[做题记录] 构造题选做

1. CF743C - Vladik and fractions (*1500)

目标：给定 $n$，构造 $x,y,z$ 满足 $x\ne y,x\ne z,y\ne z$ 且 ${\displaystyle \frac{2}{n}}={\displaystyle \frac{1}{x}}+{\displaystyle \frac{1}{y}}+{\displaystyle \frac{1}{z}}$。

Hint：${\displaystyle \frac{1}{n}}+{\displaystyle \frac{1}{n+1}}+{\displaystyle \frac{1}{n(n+1)}}={\displaystyle \frac{n+1+n+1}{n(n+1)}}={\displaystyle \frac{2(n+1)}{n(n+1)}}={\displaystyle \frac{2}{n}}$。

若 $n=1$ 时会出现 $x=1,y=2,z=2$ 的情况，无解。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
signed main () {
	int n;
	scanf ("%d", &n);
	if (n == 1) printf ("-1\n");
	else printf ("%d %d %d\n", n, n + 1, n * (n + 1));
	return 0;
}

2. CF1758D - Range = √Sum (*1800)

目标：给定 $n$，求一个长度为 $n$ 的数组 $[a_1,a_2,\dots ,a_n]$，满足：$1\le a_i\le {10}^9,1\le i\le n$ 且 ${\displaystyle \underset{i=1}{\overset{n}{max}}{a}_i-\underset{i=1}{\overset{n}{min}}{a}_i=\sqrt{\sum _{i=1}^n{a}_i}}$。

考虑奇偶分类。

当 $nmod2=0$ 时，发现 ${\displaystyle [\frac{n}{2},\frac{n}{2}+1,\dots ,n-1,n+1,\dots ,\frac{3n}{2}]}$ 满足条件。

极差：${\displaystyle \frac{3n-n}{2}}=n$。

和：

$${\displaystyle \frac{\frac{3n}{2}(\frac{3n}{2}+1)}{2}-\frac{\frac{n}{2}(\frac{n}{2}-1)}{2}-n}$$

$$\frac{\frac{9n^2}{4}+\frac{3n}{2}}{2}-\frac{\frac{n^2}{4}-\frac{n}{2}}{2}-n$$

$$\frac{9n^2}{8}+\frac{3n}{4}-\frac{n^2}{8}+\frac{n}{4}-n$$

$$Sum=n^2$$

所以满足条件。

当 $nmod2=1$ 时，先构造一个以 $n$ 为中心的公差为 $1$ 的等差数列（${\displaystyle [\frac{n+5}{2},\frac{n+7}{2},\dots ,\frac{3n+3}{2}]}$），极差为 $n-1$，和为 $n^2$。

将所有数加 $2$，可以让和变为 $n^2+2n$，极差为 $n-1$。

发现极差需要增加 $2$，那就最小值减 $1$，最大值加 $1$ 吧，现在极差为 $n+1$，和为 $n^2+2n$，现在只需要将和加 $1$ 了，那就将次大值加 $1$，保证还会小于最大值，所以不会影响极差，和为 $n^2+2n+1=(n+1{)}^2$。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
int _, n, a[300005];
signed main () {
	scanf ("%d", &_);
	while (_ --) {
		scanf ("%d", &n);
		if (n % 2 == 0) {
			for (int i = n / 2;i <= 3 * n / 2; ++ i) {
				if (i != n) printf ("%d ", i);
			}
			printf ("\n");
		}
		else {
			for (int i = 1;i <= n; ++ i) a[i] = i;
			int mid = n / 2 + 1;
			int delta = n - mid;
			for (int i = 1;i <= n; ++ i) a[i] += delta + 2;
			a[1] --, a[n] ++;
			a[n - 1] ++;
			for (int i = 1;i <= n; ++ i) printf ("%d ", a[i]);
			printf ("\n");
		}
	}
	return 0;
}

3. CF1734E - Rectangular Congruence (*2100)

目标：给定质数 $n$，求一个矩阵满足：

• 对于所有的 $1\le i,j\le n$，$0\le a_{i,j}<n$。

• 对于所有的 $1\le r_1<r_2\le n,1\le c_1<c_2\le n$，有 $(a_{r_1,c_1}+a_{r_2,c_2})modn\ne (a_{r_1,c_2}+a_{r_2,c_1})modn$。

• 对于所有的 $1\le i\le n$，$a_{i,i}=b_i$。

对第二个条件进行变形，有：$(a_{r_1,c_1}-a_{r_1,c_2})modn=(a_{r_2,c_1}-a_{r_2,c_2})modn$。

所以对于一个合法的矩阵的同一行加上同一个数依然合法。

让第 $i$ 行的差值为 $i-1$ 的倍数，可以构造出：

$\left[\begin{array}{cccc}0& 0& \dots & 0\\ 1& 2& \dots & n-1\\ ⋮& ⋮& \ddots & ⋮\\ 0& (n-1)& \dots & (n-1)\times (n-1)modn\end{array}\right]$

剩下的操作就是调整每一行使得它满足第三个条件。

第一个条件就是 $modn$。

#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/hash_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef pair < int, int > PII;
typedef int itn;
mt19937 RND_MAKER (chrono :: steady_clock :: now ().time_since_epoch ().count ());
inline ll randomly (const ll l, const ll r) {return (RND_MAKER () ^ (1ull << 63)) % (r - l + 1) + l;}
//#define int long long
const double pi = acos (-1);
//__gnu_pbds :: tree < Key, Mapped, Cmp_Fn = std :: less < Key >, Tag = rb_tree_tag, Node_Upadte = null_tree_node_update, Allocator = std :: allocator < char > > ;
//__gnu_pbds :: tree < PPS, __gnu_pbds :: null_type, less < PPS >, __gnu_pbds :: rb_tree_tag, __gnu_pbds :: tree_order_statistics_node_update > tr;
int a[355][355], n, b[355];
signed main () {
	scanf ("%d", &n);
	for (int i = 1;i <= n; ++ i) {
		for (int j = 1;j <= n; ++ j) {
			a[i][j] = (i - 1) * (j - 1) % n;
		}
	}
	for (int i = 1;i <= n; ++ i) {
		scanf ("%d", &b[i]);
		int delta = b[i] - a[i][i];
		for (int j = 1;j <= n; ++ j) a[i][j] += delta;
	}
	for (int i = 1;i <= n; ++ i) {
		for (int j = 1;j <= n; ++ j) {
			a[i][j] = (a[i][j] % n + n) % n;
			printf ("%d ", a[i][j]);
		}
		printf ("\n");
	}
	return 0;
}