题意：将一个数拆成若干数的和使得它们的最小公倍数最大
思路：一个数x可以拆成p1k1 + p2k2 + ... + pnkn形式，其中pi是质数或1。对于最小公倍数最大的情况，一定可以表示成这种形式。令dp[i][j]表示考虑前j个质数来构成i的最大公倍数，那么可以得到如下转移方程：
dp[i][j]=max{dp[i][j-1],dp[i-k][j-1]*k}，k=prime[j]t<=i
用滚动数组计算dp数组，同时用一个数组来存具体方案，在dp值更新时更新具体方案。
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#pragma comment(linker, "/STACK:10240000") #include <map> #include <set> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include <vector> #include <cstdio> #include <string> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define X first #define Y second #define pb push_back #define mp make_pair #define all(a) (a).begin(), (a).end() #define fillchar(a, x) memset(a, x, sizeof(a)) #define copy(a, b) memcpy(a, b, sizeof(a)) typedef long long ll; typedef pair<int, int> pii; typedef unsigned long long ull; #ifndef ONLINE_JUDGE void RI(vector<int>&a,int n){a.resize(n);for(int i=0;i<n;i++)scanf("%d",&a[i]);} void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R> void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?1:-1; while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T> void print(const T t){cout<<t<<endl;}template<typename F,typename...R> void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T> void print(T*p, T*q){int d=p<q?1:-1;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;} #endif template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);} template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);} const double PI = acos(-1.0); const int INF = 1e9 + 7; const double EPS = 1e-12; /* -------------------------------------------------------------------------------- */ const int maxn = 1e4 + 7; vector<int> prime; vector<int> rst[maxn]; int extra[maxn]; bool vis[maxn]; double dp[maxn]; void getprimelist() { for (int i = 2; i < maxn; i ++) { if (!vis[i]) { prime.pb(i); for (ll j = (ll)i * i; j < maxn; j += i) { vis[j] = true; } } } } void initdp() { for (int i = 0; i < maxn; i ++) dp[i] = 1.0; for (int j = 0; j < prime.size(); j ++) { for (int i = maxn - 1; i; i --) { for (ll k = prime[j]; k <= i; k *= prime[j]) { if (umax(dp[i], dp[i - k] * k)) { rst[i] = rst[i - k]; rst[i].pb(k); } } } } for (int i = 1; i < maxn; i ++) { sort(all(rst[i])); extra[i] = i; for (int j = 0; j < rst[i].size(); j ++) { extra[i] -= rst[i][j]; } } } void work(int n) { vector<int> ans; for (int i = 0; i < extra[n]; i ++) ans.pb(1); for (int i = 0; i < rst[n].size(); i ++) ans.pb(rst[n][i]); int start = 1; for (int i = 0; i < ans.size(); i ++) { for (int j = start + 1; j < start + ans[i]; j ++) printf("%d ", j); printf("%d%c", start, i == ans.size() - 1? ' ' : ' '); start += ans[i]; } } int main() { #ifndef ONLINE_JUDGE freopen("in.txt", "r", stdin); //freopen("out.txt", "w", stdout); #endif // ONLINE_JUDGE getprimelist(); initdp(); int T, n; cin >> T; while (T --) { cin >> n; work(n); } return 0; } |