Mathematical Circus-数论-分类讨论

codeforces Mathematical Circus-div2-B

题意：

给定n，k。是否能把（1--n）的数分成符合条件的（a，b）对。

条件：（a+k）*b%4==0

解：

因为：原式=(a+k)*b≡0(mod 4)

ab+b*k≡0(mod 4)

若k>=4,bk=b*(4t+k-4t)=4t*b+b*(k-4t)

上式有：ab+b*k≡ab+b*(k-4t)+b*4t≡ab+b*(k-4t)≡0(mod 4)

也就是说，我们只需要研究k=0,1,2,3的情况

k=0时，即：ab%4==0,不妨令a is odd，b is even，式子显然不成立

k=1 or 3：一种可能是：a+k is even，b is even，这是总可以成立的

k=2：1:a is even ,b is odd then a+k is even but not always is div by 4

2： when b==4t,t>=1,is right . other  b is odd,a+k is div by 4,a just is div by 2,but not div by 4.

#include<iostream>
#include<cstring>
#define maxn (int)(2e5+1e1)
using namespace std;
int main()
{
    int t; cin >> t;
    while (t--)
    {
        int n, k; cin >> n >> k;
        while (k > 4)k %= 4;
        if (k == 0)
        {
            cout << "NO\n";
            continue;
        }
        cout << "YES\n";
        if (k == 1 || k == 3)
        {
            for (int i = 1; i <= n - 1; i += 2)
            {
                cout << i << " " << i + 1 << "\n";
            }
        }
        else if (k == 2)
        {
            for (int i = 1; i <= n - 1; i += 2)
            {
                if ((i + 1) % 4 == 0)cout << i << " " << i + 1 << "\n";
                else cout << i + 1 << " " << i << "\n";
            }
        }
    }
    return 0;
}