湖南大学ACM程序设计新生杯大赛（同步赛）E-Permutation【打表+规律】

时间限制：C/C++ 1秒，其他语言2秒

空间限制：C/C++ 32768K，其他语言65536K

64bit IO Format: %lld

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

分析：当时暴力打了个10个的表。然后莽了一发。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define ll long long int
using namespace std;
const int maxn = 1e6 + 10;
const ll mod = 1e8 + 7;
int a[maxn];
int main()
{
int n;
scanf("%d", &n);
if (n == 1 || n == 2)
printf("Yes\n");
else printf("No\n");
return 0;
}