字符串算法，通过使用模式串'abaa'来有效打断回文串形成，并针对不同输入n提供了解决方案。

Description

For each integer i from 1 to n, you must print a string si of length n consisting of letters ‘a’ and ‘b’ only. The string si must contain exactly i distinct palindrome substrings. Two substrings are considered distinct if they are different as strings.

Input

The input contains one integer n (1 ≤ n ≤ 2000).

Output

You must print n lines. If for some i, the answer exists, print it in the form “i : si” where si is one of possible strings. Otherwise, print “i : NO”.

Sample Input

Sample Output

1 : NO
2 : NO
3 : NO
4 : aaaa

题意：让你构造长度为n，字符集为2，且本质不同的回文串恰好i个的字符。

分析：打表发现abaa可以很好的截断回文串，然后分类讨论构造答案。

#include<iostream>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<cstdio>
#include<set>
#include<map>
#include<vector>
#include<cstring>
#include<stack>
#include<cmath>
#include<queue>
#define INF 0x3f3f3f3f
#define eps 1e-9
#define MAXN 10005
using namespace std;
int n;
char s[MAXN],p[MAXN];
int main()
{
	scanf("%d",&n);
	if(n <= 7)
	{
		for(int i = 1;i < n;i++) cout<<i<<" : NO"<<endl;
		for(int i = 0;i < n;i++) s[i] = 'a';
		s[n] = '\0';
		cout<<n<<" : "<<s<<endl;
		return 0;
	}
	if(n == 8)
	{
		for(int i = 1;i < 7;i++) cout<<i<<" : NO"<<endl;
		cout<<7<<" : bbabaabb"<<endl;
		cout<<8<<" : "<<"aaaaaaaa"<<endl;
		return 0;
	}
	for(int i = 1;i < 8;i++) cout<<i<<" : NO"<<endl;
	p[0] = 'b',p[1] = 'b',p[2] = 'a',p[3] = 'b',p[4] = 'a',p[5] = '\0';
	for(int i = 8;i < n;i++)
	{
		int leng = strlen(p);
		p[leng] = 'a';
		p[++leng] = '\0';
		for(int j = 0;j < n;j++) s[j] = p[j % leng];
		s[n] = '\0';
		cout<<i<<" : "<<s<<endl;
	}
	for(int i = 0;i < n;i++) s[i] = 'a';
	s[n] = '\0';
	cout<<n<<" : "<<s<<endl;
}