Codeforces Round #177 (Div. 1) / 288A Polo the Penguin and Strings(贪心)-优快云博客

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... s_n, then the following inequality holds,s_i ≠ s_i + 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... x_p is lexicographically less than string y = y₁y₂... y_q, if either p < q and x₁ = y₁, x₂ = y₂, ... , x_p = y_p, or there is such number r (r < p, r < q), that x₁ = y₁, x₂ = y₂, ... , x_r = y_r and x_r + 1 < y_r + 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

大水题。

完整代码：

/*30ms,0KB*/

#include<cstdio>

int main()
{
	int n, k;
	scanf("%d%d", &n, &k);
	if (n < k)
		printf("-1");
	else if (n == k)
		for (int i = 97; i < 97 + k; ++i)
			putchar(i);
	else if (k == 1)
		printf("-1");
	else
	{
		n -= k - 2;
		if (n & 1)
		{
			n = (n - 1) >> 1;
			for (int i = 0; i < n; ++i)
			{
				putchar('a');
				putchar('b');
			}
			putchar('a');
		}
		else
		{
			n >>= 1;
			for (int i = 0; i < n; ++i)
			{
				putchar('a');
				putchar('b');
			}
		}
		for (int i = 99; i < 97 + k; ++i)
			putchar(i);
	}
	return 0;
}