对于任意输入的正整数n，请编程求出具有n个不同因子的最小正整数m。
Solution
（乍一看很简单却搞了好久？我真是太菜了）
根据因子个数计算公式
若 \(m = \prod p_i^{q_i}\), 则 \(n = \prod (q_i + 1)\)
设 \(f[i][j]\) 为只包含前 \(j\) 个质因数，因子个数为 \(i\) 的最小的数
转移类似背包： \(f[i][j]=min_{k|i} (f[i/k][j-1] \cdot p_j^{k-1})\)
这样直接做是 \(O(n \sqrt n \log n)\) ，考虑到需要枚举的 \(i\) 有且仅有 \(n\) 的因数，而约数个数的一个宽上界是 \(O(\sqrt n)\)，复杂度就压缩到了 \(O(n \log n)\)
使用高精度直接 dp 可能会复杂度爆炸，所以我们对数一下
#include <bits/stdc++.h>
using namespace std;
struct Biguint {
int a[100005], len;
Biguint() {
memset(a, 0, sizeof a);
len = 0;
}
void read() {
string str;
cin >> str;
memset(a, 0, sizeof a);
len = str.length();
for (int i = 0; i < str.size(); i++)
a[i] = str[str.length() - i - 1] - '0';
}
void print() {
for (int i = len - 1; i >= 0; i--) {
cout << a[i];
}
}
bool operator < (const Biguint& obj) {
const int* b = obj.a;
if (this->len == obj.len) {
for (int i = len - 1; i>=0; --i)
if (a[i] != b[i]) return a[i] < b[i];
return false;
}
else return this->len < obj.len;
}
bool operator > (const Biguint& obj) {
const int* b = obj.a;
if (this->len == obj.len) {
for (int i = len - 1; i>=0; --i)
if (a[i] != b[i]) return a[i] > b[i];
return false;
}
else return this->len > obj.len;
}
bool operator != (const Biguint& obj) {
return (*this < obj) | (*this > obj);
}
bool operator == (const Biguint& obj) {
return !((*this < obj) | (*this > obj));
}
bool operator <= (const Biguint& obj) {
return (*this) < obj || (*this) == obj;
}
bool operator >= (const Biguint& obj) {
return (*this) > obj || (*this) == obj;
}
Biguint operator += (const Biguint& obj) {
const int* b = obj.a;
if (obj.len > len) len = obj.len;
for (int i = 0; i < len; i++) {
a[i] += b[i];
if (a[i] >= 10) a[i + 1] += a[i] / 10, a[i] %= 10;
}
if (a[len]) ++len;
while (a[len - 1] >= 10)
a[len] += a[len - 1] / 10, a[len - 1] %= 10, ++len;
return *this;
}
Biguint operator + (const Biguint& obj) {
Biguint ret;
ret += *this;
ret += obj;
return ret;
}
Biguint operator -= (const Biguint& obj) {
const int* b = obj.a;
for (int i = 0; i < len; i++) {
a[i] -= b[i];
if (a[i] < 0) a[i + 1]--, a[i] += 10;
}
while (a[len - 1] == 0 && len > 0) --len;
return *this;
}
Biguint operator -(const Biguint& obj) {
Biguint ret;
ret += *this;
ret -= obj;
return ret;
}
Biguint operator *= (int b) {
for (int i = 0; i < len; i++)
a[i] *= b;
for (int i = 0; i < len; i++)
a[i + 1] += a[i] / 10, a[i] %= 10;
++len;
while (a[len - 1] >= 10)
a[len] += a[len - 1] / 10, a[len - 1] %= 10, ++len;
while (a[len - 1] == 0 && len > 0) --len;
return *this;
}
Biguint operator * (int b) {
Biguint ret;
ret = *this;
ret *= b;
return ret;
}
Biguint operator * (const Biguint& obj) {
const int* b = obj.a;
Biguint ret;
for (int i = 0; i < len; i++)
for (int j = 0; j < obj.len; j++)
ret.a[i + j] += a[i] * b[j];
for (int i = 0; i < len + obj.len; i++)
ret.a[i + 1] += ret.a[i] / 10, ret.a[i] %= 10;
ret.len = len + obj.len;
++ret.len;
while (ret.a[ret.len - 1])
ret.a[ret.len] += ret.a[ret.len - 1] / 10, ret.a[ret.len - 1] %= 10, ++ret.len;
while (ret.a[ret.len - 1] == 0 && ret.len > 0) --ret.len;
return ret;
}
};
ostream& operator << (ostream& os, Biguint num)
{
for (int i = num.len - 1; i >= 0; --i)
os << num.a[i];
if (num.len == 0) os << "0";
return os;
}
istream& operator >> (istream& is, Biguint& num)
{
string str;
is >> str;
memset(num.a, 0, sizeof num.a);
num.len = str.length();
for (int i = 0; i < str.length(); i++)
num.a[i] = str[str.length() - i - 1] - '0';
return is;
}
const int N = 500005;
const int p[21] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59, 61, 67, 71};
int n,g[50005][21],h[50005][21],ii[50005],top;
double lp[21]={},f[50005][21];
signed main() {
ios::sync_with_stdio(false);
cin>>n;
for(int i=1;i<=n;i++) if(n%i==0) ii[++top]=i;
for(int i=1;i<=20;i++) lp[i]=log(p[i]);
for(int i=0;i<=50000;i++) for(int j=0;j<=20;j++) f[i][j]=1e9;
f[1][0]=1;
for(int _i=2;_i<=top;_i++) {
int i=ii[_i];
int sq=sqrt(i);
for(int j=1;j<=20;j++) {
for(int k=1;k<=sq;k++) {
if(i%k==0) {
if(f[i][j]>f[i/k][j-1]+(k-1)*lp[j]) {
f[i][j]=f[i/k][j-1]+(k-1)*lp[j];
g[i][j]=i/k;
h[i][j]=k-1;
}
}
}
for(int u=1;u<=sq;u++) {
int k=i/u;
if(i%k==0) {
if(f[i][j]>f[i/k][j-1]+(k-1)*lp[j]) {
f[i][j]=f[i/k][j-1]+(k-1)*lp[j];
g[i][j]=i/k;
h[i][j]=k-1;
}
}
}
}
}
int pos=n;
Biguint ans;
ans.len=1;
ans.a[0]=1;
for(int j=20;j;--j) {
int i=pos;
for(int k=1;k<=h[i][j];k++) ans*=p[j];
pos=g[i][j];
}
cout<<ans;
}