【codeforces 534A】Exam

【题目链接】:http://codeforces.com/contest/534/problem/A

【题意】

给你n个人,要求任意两个编号相邻的人不能相邻;
让你安排座位方案,使得最多人的可以入座

【题解】

对于前4个特殊一点判断一下就好;
n=4的时候也是4->即2 4 1 3
到5之后就可以用前面n/2个位置是奇数,后面n/2个位置是偶数了
比如
1 3 5 7 2 4 6
就对应了n=7的情况

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%lld",&x)
#define ref(x) scanf("%lf",&x)

typedef pair<int, int> pii;
typedef pair<LL, LL> pll;

const int dx[9] = { 0,1,-1,0,0,-1,-1,1,1 };
const int dy[9] = { 0,0,0,-1,1,-1,1,-1,1 };
const double pi = acos(-1.0);
const int N = 5000 + 100;

int n, num, a[N];

int main()
{
    //freopen("F:\rush.txt", "r", stdin);
    while (cin >> n)
    {
        num = 0;
        if (n == 4)
        {
            puts("4");
            printf("2 4 1 3
");
        }
        else
        {
            for (int i = 1; i <= n; i += 2)
                a[++num] = i;
            for (int i = 2; i <= n; i += 2)
                if (abs(a[num] - i) > 1)
                {
                    a[++num] = i;
                }
                else
                    break;
            printf("%d
", num);
            for (int i = 1; i <= num; i++)
            {
                printf("%d", a[i]);
                if (i == num)
                    puts("");
                else
                    putchar(' ');
            }
        }
    }
    //printf("
%.2lf sec 
", (double)clock() / CLOCKS_PER_SEC);
    return 0;
}