codeforces 556B Case of Fake Numbers

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.
Sample test(s)
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.

题目大意：给你一个序列，每轮奇数位上的数字加一，偶数位上的数字减一，问经过最多n轮之后会不会出现0,1,2……n-1的情况，会的话输出Yes，否则输出No。

解题思路：模拟。

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <cstdlib>using namespace std;typedef long long ll;int num[1005];int n;int check() {    for (int i = 0; i < n; i++) {        if (num[i] != i) return 0;    }    return 1;}   int main() {    scanf("%d", &n);    for (int i = 0; i < n; i++) {        scanf("%d", &num[i]);       }    for (int S = 0; S < n; S++) {        if (check()) {            printf("Yes\n");                return 0;        }        num[0] = (num[0] + 1) % n;        num[1] = (num[1] - 1 + n) % n;        for (int i = 2; i < n; i++) {            if (i % 2 == 0) {                num[i] = (num[i] + 1) % n;              } else num[i] = (num[i] - 1 + n) % n;        }    }       printf("No\n");    return 0;}