1 0 1 0 0 0
11
ans = 10

I was trying to remove from the set first non-zero count money unit that is NOT LESS than ticket

But actually the task asks to remove just the FIRST non-zero count money unit

If act according to the former strategy you will receive WA20.

1) First of all you have to calculate maximum of tickets the teacher can buy ((amount of money) div (price of one ticket))
2) Next step you have to find minimum of tickets. You must pick up one banknote with minimum denomination and calculate amount again: in sample test you get 0 1 0 0 0 0 sequence. Then calculate amonut of money again. So, now you can get minimun of tickets ((new amount of money) div (price of one ticket) + 1).
3) So, now you have minimum and maximum of tickets. Just output the number of possible answers in the first line (tMax - tMin + 1) and the variants in ascending order (for i:=tMin to tMax do write(i,' ');

Good luck and get AC!

Please give me test...My code :

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Park1796_ {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int a[] = new int[6];
        int max = 999999999, k = 0;
        for (int i = 0; i < a.length; i++) {
            a[i] = in.nextInt();
            if (a[i] != 0 && a[i] < max) {
                max = a[i];
                k = i;
            }
        }
        int b = in.nextInt();
        int sum = 0;
        sum += a[0] = 10 * a[0];
        sum += a[1] = 50 * a[1];
        sum += a[2] = 100 * a[2];
        sum += a[3] = 500 * a[3];
        sum += a[4] = 1000 * a[4];
        sum += a[5] = 5000 * a[5];
        switch (k){
            case 0: k = 10;   break;
            case 1: k = 50;   break;
            case 2: k = 100;  break;
            case 3: k = 500;  break;
            case 4: k = 1000; break;
            case 5: k = 5000; break;
        }
        int l = 0, d = sum - k;
        int count = 0, f = 0;
        f = d / b + 1;
        while (f * b <= sum && f * b >= d){
            out.print(f + " ");
            count++;
            f++;
        }
        System.out.println(count);
        out.flush();
    }
}

Edited by author 19.02.2015 03:01

First make the sum to determinate MIN and MAX money the teacher wanted to give. Then divide it by the price of one ticket. Do it in this order. Hope I helped! :)

What is WA 3 Test. Give me please some example

Is such an input possible?

2 0 0 0 1 0
100
this means that teacher gave 1020 but the 20 that were given were unnecessary, and it's said that every note is needed.

Thanks.

test:
0 3 0 0 0 0
33

answer:
1
4

Test
0 0 0 0 0 1
499

Answer
10
1 2 3 4 5 6 7 8 9 10

i have a problem in test 6, can't find ,whats the matter?!?!
help plz!!

why is it so?

#include <iostream>
#include <cmath>
using namespace std;
int main()
{
    int value[]={ 10, 50, 100, 500, 1000, 5000 };
    int num[5],sum=0;
    int price;
    for(int i=0;i<6;i++)
    {
        cin>>num[i];
        sum+=num[i]*value[i];
    }
    cin>>price;
int i=0;
while(num[i]==0)
++i;
int lol=0,lol1=0;
while(lol*price<=sum-value[i])
    ++lol;
int j=0;
int k=0;
while((lol+k)*price<=sum)
{
    ++lol1;
    ++k;
}
cout<<lol1<<endl;
while((lol+j)*price<=sum)
{
    cout<<lol+j<<" ";
    ++j;
}
}

I don't undestand why answer 1 11?
I think answer is 6 6 7 8 9 10 11.
Thanks.

var
      i,k,j,p,min,q,x,n:integer;
      a,b:array [1..6] of integer;
      c:array [0..6] of integer;
      t:boolean;
begin
   min:=32000;
   a[1]:=10;
   a[2]:=50;
   a[3]:=100;
   a[4]:=500;
   a[5]:=1000;
   a[6]:=5000;
   n:=6;
   for i:=1 to n-1 do
   begin
      read (b[i]);
      x:=x+b[i]*a[i];
   end;
   readln (b[n]);
   readln (k);
   t:=false;
   i:=1;
   while not t do
      begin
          while  (i<=n) and (c[i]>=b[i]) do
         begin
            c[i]:=0;
                inc(i);
         end;
      if i>n then
             begin
            t:=true;
                continue;
             end;
      inc(c[i]);
      for j:=1 to n do
        p:=p+c[j]*a[j];
       if p=k then
       begin
          q:=p;
          break;
       end;
       if p>=k then
          begin
             for j:=1 to i-1 do
            c[j]:=0;

          end
       else
          i:=1;
      if  (p>=k) and (p-k<min) then
        begin
           min:=p-k;
               q:=p;
        end;
      end;
   q:=(x-q) div k;
   writeln ((x div k)-q);
   for i:=q+1 to x div k do
      write (i,' ');

end.

Can someone please tell what is in test #16?

can someone please tell what is in test #6?

Please, help with test 8!.. Don't know what do(

I have a problem in test 8. Could someone give me a hint? :(