zoj3798Abs Problem(思维)

题目链接：

huangjing

题目意思：
用1~n中的数字进行组合，得到后面减前面的一项的最大最小值。。。
思路：
多试两个就会发现从n到1的排列得到的是最小值，同理从n-1到1得到的也是最小值，那么用n-这个最小值，那么必定得到的是最大值。。。

题目：

Abs Problem

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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Alice and Bob is playing a game, and this time the game is all about the absolute value!

Alice has N different positive integers, and each number is not greater than N. Bob has a lot of blank paper, and he is responsible for the calculation things. The rule of game is pretty simple. First, Alice chooses a number a₁ from the N integers, and Bob will write it down on the first paper, that's b₁. Then in the following kth rounds, Alice will choose a number a_k (2 ≤ k ≤ N), then Bob will write the number b_k=|a_k-b_k-1| on the kth paper. |x| means the absolute value of x.

Now Alice and Bob want to kown, what is the maximum and minimum value of b_N. And you should tell them how to achieve that!

Input

The input consists of multiple test cases;

For each test case, the first line consists one integer N, the number of integers Alice have. (1 ≤ N ≤ 50000)

Output

For each test case, firstly print one line containing two numbers, the first one is the minimum value, and the second is the maximum value.

Then print one line containing N numbers, the order of integers that Alice should choose to achieve the minimum value. Then print one line containing N numbers, the order of integers that Alice should choose to achieve the maximum value.

Attention: Alice won't choose a integer more than twice.

Sample Input

2

Sample Output

1 11 22 1

代码：

#include<cstdio>#include<cmath>const int maxn=50000+10;int fabs(int x){    if(x<0)  return -x;    else return x;}int main(){    int n;    while(~scanf("%d",&n))    {        int Min=n;        for(int i=n-1;i>=1;i--)            Min=fabs(i-Min);        int Max=n-1;        for(int i=n-2;i>=1;i--)            Max=fabs(i-Max);        Max=fabs(n-Max);        printf("%d %d\n",Min,Max);        for(int i=n;i>1;i--)            printf("%d ",i);        printf("%d\n",1);        for(int i=n-1;i>=1;i--)            printf("%d ",i);        printf("%d\n",n);    }    return 0;}