CodeForces 804E. The same permutation

Seyyed and MoJaK are friends of Sajjad. Sajjad likes a permutation. Seyyed wants to change the permutation in a way that Sajjad won't like it. Seyyed thinks more swaps yield more probability to do that, so he makes MoJaK to perform a swap between every pair of positions (i, j), where i < j, exactly once. MoJaK doesn't like to upset Sajjad.

Given the permutation, determine whether it is possible to swap all pairs of positions so that the permutation stays the same. If it is possible find how to do that.

Input

The first line contains single integer n (1 ≤ n ≤ 1000) — the size of the permutation.

As the permutation is not important, you can consider ai = i, where the permutation is a1, a2, ..., an.

Output

If it is not possible to swap all pairs of positions so that the permutation stays the same, print "NO",

Otherwise print "YES", then print  lines: the i-th of these lines should contain two integers a and b (a < b) — the positions where the i-th swap is performed.

Examples input 3 output NO input 1 output YES

题意：有一个排列p，把所有点对(i, j)(i < j)提取出来，用每个点对恰好一次，swap(p_i, p_j)，使最后p和原来相同

题解：

首先我们证明交换次数必须是偶数

考虑一次交换可以拆成奇数次相邻的数的交换，一个相邻的交换逆序对数奇偶性会改变

因此C(n, 2) % 2 = 0，故n % 4 = 0或1

这提示我们按照4分组，然后构造

具体见代码

#include <bits/stdc++.h> #define xx first #define yy second #define mp make_pair #define pb push_back #define fill( x, y ) memset( x, y, sizeof x ) #define copy( x, y ) memcpy( x, y, sizeof x ) using namespace std; typedef long long LL; typedef pair < int, int > pa; inline int read() { int sc = 0, f = 1; char ch = getchar(); while( ch < '0' || ch > '9' ) { if( ch == '-' ) f = -1; ch = getchar(); } while( ch >= '0' && ch <= '9' ) sc = sc * 10 + ch - '0', ch = getchar(); return sc * f; } int main() { #ifdef wxh010910 freopen( "data.in", "r", stdin ); #endif int n = read(); if( n % 4 > 1 ) return puts( "NO" ), 0; puts( "YES" ); for( int i = n % 4 ; i < n ; i += 4 ) { for( int j = 0 ; j < i ; j++ ) printf( "%d %d\n", j + 1, i + 1 ); printf( "%d %d\n", i + 1, i + 2 ); for( int j = i - 1 ; ~j ; j-- ) printf( "%d %d\n", j + 1, i + 2 ); for( int j = 0 ; j < i ; j++ ) printf( "%d %d\n", j + 1, i + 3 ); printf( "%d %d\n", i + 3, i + 4 ); for( int j = i - 1 ; ~j ; j-- ) printf( "%d %d\n", j + 1, i + 4 ); printf( "%d %d\n", i + 1, i + 4 ); printf( "%d %d\n", i + 2, i + 3 ); printf( "%d %d\n", i + 1, i + 3 ); printf( "%d %d\n", i + 2, i + 4 ); } }