【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[]={,,,,,,,,,,,,,,,,};
 int a[],b[],c[],d[],ans[],s[];
 double f[];
 int n,ansp;
 double anse;

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

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

 void mult2(int *a,int b)
 {

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

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

 }

 void print(int *a)
 {
     for(int i=a[];i>;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=;i<=;i++) f[i]=log(prime[i]);
     anse=3e38;
     dfs(n,,n-,);
     ansp--;
     a[]=; a[]=;
     for(int i=;i<=ansp;i++)
     {
         pow(c,d,ans[i],i);
         if(i&) mult1(a,c,b);
          else mult1(b,c,a);
     }
     if(ansp&) print(b);
      else print(a);
 }

pascal

 type arr=array[-..]of longint;
 const base=;
       prime:array[..]of longint=(,,,,,,,,,
                                     ,,,,,,);

 var s,ans:array[..]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= then
  begin
   if e<anse then begin ans:=s; anse:=e; ansp:=p; end;
   exit;
  end;
  for i:= to limit do
   if n mod (i+)= then
   begin
    s[p]:=i;
    dfs(n div (i+),p+,i,e+ln(prime[p])*i);
   end;
 end;

 procedure lyk(var a:arr;b:longint);
 var i:longint;
 begin
  for i:= to a[-] do a[i]:=a[i]*b;
  for i:= to a[-] do
  begin
   inc(a[i+],a[i] div base);
   a[i]:=a[i] mod base;
  end;
  if a[a[-]+]> then inc(a[-]);
 end;

 procedure zhw(var a,b,c:arr);
 var i,j,k:longint;
 begin
  fillchar(c,sizeof(c),);
  for i:= to a[-] do
   for j:= to b[-] do
   begin
    k:=i+j;
    inc(c[k],a[i]*b[j]);
    inc(c[k+],c[k] div base);
    c[k]:=c[k] mod base;
   end;
   c[-]:=a[-]+b[-];
   if c[c[-]+]> then inc(c[-]);
 end;

 procedure pow(var x,y:arr;p:longint);
 begin
  if p= then
  begin
   fillchar(x,sizeof(x),);
   x[]:=prime[n];
  end
   else
   begin
    pow(y,x,p>>);
    zhw(y,y,x);
    if p and = then lyk(x,prime[n]);
   end;
 end;
 procedure print(var a:arr);
 var i:longint;
 begin
  write(a[a[-]]);
  for i:=a[-]- downto  do
   write(a[i] div ,a[i] div  mod ,a[i] div  mod ,a[i] mod );
  writeln;
 end;

 begin

  readln(n);
  anse:=3e38;
  dfs(n,,n-,);
  dec(ansp);
  fillchar(a,sizeof(a),);
  a[]:=;
  for n:= to ansp do
  begin
   pow(c,d,ans[n]);
   if n and = then zhw(a,c,b)
    else zhw(b,c,a);
  end;

  if ansp and = then print(b)
   else print(a);

 end.