Codeforces 110B-Lucky String（技能）

最新推荐文章于 2020-03-10 22:12:19 发布

转载最新推荐文章于 2020-03-10 22:12:19 发布 · 121 阅读

本文提供了一种方法，用于生成长度为n的幸运字符串，其中每个字母的相邻位置之差为4或7。实现了一个循环输出abcd模式的简单算法。

Petya loves lucky numbers. We all know that lucky numbers are the positive integers whose decimal representations contain only the lucky digits 4 and 7. For example, numbers 47, 744, 4 are lucky and 5, 17,467 are not.

Petya recently learned to determine whether a string of lowercase Latin letters is lucky. For each individual letter all its positions in the string are written out in the increasing order. This results in 26 lists of numbers; some of them can be empty. A string is considered lucky if and only if in each list the absolute difference of any two adjacent numbers is a lucky number.

For example, let's consider string "zbcdzefdzc". The lists of positions of equal letters are:

b: 2
c: 3, 10
d: 4, 8
e: 6
f: 7
z: 1, 5, 9
Lists of positions of letters a, g, h, ..., y are empty.

This string is lucky as all differences are lucky numbers. For letters z: 5 - 1 = 4, 9 - 5 = 4, for letters c:10 - 3 = 7, for letters d: 8 - 4 = 4.

Note that if some letter occurs only once in a string, it doesn't influence the string's luckiness after building the lists of positions of equal letters. The string where all the letters are distinct is considered lucky.

Find the lexicographically minimal lucky string whose length equals n.

Input

The single line contains a positive integer n (1 ≤ n ≤ 105) — the length of the sought string.

Output

Print on the single line the lexicographically minimal lucky string whose length equals n.

Sample test(s)

input
5

output
abcda

input
3

output
abc题意： 要求生成字符串：每一个字母的出现的相邻位置之差为4或7.事实上仅仅须要4个字母就可以 abcdancdabcd....循环输出就可以。
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cctype>
#include <cstdlib>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
using namespace std;
const int INF = 0x3f3f3f3f;
#define LL long long
char s[1000000];
int main()
{
	int n;
	while(~scanf("%d",&n))
	{
		for(int i=0;i<n;i++)
		{

			if((i+1)%4!=0)
				s[i]='a'+(i+1)%4-1;
			else
				s[i]='d';
		}
		s[n]='\0';
		puts(s);
	}
	return 0;
}