AtCoder Regular Contest 149 C.Avoid Prime Sum(构造题,补写法)

题目

构造一个n*n(n<=1e3)的矩阵，恰用到1到n*n这n*n个数，

使得相邻两项之和不是质数

思路来源

官方题解/zhoukangyang代码/SSerxhs代码/Heltion代码

题解

自己虽然搞过去了，但是很复杂，多学学别人怎么写的

首先想到是奇数分一堆偶数分一堆，特判n=3后，分奇偶直接搞也可做

简洁的做法是特判n=3,4,5，然后n>=6用奇数和偶数中3的倍数做两堆数的分隔

此时奇数里3的倍数，偶数里3的倍数已经能独占一行了

代码1

中规中矩，n=3,4,5构造，n>=6按规则排一下序，很简洁

#include<bits/stdc++.h>
#define L(i, j, k) for(int i = (j); i <= (k); ++i)
#define R(i, j, k) for(int i = (j); i >= (k); --i)
#define ll long long
#define ull unsigned long long
#define vi vector <int>
#define sz(a) ((int) (a).size())
#define me(f, x) memset(f, x, sizeof(f))
using namespace std;
const int N = 1e6 + 7, mod = 998244353, inv3 = (mod + 1) / 3;
int n, m, w[N], ord[N];
int main () {ios :: sync_with_stdio(false);cin.tie (0); cout.tie (0);cin >> n;if(n == 3) {cout << "5 4 8\n";cout << "3 6 2\n";cout << "1 9 7\n";return 0;}if(n == 4) {cout << "15 11 16 12\n";cout << "13 3 6 9\n";cout << "14 7 8 1\n";cout << "4 2 10 5\n";return 0;}if(n == 5) {cout << "1 3 5 7 9\n";cout << "11 13 15 17 19\n";cout << "23 25 21 4 14\n";cout << "10 8 12 24 22\n";cout << "18 20 2 16 6\n";return 0;}L(i, 1, n * n) {if(i % 2 == 0) w[i] = 1e9 - (i % 3 == 0);else w[i] = (i % 3 == 0);ord[i] = i;}sort(ord + 1, ord + n * n + 1, [&] (int x, int y) {return w[x] < w[y];});L(i, 1, n * n) cout << ord[i] << " \n"[i % n == 0];return 0;
}
/*
6 2 5
5 2 3 1 4 6
*/

