[Codeforces 482A] Diverse Permutation

[题目链接]

https://codeforces.com/contest/482/problem/A

[算法]

首先构造一个(k + 1)个数的序列 ， 满足它们的差为1-k

对于i > k + 1,令Ai = i

时间复杂度 : O(N)

[代码]

#include<bits/stdc++.h>
using namespace std;

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void read(T &x)
{
    T f = ; x = ;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (x << ) + (x << ) + c - '';
    x *= f;
}
int main()
{

        int n , k;
        read(n); read(k);
        printf("");
        int value = k , now = ;
        for (int i = ; i <= k; i++)
        {
                if (i & ) now += value;
                else now -= value;
                printf(" %d",now);
                value--;
        }
        for (int i = k + ; i <= n; i++) printf(" %d",i);
        printf("\n");

        return ;

}