NOIP 模拟 $89\; \rm 子集$

题解 \(by\;zj\varphi\)

每次直接来回一条龙选，如果 \(\frac{n}{k}\) 是奇数，那么将连续的三行取出来，凑一下。

具体实现就是对其中两行凑出来公差为 \(1\) 的等差数列，可以先算出来中位数，再两两配对。

记得特判无解，但 \(n=1,k=1\) 时是有解的。

Code

#include<bits/stdc++.h>
#define ri signed
#define pd(i) ++i
#define bq(i) --i
#define func(x) std::function<x>
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    #define debug1(x) std::cerr << #x"=" << x << ' '
    #define debug2(x) std::cerr << #x"=" << x << std::endl
    #define Debug(x) assert(x)
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            bool f=false;x=0;char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    using ll=long long;
    static const int N=1e6+7;
    int st[N][2],T,n,k;
    inline int main() {
        FI=freopen("subset.in","r",stdin);
        FO=freopen("subset.out","w",stdout);
        cin >> T;
        for (ri z(1);z<=T;pd(z)) {
            cin >> n >> k;
            if (n==1) {printf("Yes\n1\n");continue;}
            ll al=1ll*n*(n+1)>>1;
            if (al%k) {printf("No\n");continue;}
            int nm=n/k,cnt=0;
            if (nm&1) {
                if (nm==1) {printf("No\n");continue;}
                printf("Yes\n");
                const int bs=k<<1;
                int hf=n+n-bs+1+(k>>1),lim=n-bs,bg=0;
                for (ri i(k);i>0;i-=2) {
                    st[++bg][0]=lim+i;
                    st[bg][1]=hf-st[bg][0];
                    --hf;
                }
                for (ri i(k-1);i>0;i-=2) {
                    st[++bg][0]=lim+i;
                    st[bg][1]=hf-st[bg][0];
                    --hf;
                }
                for (ri i(1);i<=k;pd(i)) {
                    const int tmp=bs-(i<<1)+1;
                    for (ri j(i);j<=lim-k;j+=bs) printf("%d %d ",j,j+tmp);
                    printf("%d %d %d\n",lim-k+i,st[i][0],st[i][1]);
                }
            } else {
                printf("Yes\n");
                const int bs=k<<1;
                for (ri i(1);i<=k;pd(i)) {
                    const int tmp=bs-(i<<1)+1;
                    for (ri j(i);j<=n;j+=bs) printf("%d %d ",j,j+tmp);
                    printf("\n");
                }
            }
        }
        return 0;
    }
}
int main() {return nanfeng::main();}