【BZOJ1225】求正整数（数论）

题意：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。

n<=50000

思路：记得以前好像看的是maigo的题解

n即为将m分解为质数幂次的乘积后的次数+1之积

经检验只需要取前16个质数

其次幂次的数据单调不增

乘积大小比较时候表示为ln之和，这样比较巧妙的避开了大整数比较

加了这几个优化跑的飞快

注意需要加高精

C++

#include<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<map>
#include<set>
#include<cmath>
int prime[17]={0,2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53};
int a[50000],b[50000],c[50000],d[50000],ans[200],s[200];
double f[20];
int n,ansp;
double anse;

void dfs(int n,int p,int limit,double e)
{
    if(n==1) 
    {
        if(e<anse) 
        {
            for(int i=0;i<=200;i++) ans[i]=s[i];
            anse=e;
            ansp=p;
        }
        exit;
    }
    for(int i=1;i<=limit;i++)
     if(!(n%(i+1)))
     {
         s[p]=i;
         dfs(n/(i+1),p+1,i,e+f[p]*i);
     }
}

void mult1(int *a,int *b,int *c)
{
    for(int i=1;i<=c[0];i++) c[i]=0;
    c[0]=0;
    for(int i=1;i<=a[0];i++)
     for(int j=1;j<=b[0];j++)
     {
         int k=i+j-1;
         c[k]+=a[i]*b[j];
         c[k+1]+=c[k]/10;
         c[k]%=10;
     }
     c[0]=a[0]+b[0];
     if(!c[c[0]]) c[0]--;
}

void mult2(int *a,int b)
{

    for(int i=1;i<=a[0];i++) a[i]*=b;
    for(int i=1;i<=a[0]-1;i++)
    {
        a[i+1]+=a[i]/10;
        a[i]%=10;
    }
    while(a[a[0]]>9)
    {
        a[a[0]+1]=a[a[0]]/10;
        a[a[0]]%=10;
        a[0]++;
    }
}

void pow(int *x,int *y,int k,int p)
{
    if(k==1)
    {
      for(int i=1;i<=x[0];i++) x[i]=0;
      x[0]=0; 
      if(prime[p]<10)
      {
          x[0]=1; x[1]=prime[p];
      }
       else
       {
           x[0]=2; x[2]=prime[p]/10; x[1]=prime[p]%10;
       }
    }
     else 
     {
         pow(y,x,k>>1,p);
         mult1(y,y,x);
         if(k&1) mult2(x,prime[p]);
     }
   
}

void print(int *a)
{
    for(int i=a[0];i>0;i--) printf("%d",a[i]);
    printf("\n");
}

int main()
{
    freopen("bzoj1225.in","r",stdin);
    freopen("bzoj1225.out","w",stdout);
    scanf("%d",&n);
    for(int i=1;i<=17;i++) f[i]=log(prime[i]);
    anse=3e38;
    dfs(n,1,n-1,0);
    ansp--;
    a[0]=1; a[1]=1;
    for(int i=1;i<=ansp;i++)
    {
        pow(c,d,ans[i],i);
        if(i&1) mult1(a,c,b);
         else mult1(b,c,a);
    }
    if(ansp&1) print(b);
     else print(a); 
}

pascal

type arr=array[-1..7000]of longint;
const base=10000;
      prime:array[1..16]of longint=(2,3,5,7,11,13,17,19,23,
                                    29,31,37,41,43,47,53);
  
var s,ans:array[0..200]of longint;
    n,ansp:longint;
    anse:real;
    a,b,c,d:arr;
    bool:boolean;
  
procedure dfs(n,p,limit:longint;e:real);
var i:longint;
begin
 if n=1 then
 begin
  if e<anse then begin ans:=s; anse:=e; ansp:=p; end;
  exit;
 end;
 for i:=1 to limit do
  if n mod (i+1)=0 then
  begin
   s[p]:=i;
   dfs(n div (i+1),p+1,i,e+ln(prime[p])*i);
  end;
end;
  
procedure lyk(var a:arr;b:longint);
var i:longint;
begin
 for i:=0 to a[-1] do a[i]:=a[i]*b;
 for i:=0 to a[-1] do
 begin
  inc(a[i+1],a[i] div base);
  a[i]:=a[i] mod base;
 end;
 if a[a[-1]+1]>0 then inc(a[-1]);
end;
  
procedure zhw(var a,b,c:arr);
var i,j,k:longint;
begin
 fillchar(c,sizeof(c),0);
 for i:=0 to a[-1] do
  for j:=0 to b[-1] do
  begin
   k:=i+j;
   inc(c[k],a[i]*b[j]);
   inc(c[k+1],c[k] div base);
   c[k]:=c[k] mod base;
  end;
  c[-1]:=a[-1]+b[-1];
  if c[c[-1]+1]>0 then inc(c[-1]);
end;
  
procedure pow(var x,y:arr;p:longint);
begin
 if p=1 then
 begin
  fillchar(x,sizeof(x),0);
  x[0]:=prime[n];
 end
  else
  begin
   pow(y,x,p>>1);
   zhw(y,y,x);
   if p and 1=1 then lyk(x,prime[n]);
  end;
end;
procedure print(var a:arr);
var i:longint;
begin
 write(a[a[-1]]);
 for i:=a[-1]-1 downto 0 do
  write(a[i] div 1000,a[i] div 100 mod 10,a[i] div 10 mod 10,a[i] mod 10);
 writeln;
end;
  
begin
  
 readln(n);
 anse:=3e38;
 dfs(n,1,n-1,0);
 dec(ansp);
 fillchar(a,sizeof(a),0);
 a[0]:=1;
 for n:=1 to ansp do
 begin
  pow(c,d,ans[n]);
  if n and 1=1 then zhw(a,c,b)
   else zhw(b,c,a);
 end;
  
 if ansp and 1=1 then print(b)
  else print(a);
  
end.