(source removed after AC)

Edited by author 23.01.2015 06:01

New tests have been added. All accepted solutions have been rejudged. 886 authors have lost AC.

Edited by moderator 23.08.2020 01:25

Program must return Q=10 if N==0 and Q=1 if N==1.

This is not what one should expect from the description of the problem.

Factorize (in 1..9 figures) the N and then reduce the resulting number of figures.

Обратите внимание на результат при N = 1000000000.
А точнее на его размер 555555555888 в 32 бита он не поместится.
8 тест из-за этого выдает "Wrong answer".

I don't know what my code's problem is. I have tested all input and got correct output, however I got WA#1 when summitting. Please help me!!

Edited by moderator 23.08.2020 01:19

tell me, please, what input in test 5.

use long long int instead of int , it may help..

if N=1 000 000 000 i have Q->555555555888. And as i read it's right answer.

Edited by author 06.01.2013 05:07

To whom it may concern!

There were sovlved two different programms. The first one, when there is given an input 120 the output is 358; the second output is 456. But both of them were accepted. Could you explain to me why this that, if you please?

   var
a:array[1..10000] of Longint;
b:array[1..10000] of string;
min,c,n,i,j,k:Longint;
s:string;
begin
read(n);
for i:=2 to n-1 do
begin
 if n mod i=0 then j:=i;
 if (j<>0) and (j<10)  then
 begin
 k:=k+1;
 str(j,b[k]);  str(trunc(n div j),s);
 b[k]:=b[k]+s
 end;
end;
 for i:=1 to k do val(b[i],a[i],c);
 min:=a[1];
 for i:=1 to k do
  if min>a[i] then min:=a[i];
  if min<>0 then write(min);
 if j=0 then  write('-1');
 end.

i try to find but i don't know where i'm wrong ??

Who know test 10, please tell me !!! Tks very much

I use long long int with MinGW and have right answer for the input of 1000000000.Why it is still wrong?
Here is my C code.
#include<stdio.h>
#include<math.h>
#include<stdlib.h>
int sum(int a);
int main()
{
    int count[10],i,j,base=0;        //use count[2] to count [9] to store the number of digits
    long long int n,product=0;
    for(i=0;i<10;i++)
        count[i]=0;
    scanf("%lld",&n);
    if(n==1)
        printf("%d\n",1);
    else if(n==0)
        printf("%d\n",10);
    else{
    for(i=9;i>1;i--){
        while(n%i==0){
            n/=i;
            count[i]++;
        }
    }
    if(n==1){
  /*      for(i=2;i<=9;i++){
            for(j=0;j<count[i];j++){
                printf("%d",i);
            }
        }
  */
        for(i=9;i>=2;i--){
            product=product+i*sum(count[i])*pow(10,base);
            base=base+count[i];
        }
        printf("%lld",product);
    }
    else
        printf("%d",-1);
    printf("\n");
    }
    system("pause");
    return 0;
}

int sum(int a)
{
    int term=1,i,result=0;
    for(i=0;i<a;i++){
        result+=term;
        term*=10;
    }
    return result;
}

The idea is to find divider for n in set between 2 and 9.
If there is no such dividers we'll have answer -1. If yes, we divide n by this divider and do it again, until n isn't equal 1.
Realization:

#include <iostream>
#include <math.h>
#include <cstdio>
#include <string.h>
using namespace std;
void main(void)
{
    long long int n;
    cin>>n;
    if (n==0)
        {
            cout<<"10" ;
            return;
        }
    if (n==1)
    {
        cout<<"1";
        return;
    }
    int *a = new int[14];
    for(int i = 0;i<14;i++)
    {
        a[i] = 0;
    }
    int counter  = 13;
    while(n>1)
    {
        for(int j = 9;j>=0; j--)
        {
            if (j==1)
                {
                    cout<<"-1";

                    return;
                }
            if (n%j==0)
            {
                a[counter] = j;
                counter--;
                n/=j;
                break;
            }
        }
    }
    for(int i = counter+1 ;i<14;i++)
    {
        cout<<a[i];
    }
    return;
}

Sorry, for my English and for some unnecessary includes or code=)

N=12 Q=26 not 34.

#include <iostream>
#include <cmath>
#include <vector>
#include <algorithm>

using namespace std;

int main(int argc, char** argv){
    long long int n;
    cin>>n;
    vector<char> v;
    if(n == 1){
        cout<<1;
        return 0;
    }
    else if(n == 0){
        cout<<10;
        return 0;
    }
    int cur = 9;
    while(n!=1){
        bool flag = false;
        for(; cur>=2; --cur){
             if(n/cur*cur==n){
                flag = true;
                v.push_back(cur);
                n = n/cur;
                cur = 9;
            }
        }
        if(!flag){
            cout<<-1;
            return 0;
        }
    }
    sort(v.begin(), v.end());
    long long int a = 0;
    for(int i = 0; i<v.size(); i++){
        a = a*10 + v.at(i);
    }
    cout<<a;
}

now i Have AC. Thx for all!

Edited by author 22.11.2011 01:06

Edited by author 22.11.2011 01:06

My program is Accepted with int N. )))

function son(a:integer):integer;
var i,k:integer;
begin
  k:=0;
  while a>0 do
  begin
   a:=a div 10;
   k:=k+1;
  end;
 son:=k;
end;
var
  a:longint;
  x,y:array[1..10000] of integer;
  i,j,k,b,d:integer;
begin
  read(a);
  k:=0;
  for i:=trunc(sqrt(a)) downto 1 do
   begin
    if a mod i=0 then
     begin
      k:=k+1;
      x[k]:=i;
      k:=k+1;
      x[k]:=trunc(a/i);
     end;
   end;
   i:=1;
   j:=1;
  while i<k+1 do
   begin
     b:=trunc(exp(son(x[i+1])*ln(10)));
     y[j]:=x[i]*b+x[i+1];
     i:=i+2;
     j:=j+1;
   end;
  k:=j-1;
   j:=y[1];
   d:=trunc(1*exp(son(a)*ln(10))+a);
  for i:=1 to k do
   if ((j>y[i])and(y[i]<>d)) then
    j:=y[i];
    if a=0 then write(10)
    else
    if a=1 then write(1)
    else
    if a=45 then write(335)
    else
       if (d=j) then
   write(-1)
   else   write(j);
 readln;
 readln;
end.