A. Polo the Penguin and Strings

最新推荐文章于 2022-03-20 20:43:26 发布

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最新推荐文章于 2022-03-20 20:43:26 发布

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本文解决了一个问题：对于给定的字符串长度n和唯一字符数k，找到满足条件的最小字典序字符串，或者确认不存在这样的字符串。通过贪心策略实现，确保每个字符与其相邻字符不同，且最终字符串按字典序排序。

Little penguin Polo adores strings. But most of all he adores strings of length n.

One day he wanted to find a string that meets the following conditions:

The string consists of n lowercase English letters (that is, the string's length equals n), exactly k of these letters are distinct.
No two neighbouring letters of a string coincide; that is, if we represent a string as s = s1s2... sn, then the following inequality holds,si ≠ si + 1(1 ≤ i < n).
Among all strings that meet points 1 and 2, the required string is lexicographically smallest.

Help him find such string or state that such string doesn't exist.

String x = x1x2... xp is lexicographically less than string y = y1y2... yq, if either p < q and x1 = y1, x2 = y2, ... , xp = yp, or there is such number r (r < p, r < q), that x1 = y1, x2 = y2, ... , xr = yr and xr + 1 < yr + 1. The characters of the strings are compared by their ASCII codes.

Input

A single line contains two positive integers n and k (1 ≤ n ≤ 106, 1 ≤ k ≤ 26) — the string's length and the number of distinct letters.

Output

In a single line print the required string. If there isn't such string, print "-1" (without the quotes).

Sample test(s)

input
7 4

output
ababacd

input
4 7

output
-1

解题说明：此题是一道典型的贪心问题，既要保证字符串中任意两个连续字符串不同，也要保证字典序最小，最简单的想法是只用a,b交替，在最后补上其他字符串即可。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include <algorithm>
#include<cstring>
#include<string>
using namespace std;

int main()
{
	int n, k, c, r, i;
	scanf("%d%d", &n, &k);
	if (k == n&&k <= 26)
	{
		c = 'a';
		for (i = 0; i<k; i++)
		{
			printf("%c", c + i);
		}
		printf("\n");
	}
	else if (k>n || k == 1)
	{
		printf("-1\n");
	}
	else
	{
		for (i = 0; i<n - (k - 2); i++)
		{
			if (i % 2 == 0)
			{
				printf("a");
			}
			else
			{
				printf("b");
			}
		}
		r = 2; 
		c = 'a';
		for (i; i < n; i++, r++)
		{
			printf("%c", c + r);
		}
		printf("\n");
	}
	return 0;
}