3-palindrome CodeForces - 805B (思维)

最新推荐文章于 2021-05-26 22:03:33 发布

转载最新推荐文章于 2021-05-26 22:03:33 发布 · 87 阅读

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原文链接：http://www.cnblogs.com/qieqiemin/p/10246825.html

本文探讨了如何构造一个由字符'a'、'b'和'c'组成的字符串，该字符串长度为N，且所有长度为3的子串都不是回文串，同时最小化字符'c'的使用。通过分析，发现仅使用'a'和'b'即可满足条件，通过交替重复模式避免产生回文串。

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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

Output

aa

Input

Output

bba

Note

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

题意：给你一个整数N，让你输出一个字符串，只包括abc三个字符串，并且它的所有长度为3的子串不能使回文串，并且c字符要求尽可能的少。

思路：既然要求c字符出现的尽可能的少，那么设想下是否可以不用c字符来构造出这个字符串。

通过思考可以发现，仅用两个字符ab，可以通过连续两个相同的字符来构造出整个字符串

即aabb，aabbaa，aabba，aab，aabbaabb，这样的。并且一定符合条件。

我们只需要用代码实现出来就好了。

我的AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <vector>
#define rep(i,x,n) for(int i=x;i<n;i++)
#define repd(i,x,n) for(int i=x;i<=n;i++)
#define pii pair<int,int>
#define pll pair<long long ,long long>
#define gbtb ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define MS0(X) memset((X), 0, sizeof((X)))
#define MSC0(X) memset((X), '\0', sizeof((X)))
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define gg(x) getInt(&x)
using namespace std;
typedef long long ll;
inline void getInt(int* p);
const int maxn=1000010;
const int inf=0x3f3f3f3f;
/*** TEMPLATE CODE * * STARTS HERE ***/

int main()
{
    ll l,r;
    cin>>l>>r;
    if(l==r)
    {
        cout<<l<<endl;
    }else
    {
        if(r-l>=2)
        {
            cout<<2<<endl;
        }else
        {
            cout<<l<<endl;
        }
    }
    return 0;
}

inline void getInt(int* p) {
    char ch;
    do {
        ch = getchar();
    } while (ch == ' ' || ch == '\n');
    if (ch == '-') {
        *p = -(getchar() - '0');
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 - ch + '0';
        }
    }
    else {
        *p = ch - '0';
        while ((ch = getchar()) >= '0' && ch <= '9') {
            *p = *p * 10 + ch - '0';
        }
    }
}

转载于:https://www.cnblogs.com/qieqiemin/p/10246825.html