E. Yet Another Ball Problem

 http://codeforces.com/contest/1118/problem/E

题意：在1~k中选两个一对，使得：1.相邻两对相同位置不能相同  (1,3),(1,4) ;2.任何两对不能相同(1,2),(1,2);3.每对两个数不能相同(1,1);

思路：男的1至k，1至k地标，女的2....k,1;    3....k,1,2;  4....k,1,2,3等等

#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<set>
#include<vector>
#include<queue>
#include<cmath>
#include<cstring>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<map>
#include<stack>
#include<string>
using namespace std;

#define sfi(i) scanf("%d",&i)
#define pri(i) printf("%d\n",i)
#define sff(i) scanf("%lf",&i)
#define ll long long
#define mem(x,y) memset(x,y,sizeof(x))
#define INF 0x3f3f3f3f
#define eps 1e-6
#define PI acos(-1)
#define lowbit(x) ((x)&(-x))
#define zero(x) (((x)>0?(x):-(x))<eps)
#define fl() printf("flag\n")
ll gcd(ll a,ll b){while(b^=a^=b^=a%=b);return a;}
const int maxn=2e5+9;
const int mod=1e9+7;

template <class T>
inline void sc(T &ret)
{
    char c;
    ret = 0;
    while ((c = getchar()) < '0' || c > '9');
    while (c >= '0' && c <= '9')
    {
        ret = ret * 10 + (c - '0'), c = getchar();
    }
}

int main()
{
    ll n,k;
    cin>>n>>k;
    if(n>k*(k-1))
    {
        puts("NO");
        return 0;
    }
    puts("YES");
    ll cnt=0;
    for(int i=0;i<n;i++)
    {
        if(i%k==0) cnt++;
        cout<<i%k+1<<" "<<(cnt+i%k)%k+1<<endl;
    }
    return 0;
}