链接
题解
枚举较小的边, 求出最优秀的边长,然后开始构造
基本思想就是沿着对角线方向填数字,我是先把出现个数等于短边长度的那些数字沿着某些对角线填满,然后剩下的直接按顺序填就行了
代码
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#define iinf 0x3f3f3f3f
#define linf (1ll<<60)
#define eps 1e-8
#define maxn 1000010
#define maxe 1000010
#define cl(x) memset(x,0,sizeof(x))
#define rep(_,__) for(_=1;_<=(__);_++)
#define em(x) emplace(x)
#define emb(x) emplace_back(x)
#define emf(x) emplace_front(x)
#define fi first
#define se second
#define de(x) cerr<<#x<<" = "<<x<<endl
using namespace std;
using namespace __gnu_pbds;
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
ll read(ll x=0)
{
ll c, f(1);
for(c=getchar();!isdigit(c);c=getchar())if(c=='-')f=-f;
for(;isdigit(c);c=getchar())x=x*10+c-0x30;
return f*x;
}
struct Array2
{
ll n, m, a[maxn];
void init(ll N, ll M)
{
n=N, m=M;
for(ll i=0;i<n*m;i++)a[i]=0;
}
ll index(ll i, ll j){return i*m+j;}
ll& at(ll i, ll j){return a[index(i,j)];}
ll* operator[](ll i){return a+i*m;}
}ans;
ll n, a[maxn];
map<ll,ll> cnt;
vector<pll> v;
int main()
{
ll i, j, r, c, tot;
n=read();
rep(i,n)a[i]=read(), cnt[a[i]]++;
for(auto pr:cnt)v.emb(pr);
r=c=1;
for(i=1;i*i<=n;i++)
{
tot=0;
for(auto pr:v)tot+=min(pr.second,i);
ll t=tot/i;
if(t>=i and i*t>r*c)r=i, c=t;
}
vector<ll> lis;
ans.init(r,c);
ll p=0;
for(auto pr:v)
{
if(r<=pr.second)
{
for(i=0;i<r;i++)ans[i][(p+i)%c]=pr.first;
p++;
}
else rep(i,pr.second)lis.emb(pr.first);
}
printf("%lld\n%lld %lld\n",r*c,r,c);
for(j=0;j<c;j++)
{
for(i=0;i<r;i++)
{
if(ans[i][(j+i)%c]==0)
{
ans[i][(j+i)%c]=lis.back();
lis.pop_back();
}
}
}
for(i=0;i<r;i++)for(j=0;j<c;j++)printf("%lld%c",ans[i][j],j==c-1?char(10):char(32));
return 0;
}