代码2

n=3,4,5直接暴力随机置换至有解

//这回只花了114514min就打完了。
//真好。记得多手造几组。ACM拍什么拍。
#include "bits/stdc++.h"
using namespace std;
template<typename typC,typename typD> istream &operator>>(istream &cin,pair<typC,typD> &a) { return cin>>a.first>>a.second; }
template<typename typC> istream &operator>>(istream &cin,vector<typC> &a) { for (auto &x:a) cin>>x; return cin; }
template<typename typC,typename typD> ostream &operator<<(ostream &cout,const pair<typC,typD> &a) { return cout<<a.first<<' '<<a.second; }
template<typename typC,typename typD> ostream &operator<<(ostream &cout,const vector<pair<typC,typD>> &a) { for (auto &x:a) cout<<x<<'\n'; return cout; }
template<typename typC> ostream &operator<<(ostream &cout,const vector<typC> &a) { int n=a.size(); if (!n) return cout; cout<<a[0]; for (int i=1; i<n; i++) cout<<' '<<a[i]; return cout; }
template<typename typC> bool cmin(typC &x,const typC &y) { if (x>y) { x=y; return 1; } return 0; }
template<typename typC> bool cmax(typC &x,const typC &y) { if (x<y) { x=y; return 1; } return 0; }
#if !defined(ONLINE_JUDGE)&&defined(LOCAL)
#include "my_header\debug.h"
#else
#define dbg(...) ;
#define dbgn(...) ;
#endif
typedef unsigned int ui;
typedef long long ll;
#define all(x) (x).begin(),(x).end()
// void inc(ui &x,const ui &y) { if ((x+=y)>=p) x-=p; }
// void dec(ui &x,const ui &y) { if ((x+=p-y)>=p) x-=p; }
namespace pr
{typedef long long ll;typedef __int128 lll;typedef pair<ll,int> pa;ll ksm(ll x,ll y,const ll p){ll r=1;while (y){if (y&1) r=(lll)r*x%p;x=(lll)x*x%p; y>>=1;}return r;}namespace miller{const int p[7]={2,3,5,7,11,61,24251};ll s,t;bool test(ll n,int p){if (p>=n) return 1;ll r=ksm(p,t,n),w;for (int j=0; j<s&&r!=1; j++){w=(lll)r*r%n;if (w==1&&r!=n-1) return 0;r=w;}return r==1;}bool prime(ll n){if (n<2||n==46'856'248'255'981ll) return 0;for (int i=0; i<7; ++i) if (n%p[i]==0) return n==p[i];s=__builtin_ctz(n-1); t=n-1>>s;for (int i=0; i<7; ++i) if (!test(n,p[i])) return 0;return 1;}}using miller::prime;mt19937_64 rnd(chrono::steady_clock::now().time_since_epoch().count());namespace rho{void nxt(ll &x,ll &y,ll &p) { x=((lll)x*x+y)%p; }ll find(ll n,ll C){ll l,r,d,p=1;l=rnd()%(n-2)+2,r=l;nxt(r,C,n);int cnt=0;while (l^r){p=(lll)p*llabs(l-r)%n;if (!p) return gcd(n,llabs(l-r));++cnt;if (cnt==127){cnt=0;d=gcd(llabs(l-r),n);if (d>1) return d;}nxt(l,C,n); nxt(r,C,n); nxt(r,C,n);}return gcd(n,p);}vector<pa> w;vector<ll> d;void dfs(ll n,int cnt){if (n==1) return;if (prime(n)) return w.emplace_back(n,cnt),void();ll p=n,C=rnd()%(n-1)+1;while (p==1||p==n) p=find(n,C++);int r=1; n/=p;while (n%p==0) n/=p,++r;dfs(p,r*cnt); dfs(n,cnt);}vector<pa> getw(ll n){w=vector<pa>(0); dfs(n,1);if (n==1) return w;sort(w.begin(),w.end());int i,j;for (i=1,j=0; i<w.size(); i++) if (w[i].first==w[j].first) w[j].second+=w[i].second; else w[++j]=w[i];w.resize(j+1);return w;}void dfss(int x,ll n){if (x==w.size()) return d.push_back(n),void();dfss(x+1,n);for (int i=1; i<=w[x].second; i++) dfss(x+1,n*=w[x].first);}vector<ll> getd(ll n){getw(n); d=vector<ll>(0); dfss(0,1);sort(d.begin(),d.end());return d;}}using rho::getw,rho::getd;using miller::prime;
}
using pr::getw,pr::getd;
const int N=1e6+5;
int main()
{ios::sync_with_stdio(0); cin.tie(0);cout<<setiosflags(ios::fixed)<<setprecision(15);int n,i,j,k;cin>>n;int m=n*n;vector a(n,vector(n,0));if (n>=6){for (i=j=0,k=1; k<=m; k+=2) if (k%3){a[i][j]=k;if (++j==n) ++i,j=0;}for (k=1; k<=m; k+=2) if (k%3==0){a[i][j]=k;if (++j==n) ++i,j=0;}for (k=2; k<=m; k+=2) if (k%3==0){a[i][j]=k;if (++j==n) ++i,j=0;}for (k=2; k<=m; k+=2) if (k%3){a[i][j]=k;if (++j==n) ++i,j=0;}}else{for (i=j=0,k=1; k<=m; k+=2) if (k%3){a[i][j]=k;if (++j==n) ++i,j=0;}for (k=1; k<=m; k+=2) if (k%3==0){a[i][j]=k;if (++j==n) ++i,j=0;}for (k=2; k<=m; k+=2) if (k%3==0){a[i][j]=k;if (++j==n) ++i,j=0;}for (k=2; k<=m; k+=2) if (k%3){a[i][j]=k;if (++j==n) ++i,j=0;}mt19937 rnd(345);while (1){bool flg=0;for (i=0; i<n; i++) for (j=0; j<n; j++) if (j+1<n&&pr::prime(a[i][j]+a[i][j+1])||i+1<n&&pr::prime(a[i][j]+a[i+1][j])){swap(a[i][j],a[rnd()%n][rnd()%n]);flg=1;}if (!flg) break;}}for (auto &v:a) cout<<v<<'\n';/*for (i=0; i<n; i++) for (j=0; j<n; j++){dbg(i,j);if (j+1<n) assert(!pr::prime(a[i][j]+a[i][j+1]));if (i+1<n) assert(!pr::prime(a[i][j]+a[i+1][j]));}*/
}

代码3

暴力：n为奇数，把中心一填1 7 8 2一填，然后解决一行分界线；n为偶数，解决一行分界线

奇 1 奇

奇 7 8

偶 2 偶

#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int p(int n) {for (int i = 2; i * i <= n; i += 1)if (n % i == 0)return 1;return 0;
}
int main() {ios::sync_with_stdio(false);cin.tie(nullptr);cout << fixed << setprecision(20);int N;cin >> N;vector A(N, vector<int>(N));array<set<int>, 2> S;for (int i = 1; i <= N * N; i += 1)S[i % 2].insert(i);vector<pair<int, int>> vp;if (N % 2 == 1) {A[N / 2][N / 2] = 7;A[N / 2][N / 2 + 1] = 8;A[N / 2 + 1][N / 2] = 2;A[N / 2 - 1][N / 2 + 1] = 1;S[1].erase(1);S[1].erase(7);S[0].erase(2);S[0].erase(8);for (int i = 0; i < N / 2; i += 1)vp.emplace_back(N / 2, i);for (int i = N / 2 + 2; i < N; i += 1)vp.emplace_back(N / 2 - 1, i);}else for (int i = 0; i < N; i += 1)vp.emplace_back(N / 2 - 1, i);int n = N * N * 2;for (auto [x, y] : vp) {int si = -1, sj = -1;for (int i : S[0])if (si == -1) {for (int j : S[1])if (p(i + j)) {si = i;sj = j;break;}}else break;S[0].erase(si);S[1].erase(sj);A[x][y] = sj;A[x + 1][y] = si;}for (auto& Ai : A) {for (int& Aij : Ai) {if (Aij == 0) {int k = not S[1].empty();Aij = *S[k].begin();S[k].erase(Aij);}cout << Aij << " ";}cout << "\n";}
}

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AtCoder Regular Contest 149 C.Avoid Prime Sum(构造题,补写法)

题目

思路来源

题解

代码1

代码2

代码3

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