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作者：stxy-ferryman
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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.
Example
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.
题意：有n个齿轮，依次相接，齿轮示数为ai，按下某个不可名状的东西以后，奇数齿轮顺时针旋转（示数+1），偶数齿轮逆时针旋转（示数-1），如果能转出0，1，2….n-1则输出yes否则输出no
题解：转一遍，检验一遍，对则输出yes，如果转完n次仍不能成功，那就没有什么价值继续了，输出no
当然，有o(n)算法http://paste.ubuntu.com/11788578/
代码：
var
b:boolean;
n,i,j,k:longint;
a:array[..] of longint;
function jy(n:longint):boolean;
begin
for i:= to n do
if a[i]<>i- then exit(false);
exit(true);
end;
begin
readln(n);
for i:= to n do
read(a[i]);
b:=true;
while (not jy(n)) do
begin
for i:= to n do
begin
if i mod = then inc(a[i])
else dec(a[i]);
if a[i]< then a[i]:=n-;
if a[i]>n- then a[i]:=;
end;
inc(k);
if k>n then begin b:=false; break; end;
end;
if b then writeln('Yes')
else writeln('No');
end.
