Permutation p is an ordered set of integers p1, p2, ..., pn, consisting of n distinct positive integers not larger than n. We'll denote as n the length of permutation p1, p2, ..., pn.
Your task is to find such permutation p of length n, that the group of numbers |p1 - p2|, |p2 - p3|, ..., |pn - 1 - pn| has exactly k distinct elements.
The single line of the input contains two space-separated positive integers n, k (1 ≤ k < n ≤ 105).
Print n integers forming the permutation. If there are multiple answers, print any of them.
3 2
1 3 2
3 1
1 2 3
5 2
1 3 2 4 5
By |x| we denote the absolute value of number x.
题意:
给出 N 和 K,代表有一个 1 ~ N 的数列,问如何排序,能使两者之间差值的绝对值相差数的种数有 K 个。任意输出一个满足的序列即可。
思路:
构造数列。把数列写出来会发现,一个数列最多两两之间相差值也就只是 n - 1 个,这个数列可以通过 第一项 +(n-1) = 第二项,第二项 -(n-2) = 第三项,……这样子构成下去。
既然知道这样的话,就很容易构造出数列了,按题意要出现多少个 K,那么就出现这个数列的前 K 项,剩下的递增或者递减下去就行了。
AC:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std; int num[100005]; int main() { int n, k; scanf("%d%d", &n, &k); int temp = 1, from = n - 1; num[1] = 1; for (int i = 2; i <= n; ++i) { int res = temp * from; num[i] = num[i - 1] + res; --from; temp *= -1; } for (int i = 1; i <= k; ++i) { printf("%d ", num[i]); } int ans = num[k]; if (k % 2) { for (int i = k + 1; i <= n; ++i) printf("%d ", ++ans); } else { for (int i = k + 1; i <= n; ++i) printf("%d ", --ans); } printf("\n"); return 0; }
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