【66.47%】【codeforces 556B】Case of Fake Numbers

time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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.

【题目链接】:http://codeforces.com/contest/556/problem/B

【题解】

当英语的阅读题能吓死不少人。。。
读懂题意后其实很简单了；
第一个齿轮它的活跃齿肯定是0;看看第一个齿轮要转几次;
则其他齿轮也只能转相应的次数；
如果转了相应的次数某个齿轮i不能变成i-1;则NO;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)
#define pri(x) printf("%d",x)
#define prl(x) printf("%I64d",x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 1e3+10;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n;
int a[MAXN];

void cl(int tag,int &t)
{
    if (tag==1)
    {
        t++;
        if (t>n-1) t = 0;
    }
    else
    {
        t--;
        if (t<0) t = n-1;
    }
}

int main()
{
    //freopen("F:\rush.txt","r",stdin);
    rei(n);
    rep1(i,1,n)
        rei(a[i]);
    int now = 1;
    int quan = 0;
    while (a[1]!=0)
        {
             cl(now,a[1]);
             quan++;
        }
    rep1(i,2,n)
    {
        now*=-1;
        int cnt = 0;
        while (a[i]!=i-1)
        {
            cl(now,a[i]);
            cnt++;
            if (cnt>quan)
            {
                puts("No");
                return 0;
            }
        }
        if (cnt < quan)
        {
            puts("No");
            return 0;
        }
    }
    puts("Yes");
    return 0;
}