CF - 805B. 3-palindrome - 贪心+构造

1.题目描述：

B. 3-palindrome

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

In the beginning of the new year Keivan decided to reverse his name. He doesn't like palindromes, so he changed Naviek to Navick.

He is too selfish, so for a given n he wants to obtain a string of n characters,
each of which is either 'a', 'b' or 'c',
with no palindromes of length 3 appearing in the string as a substring. For
example, the strings "abc" and "abca" suit him, while
the string "aba" doesn't. He also want the number of letters 'c'
in his string to be as little as possible.

Input

The first line contains single integer n (1 ≤ n ≤ 2·105) —
the length of the string.

Output

Print the string that satisfies all the constraints.

If there are multiple answers, print any of them.

Examples

input

2

output

aa

input

3

output

bba

Note

A palindrome is a sequence of characters which reads the same backward and forward.

2.题意概述：

要求你用'a','b','c'三种字符构造一个长度为n个字符串，且任意连续三个字符不是回文串，且c出现次数最少。

3.解题思路：

考虑先全部构造长度为n的全'a'串，从左边开始贪心地维护前i个字符串不是回文串：如果ch[i - 2] == ch[i]，那么把ch[i]改成'b'就好了。这里还有个迷雾就是‘c’出现次数最少，这里容易发现，这样构造‘c’字符一定不会出现。

4.AC代码： 

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 200100
#define lson root << 1
#define rson root << 1 | 1
#define lent (t[root].r - t[root].l + 1)
#define lenl (t[lson].r - t[lson].l + 1)
#define lenr (t[rson].r - t[rson].l + 1)
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e exp(1.0)
using namespace std;
const int mod = 1e9 + 7;
typedef long long ll;
typedef unsigned long long ull;
char ans[maxn];
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
freopen("out.txt", "w", stdout);
long _begin_time = clock();
#endif
int n;
while (~scanf("%d", &n))
{
fill(ans, ans + n, 'a');
for (int i = 2; i < n; i++)
if (ans[i] == ans[i - 2])
ans[i] = 'b';
puts(ans);
}
#ifndef ONLINE_JUDGE
long _end_time = clock();
printf("time = %ld ms.", _end_time - _begin_time);
#endif
return 0;
}