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.
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.
In a single line print the required string. If there isn't such string, print "-1" (without the quotes).
7 4
ababacd
4 7
-1
解题说明:此题是一道典型的贪心问题,既要保证字符串中随意两个连续字符串不同,也要保证字典序最小,最简单的想法是仅仅用a,b交替。在最后补上其它字符串就可以。
#include#include #include #include #include #include #include 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 n || k == 1) { printf("-1\n"); } else { for (i = 0; i