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.

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.

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

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.

   题意：有n个转盘，在位置上分别为1-n，每一个转盘上在一圈分别写着0 -- n-1，

如今有一个button，每一次按这个button。每一个转盘都会发生一些变化，详细的变化是：

1.转盘在奇数位的。转盘上面的数字+1（也就是顺时针旋转）。

2.转盘在偶数位的，转盘上面的数字-1（也就是逆时针旋转）。

问能不能在摁下n次button后，使得第i个转盘上面的数字就是i-1。

    思路：推断一下第一个转盘为0时。其它的转盘是不是满足第i个转盘的数字为i-1。不满足直接输出No。

点击打开链接

#include<iostream>

#include<algorithm>

#include<stdio.h>

#include<string.h>

#include<stdlib.h>

#include<stack>

using namespace std;

int n;

int a[1005];

int main() {

    while(scanf("%d",&n)!=EOF) {

        for(int i=1; i<=n; i++) {

            scanf("%d",&a[i]);

        }

        while(a[1] != 0) {

            for(int i=1; i<=n; i++) {

                if(i%2 == 1) {

                    a[i] += 1;

                    a[i] = a[i]%n;

                } else {

                    a[i] -= 1;

                    if(a[i]<0){

                        a[i] = n + a[i];

                    }

                }

            }

        }

        int flag = 0;

        for(int i=1;i<=n;i++){

            if(a[i] != i-1)

            {

                flag = 1;

                break;

            }

        }

        if(flag == 1){

            printf("No\n");

        }

        else{

            printf("Yes\n");

        }

    }

    return 0;

}