湖南大学ACM程序设计新生杯大赛（同步赛）E

题目描述

A mod-dot product between two arrays with length n produce a new array with length n. If array A is a1,a2,...,an and array B is b1,b2,...bn, then A mod-dot B produce an array C c1,c2,...,cn such that c1 =  a1*b1%n, c2 = a2*b2%n,...,ci = ai*bi%n,..., cn = an*bn%n.

i.e. A = [2,3,4] and B = [5,2,2] then A mod-dot B = [1,0,2].

A permutation of n is an array with length n and every number from 0 to n-1 appears in the array by exactly one time.

i.e. A = [2,0,1] is a permutation of 3, and B = [3,4,1,2,0] is a permutation of 5, but C = [1,2,2,3] is NOT a permutation of 4.

Now comes the problem: Are there two permutaion of n such that their mod-dot product is also a permutation of n?

输入描述:

The only line with the number n (1 <= n <= 1000)

输出描述:

If there are such two permutation of n that their mod-dot product is also a permutation of n, print "Yes" (without the quote). Otherwise print "No" (without the quote).

示例1

输入

2

输出

Yes

说明

A = [0,1] and B = [0,1]. Then A mod-dot B = [0,1]

示例2

输入

997

输出

No

备注:

1 <= n <= 1000

题解

规律。

写了个暴力，算了$1$到$11$的答案，发现只有$1$和$2$有解，所以猜了一发。。

#include <cstdio>
#include <algorithm>
using namespace std;
 
 
int main() {
  int n;
  scanf("%d", &n);
  if(n == 1 || n == 2) printf("Yes
");
  else printf("No
");
  return 0;
}