【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\n");

        }

        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\n", num);

            for (int i = 1; i <= num; i++)

            {

                printf("%d", a[i]);

                if (i == num)

                    puts("");

                else

                    putchar(' ');

            }

        }

    }

    //printf("\n%.2lf sec \n", (double)clock() / CLOCKS_PER_SEC);

    return 0;

}