高精度+搜索+质数 BZOJ1225 [HNOI2001] 求正整数

// 高精度+搜索+质数 BZOJ1225 [HNOI2001] 求正整数
// 思路：
// http://blog.csdn.net/huzecong/article/details/8478689
// M=p1^(t1)*p2^(t2)*p3^(t3)....
// N=(t1+1)*(t2+1)*(t3+1)*(t4+1)...
// 所以t最大到16，就可以暴力搜索了


#include <bits/stdc++.h>
using namespace std;
#define LL long long
const double inf = 123456789012345.0;
const LL MOD =100000000LL; 
const int N =1e5+10; 
#define clc(a,b) memset(a,b,sizeof(a))
const double eps = 1e-7;
void fre(){freopen("in.txt","r",stdin);}
void freout() {freopen("out.txt","w",stdout);}
inline int read() {int x=0,f=1;char ch=getchar();while(ch>'9'||ch<'0') {if(ch=='-') f=-1;ch=getchar();}while(ch>='0'&&ch<='9') {x=x*10+ch-'0';ch=getchar();}return x*f;}

int n;
const int M=16;
const int p[16]= {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47}; 
double Log[16];
class bignum  {  
    public:  
        LL num[N];  
        int tot;  
        bignum() {}  
        bignum(LL x) {  
            clear();  
            while (x) num[tot++] = x % MOD, x /= MOD;  
        }  
        void clear(){  
            tot = 0;  
            for (int i = 0; i < N; ++i) num[i] = 0LL;  
        } 
        void operator *= (const int &x){  
            for (int i = 0; i < tot; ++i) num[i] *= x;  
            for (int i = 0; i < tot; ++i)  
                if (num[i] >= MOD) {  
                    num[i + 1] += num[i] / MOD;  
                    num[i] %= MOD;  
                }  
            while (num[tot]) {  
                if (num[tot] >= MOD) {  
                    num[tot + 1] += num[tot] / MOD;  
                    num[tot] %= MOD;  
                }  
                ++tot;  
            }  
        }  
        void print()  {  
            printf("%lld", num[tot - 1]);  
            for (int i = tot - 2; i >= 0; --i)  
            printf("%08lld", num[i]);  
        }  
}ans(1);  

int t[16],ct[16];
double mn;
int cnt;

void dfs(int d,int x,int m,double tem){
    if(tem>mn) return;
    if(x==1){
        if(tem<mn){
            mn=tem,cnt=d-1;
            for(int i=1;i<d;i++) t[i]=ct[i];
        }
    }
    for(int i=1;i*i<=x&&i<=m;i++){
        if(!(x%i)){
            if(i!=1){
                ct[d]=i;
                dfs(d+1,x/i,i,tem+(double)Log[d]*(i-1));
            }
            if(x/i<=m&&x/i!=i){
                ct[d]=x/i;
                dfs(d+1,i,x/i,tem+(double)Log[d]*(x/i-1));
            }
        }
    }
}

int main(){
    scanf("%d",&n);
    for(int i=1;i<=15;i++) Log[i]=(double)log(p[i]);
    mn=inf;
    cnt=0;
    dfs(1,n,n,0.0);
    for(int i=1;i<=cnt;i++){
        for(int j=t[i]-1;j>0;j--){
            ans*=p[i];
        }
    }
    ans.print();
    return 0;
}