给你一个数N,找出一个最小的可以拆分成N种乘积表达形式的数x
比如N=2,6可以拆成2x3或者1x6两种,但不是最小的,最小的是4可以拆成1x4,2x2两种
首先可以肯定的是x必然有N*2或者是N*2-1(完全平方的情况)个约数
利用求反素数的过程求出约数为N*2和N*2-1个的最小的数
#include <cstdio> #include <sstream> #include <fstream> #include <cstring> #include <iostream> #include <algorithm> #include <map> #include <cctype> #include <ctime> #include <set> #include <climits> #include <vector> #include <queue> #include <stack> #include <cstdlib> #include <cmath> #include <string> #include <list> #define INPUT_FILE "in.txt" #define OUTPUT_FILE "out.txt" using namespace std; typedef unsigned long long LL; const int INF = INT_MAX / 2; const LL maxval = 1e18 + 1; const int maxn = 75 + 5; const int maxcnt = 160; LL cnt[maxn * 2]; int prime[20] = {2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71}; int times[20]; void dfs(LL curnum,LL curcnt,int nowt) { if(curnum > maxval) return; if(curcnt <= maxcnt) { cnt[curcnt] = min(cnt[curcnt],curnum); } for(int i = 1;i <= maxcnt;i++) { if(nowt == 0 || i <= times[nowt - 1]) { curnum *= prime[nowt]; if(curnum > maxval) break; curcnt = curcnt / i * (i + 1); times[nowt] = i; dfs(curnum,curcnt,nowt + 1); } else break; } } int main() { for(int i = 1;i <= maxcnt;i++) cnt[i] = maxval; dfs(1,1,0); int N; while(cin >> N,N) { LL a1 = cnt[N * 2 - 1],a2 = cnt[N * 2],sa = sqrt(a1); if(sa * sa == a1 && a1 < a2) cout << a1 << endl; else cout << a2 << endl; } return 0; }
HDU 4228 Flooring Tiles 反素数的应用
时间: 2024-10-10 22:36:56