Ural 2037. Richness of binary words 打表找规律 构造

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
" , 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
" , 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;
}