CodeForces 556 --Case of Fake Numbers

B. Case of Fake Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Andrewid the Android is a galaxy-famous detective. He is now investigating a case of frauds who make fake copies of the famous Stolp's gears, puzzles that are as famous as the Rubik's cube once was.

Its most important components are a button and a line of n similar gears. Each gear has n teeth
containing all numbers from 0 to n - 1 in
the counter-clockwise order. When you push a button, the first gear rotates clockwise, then the second gear rotates counter-clockwise,
the the third gear rotates clockwise an so on.

Besides, each gear has exactly one active tooth. When a gear turns, a new active tooth is the one following after the current active tooth according to the direction of the rotation. For example, if n = 5,
and the active tooth is the one containing number 0, then clockwise rotation makes the tooth with number 1 active,
or the counter-clockwise rotating makes the tooth number 4 active.

Andrewid remembers that the real puzzle has the following property: you can push the button multiple times in such a way that in the end the numbers on the active teeth of the gears from first to last form sequence 0, 1, 2, ..., n - 1.
Write a program that determines whether the given puzzle is real or fake.

Input

The first line contains integer n (1 ≤ n ≤ 1000)
— the number of gears.

The second line contains n digits a1, a2, ..., an (0 ≤ ai ≤ n - 1)
— the sequence of active teeth: the active tooth of the i-th gear contains number ai.

Output

In a single line print "Yes" (without the quotes), if the given Stolp's gears puzzle is real, and "No" (without
the quotes) otherwise.

Examples

input

3
1 0 0

output

Yes

input

5
4 2 1 4 3

output

Yes

input

4
0 2 3 1

output

No

Note

In the first sample test when you push the button for the first time, the sequence of active teeth will be 2 2 1, when you push it for the second
time, you get 0 1 2.

这么长的题面粘过来真是不太好。此题题意真是迷+迷，开始没看懂题，愣是不造这题要干嘛，直到有人抢了一血才尝试去动他。

题意：有n个齿轮，每个齿轮有n个齿。编号逆时针从0到n-1。第一个齿轮顺时针转带动第二个齿轮，于是第二个齿轮逆时针转又带动第三个齿轮，and so on..现在给定你一组序列作为初始序列判断经过若干次转动后是否能回到0 1 2 ...n-1。可以转任意次。

思路：首先看懂题，知道样例怎么来的这题基本就没什么问题了。我们可以逆着来推，假设初始序列为0 1 2...n-1，判断经过若干次转动后是否能到达给定状态，其实这n个齿轮转动一周后还是会回到初始状态，我们只需判断中间是否有某个状态符合条件。数据只有1000，所以两层循环即可。

const int N=1e5+10;
int a[1001],b[N];
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++) scanf("%d",&a[i]);
        int f=0;
        for(int i=0;i<n;i++) b[i]=i;//初始状态。
        for(int i=0;i<=n&&!f;i++)
        {
            for(int i=0;i<n;i++)//编号为偶数的顺时针转
                if(i%2==0) b[i]=(b[i]+1)%n;
            else b[i]=(b[i]-1+n)%n;
            int x=0;
            for(int i=0;i<n&&!x;i++)
                if(a[i]!=b[i]) x=1;
            if(!x) f=1;
        }
        if(f) printf("Yes\n");
        else printf("No\n");
    }
    return 0;
}

大水题，，要敢于尝试和快速理解题意啊。。。