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\n");
else printf("No\n");
return 0;
}
