B. Case of Fake Numbers( Codeforces Round #310 (Div. 2) 简单题)
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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.
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.
31 0 0
Yes
54 2 1 4 3
Yes
40 2 3 1
No
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,每个转盘上在一圈分别写着0 -- n-1,
现在有一个按钮,每一次按这个按钮,每个转盘都会发生一些变化,具体的变化是:
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次按钮后,使得第i个转盘上面的数字就是i-1。
思路:判断一下第一个转盘为0时,其他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;}
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