2037. Richness of binary words

题目连接：

http://acm.timus.ru/problem.aspx?space=1&num=2037

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

4

Sample Output

1 : NO

2 : NO

3 : NO

4 : aaaa

Hint

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

无解输出no

题解：

打表找规律，比较容易发现答案其实就是aababb->aaababb->aaaababb的循环，这样的。

然后输出就好了。

代码

#include <bits/stdc++.h>

using namespace std;

int N;

void solve_special(){

    if( N <= 7 ){

		for(int i = 1 ; i < N ; ++ i) printf("%d : NO\n" , i);

		printf("%d : ",N);

		for(int j = 1 ; j <= N ; ++ j) putchar('a');

		puts("");

	}else if( N == 8 ){

		cout << "1 : NO" << endl;

		cout << "2 : NO" << endl;

		cout << "3 : NO" << endl;

		cout << "4 : NO" << endl;

		cout << "5 : NO" << endl;

		cout << "6 : NO" << endl;

		cout << "7 : aababbaa" << endl;

		cout << "8 : aaaaaaaa" << endl;

	}else if( N == 9 ){

		cout << "1 : NO" << endl;

		cout << "2 : NO" << endl;

		cout << "3 : NO" << endl;

		cout << "4 : NO" << endl;

		cout << "5 : NO" << endl;

		cout << "6 : NO" << endl;

		cout << "7 : NO" << endl;

		cout << "8 : aaababbaa" << endl;

		cout << "9 : aaaaaaaaa" << endl;

	}else if( N == 10 ){

		cout << "1 : NO" << endl;

		cout << "2 : NO" << endl;

		cout << "3 : NO" << endl;

		cout << "4 : NO" << endl;

		cout << "5 : NO" << endl;

		cout << "6 : NO" << endl;

		cout << "7 : NO" << endl;

		cout << "8 : aaababbaaa" << endl;

		cout << "9 : aaaababbaa" << endl;

		cout << "10 : aaaaaaaaaa" << endl;

	}

}

void solve(){

	int target = (N-11)/2 + 2 ;

	for(int i = 1 ; i < 8 ; ++ i){

		printf("%d : NO\n" , i);

	}

	int cur = 2;

	for(int i = 8 ; i < N ; ++ i){

		int len = 0 , flag = 0 , rs = cur;

		printf("%d : " , i);

		while( len < N ){

			if( flag == 0 ){

				putchar('a');

				-- rs;

			}else if( flag == 1 ){

				if( rs == 3 ) putchar('a');

				else putchar('b');

				-- rs;

			}

			if( rs == 0 ){

				flag ^= 1;

				if( flag == 0 ) rs = cur;

				else rs = 4;

			}

			++ len;

		}

		if( cur == target ) cur += 2;

		else ++ cur;

		puts("");

	}

	printf("%d : ",N);

	for(int i = 1 ; i <= N ; ++ i) putchar('a');

	puts("");

}

int main(int argc,char *argv[]){

	//freopen("KO.txt","w",stdout);

	scanf("%d",&N);

	if( N < 10 ) solve_special();

	else if( N == 10 ){

		cout << "1 : NO" << endl;

		cout << "2 : NO" << endl;

		cout << "3 : NO" << endl;

		cout << "4 : NO" << endl;

		cout << "5 : NO" << endl;

		cout << "6 : NO" << endl;

		cout << "7 : NO" << endl;

		cout << "8 : aaababbaaa" << endl;

		cout << "9 : aaaababbaa" << endl;

		cout << "10 : aaaaaaaaaa" << endl;

	}else solve();

	return 0;

}