ZOJ3798 Abs Problem

题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=3798

学会了对于序列用next_permutation暴力打表

可以先打表找规律

#include<cstdio>
#include<algorithm>
#define INF 0x3f3f3f3f//
#define N 100
using namespace std;
int seq[N];//保存当前序列
int mi[N], ma[N];//保存最小值最大值相应序列
int calc(int n)//总排序数
{
    int p = 1;
    for (int i = 1; i <= n; ++i)
        p *= i;
    return p;
}
int main()
{
    int n;
    while (~scanf("%d", &n))
    {
        for (int i = 1; i <= n; ++i)
            seq[i] = i;
        int k = calc(n), MIN = INF, MAX = -INF;
        for (int i = 0; i < k; ++i)
        {
            int b;
            next_permutation(seq + 1, seq + n + 1);
            for (int i = 1; i <= n; ++i)
            {
                if (i == 1) b = seq[i];
                else b = abs(b - seq[i]);
            }
            if (b <= MIN)
            {
                MIN = b;
                for (int i = 1; i <= n; ++i)
                    mi[i] = seq[i];
            }
            if (b>=MAX)
            {
                MAX = b;
                for (int i = 1; i <= n; ++i)
                    ma[i] = seq[i];
            }
        }
        printf("%d %d\n", MIN, MAX);
        for (int j = 1; j <= n; ++j)
            printf("%d ", mi[j]);
        puts("");
         for (int j = 1; j <= n; ++j)
             printf("%d ", ma[j]);
         puts("");
     }
     return 0;
}

规律：逆序时最小，将逆序最大放到末位时最大

n%4==3或0时，最小值为0，其余都为1

举个例子：序列7 6 5 4 3 2 1

此时该序列为最小值，会发现：从7开始每4个数的绝对值为0，每两个数为1，所以若n%4==3时，剩下的最后3个为3 2 1，易得最小值为0，n%4==0、1、2时同理。

最大值MAX(n)=n-MIN(n-1)

AC代码：

 1 #include<cstdio>
 2 #define N 50005
 3 int getMin(int n)
 4 {
 5     if (n % 4 == 3 || n % 4 == 0)
 6         return 0;
 7     return 1;
 8 }
 9 int main()
10 {
11     int n;
12     while (~scanf("%d", &n))
13     {
14         printf("%d %d\n", getMin(n), n - getMin(n - 1));
15         for (int i = n; i > 0; --i)
16         {
17             if (i > 1)printf("%d ", i);
18             else printf("%d\n", i);
19         }
20         for (int i = n-1; i > 0; --i)
21             printf("%d ", i);
22         printf("%d\n", n);
23     }
24     return 0;
25 }