【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; }
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