Prime & 反素数plus

题意:

　　求因数个数为n的最小正整数k. n<=10^9输出其唯一分解形式

SOL:

　　模拟题,一眼看过去有点惊讶...这不是我刚看过的反素数吗...

　　咦数据怎么这么大,恩搞个高精吧...

　　于是T了...

　　真是丝帛...因为这题不用输出答案而是输出质因子与指数,那么高精也没什么卵用...

　　想想我们在反素数的时候除了记录还要做一件什么事呢...比较答案与当前搜索的大小...但这里是在太大了,所以就要找一个更小的通用比较方法...

　　傻逼想到了高精,帅的人都用了log

　　log由于其良好的性质log(a*b)=log(a)+log(b).

　　于是balabalabalabala....

Code:

　　

/*==========================================================================
# Last modified: 2016-03-18 08:32
# Filename: t1.cpp
# Description:
==========================================================================*/
#define me AcrossTheSky
#include <cstdio>
#include <ctime>
#include <cmath>
#include <string>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
   
#include <set>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <deque>
 
#define lowbit(x) (x)&(-x)
#define FOR(i,a,b) for((i)=(a);(i)<=(b);(i)++)
#define FORP(i,a,b) for(int i=(a);i<=(b);i++)
#define FORM(i,a,b) for(int i=(a);i>=(b);i--)
#define ls(a,b) (((a)+(b)) << 1)
#define rs(a,b) (((a)+(b)) >> 1)
#define getlc(a) ch[(a)][0]
#define getrc(a) ch[(a)][1]
#define pb push_back
#define find(a,b) lower_bound((a).begin(), (a).end(), (b))-(a).begin()
 
#define INF 10000000000
#define maxn 3000
#define maxm 100000
#define pi 3.1415926535898
#define _e 2.718281828459 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
template<class T> inline
void read(T& num) {
    bool start=false,neg=false;
    char c;
    num=0;
    while((c=getchar())!=EOF) {
        if(c=='-') start=neg=true;
        else if(c>='0' && c<='9') {
            start=true;
            num=num*10+c-'0';
        } else if(start) break;
    }
    if(neg) num=-num;
}
/*==================split line==================*/
const double inf=1e18;
const double eps=0.00000001;
#define mx 107
int p[52]={0,2,3,5,7,11,
           13,17,19,23,29,
           31,37,41,43,47,
           53,59,61,67,71,
           73,79,83,89,97,
           101,103,107,109,113,
           127,131,137,139,149,
           151,157,163,167,173,
           179,181,191,193,197,
           199};
int n,b[mx],c[mx];
double ans;
void dfs(int x,double t,int num,int m)
{
    if (n%num) return;
    if (num>n) return;
    if (num==n&&ans>t)
    {
        for (int i=1;i<=x+1;i++) b[i]=c[i];
        ans=t;
        return;
    }
    else if (num==n) return;
    if (ans-log(p[x])<t) return;
    int d=n/num;
    for (int i=1;i*i<=d;i++)
    {
        if (d%i==0)
        {
            if (i-1<=m&&i!=1)
            {
                c[x]=i-1;
                dfs(x+1,t+log(p[x])*(i-1),num*i,i-1);
                c[x]=0;
            }
            if (i*i!=d&&d/i-1<=m)
            {
                c[x]=d/i-1;
                dfs(x+1,t+log(p[x])*(d/i-1),num*(d/i),d/i-1);
                c[x]=0;
            }
        }
    }
}  
int main()
{
    read(n);
    ans=inf;
    dfs(1,0,1,n);
    if (n==1) printf("1^1"); else printf("%d^%d",p[1],b[1]);
    for (int i=2;i<=45;i++)
    {
        if (!b[i]) break;
        printf("*%d^%d",p[i],b[i]);
    }
    printf("
");
    return 0; 
}