cf1276 C. Beautiful Rectangle

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题解

枚举较小的边， $O( n \sqrt n)$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;
}