как стать богатым в беларуси?
если у нас ничего нету

Text says "Pierre intends to use one-digit integers supplemented with a leading zero and two-digit integers only." but it seems like you cannot have one digit numbers in the original sequence.

For example, for test case 2, if you output "10 00" (which would translate to sequence 01 00, which ARE two consecutive numbers), you'll get wrong answer. Moral of the story, start checking for numbers from 10.

I have WA#1, but my program is correct.
Give me some tests And answer for this tests:
Input  1 : 120
My answer : 15

Input  2 : 1081
My answer : 46

Input  3 : 1073845
My answer : 1465

Input  4 : 500500
My answer : 1000

Input  5 : 11325
My answer : 150

Input  6 : 17391
My answer : 186

Input  7 : 122750946
My answer : 15668


Input  8 : 906531
My answer : 1346

Input  9 : 154290
My answer : 555

Input  10 : 7620753696
My answer : 123456

Thanks Advanced !!!!

вообще не могу понять что не так. проходят пробные тесты,проходят тесты которые нашла в обсуждениях.
если не трудно может покидаете тесты, плииз...

var a,a1,a2,s1,s,ap,al,sm1p,sm2p,sm3p,smap,sm1l,sm2l,sm3l,smal,s2,s3:Integer;
begin
ReadLn(a);
a1:=a+1;
a2:=a-1;

ap:=a1 div 1000;
al:=a1 mod 1000;

sm1p:=ap div 100;
sm2p:=ap mod 100 div 10;
sm3p:=ap mod 10;
smap:=sm1p+sm2p+sm3p;

sm1l:=al div 100;
sm2l:=al mod 100 div 10;
sm3l:=al mod 10;
smal:=sm1l+sm2l+sm3l;

if smal=smap then Writeln('YES')
     else
     begin
     ap:=a2 div 1000;
     al:=a2 mod 1000;

     sm1p:=ap div 100;
     sm2p:=ap mod 100 div 10;
     sm3p:=ap mod 10;
     smap:=sm1p+sm2p+sm3p;

     sm1l:=al div 100;
     sm2l:=al mod 100 div 10;
     sm3l:=al mod 10;
     smal:=sm1l+sm2l+sm3l;
     if smal=smap then Writeln('YES')
             else
                  WriteLn('NO');
     end;
end.

Hello guys,

those, who have wa on 7'th case may try to change output precision to some big number. As for me, it worked on 15 signs.

I have no idea why this happens ^,^

If ever dear admins may be interested in case, please compare:
id 4772942 has wa7
id 4772947 has ok (line 167).

Best regards, mq.

Passed all the test given in the discussion
I ran 1 bfs from 'r' and another bfs from 's' where i also kept data for those nodes who can have multiple parents.
than I while restoring path from destination to source I selected the node who has higher level from 'r' (that selection occurs if only the node has possibility multiple parents)
returned the ans which is the minimum among the final path

I got AC with 0.015s,216Kb
I had read the problem several time until understand.
N is the length of the digit number.
It will be something like that
a1 a2 a3 ...an
for each ai in a1 a2 a3..an, 0 <= ai < K.
The output number i used was interger (compiler C++ 2010). (no big number).

hope it can help you.
And sorry for my poor English.

Edited by author 25.04.2014 23:05

https://ideone.com/Ocpa4V

I can't figure out what's wrong with my code. I used 'Queue' data structure.

Edited by author 23.01.2018 16:02

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<string>
#include<cmath>
using namespace std;



vector<long long > C(20000,0);
void mult(vector<long long> A,vector<long long> B)
{
    int d=0;
    for(int i=0;i<300;i++)
    {
        for(int j=0;j<300;j++)
        {
            C[i+j]+=B[i]*A[j];
            C[i+j+1]=C[i+j+1]+C[i+j]/100000000;
            C[i+j]=C[i+j]%100000000;
        }
    }
    C[0]++;
    for(int i=0;i<6000;i++)
    {
        C[i+1]=C[i+1]+C[i]/100000000;
        C[i]=C[i]%100000000;
    }
}
vector<long long> D(20000,0);
void mult1(vector<long long> A,vector<long long> B)
{
    int d=0;
    for(int i=0;i<300;i++)
    {
        for(int j=0;j<300;j++)
        {
            D[i+j]+=B[i]*A[j];
            D[i+j+1]=D[i+j+1]+D[i+j]/100000000;
            D[i+j]=D[i+j]%100000000;
        }
    }
    for(int i=0;i<6000;i++)
    {
        D[i+1]=D[i+1]+D[i]/100000000;
        D[i]=D[i]%100000000;
    }
}
void write(vector<long long> A)
{
    bool f=false;
    for(int i=9999;i>=0;i--)
    {
        if(A[i]!=0)
            f=true;
        if(f)
            cout<<A[i];
    }
    cout<<"\n";
}
int main()
{
#ifdef _DEBUG
    freopen("input.txt","rt",stdin);
    freopen("output.txt","wt",stdout);
#endif
    int n;
    cin>>n;

    vector<long long> A(10000,0);
    vector<long long> B(10000,0);
    A[0]=1;
    B[0]=1;
    for(int i=0;i<n;i++)
    {
        mult(A,B);
        mult1(A,B);
        for(int i=0;i<4000;i++)
        {
            B[i]=D[i];
        }
        for(int i=0;i<4000;i++)
        {
            A[i]=C[i];
        }
        for(int i=0;i<4000;i++)
        {
            C[i]=0;
            D[i]=0;
        }
        write(A);
    }
}

Вот код, скорее всего он не пройдет макс тест, но при запуске на первом тесте он выдает мне в студии ответ правильный, а тут нет

I don't understand!!! How given (2, aa) answer is 6. I think that must be 4!!!  Please help, who know that!!!!! (Я не понел Как 2 и аа могут дать ответ 6 Я думаю что должно быть 4 Помогите------------)

I am getting WA 5   Can anyone provide me the test case 5..  or hint to test case 5..

var a,b,c,d,cen,k:integer;
begin
read(a,b,c,d);
while k<>1 do begin
if (a+b>c) then begin
k:=1;cen:=c;end
else a:=a+b;
if(c-d<a)then begin
k:=1;cen:=a;end
else c:=c-d;end;
writeln(cen);
end.

try to change your variables to double, and don't use int.

I think it's clear for the system of linear equations, that if M<N, than it's impossible. If M=N then it may be possible (and I know how to check). How about case M>N? We need to select N pairs that get an unambiguous solution?

http://acm.timus.ru/problem.aspx?space=1&num=1513
It seems that those are the same problem.  In 1513,  just b=INF.
But that problem is of rating of 100, and Lemon Tale is of rating of around 500.

I'd like to thank authors for this problem - it became one of my favorite on Timus!
Thanks again for very interesting problem!
:-)

Unfortunately entirely deterministic version of this algorithm fails on Test #10. So I'm very curious about this Test #10. Because I've done a lot of testing in MATLAB of varoius modification of Floyd-Steinberg dithering algo, including zig-zag proseccing and edge attending. In all cases constraint in statement can be hardened from 20 to 10 and even less. On normal images error is about 6-8.

I guess Test #10 some kind of periodic image, that has bad compatibility with that unnatural restriction put in the problem. I mean it square form and discontinuous nature. I think dithering yield much more better image than that which can be produced by uniform minimization of constrain function.

I solve this issue whit Test #10 by adding a little bit of randomness in my dithering algo. I mean in place of this threashold function:

int quantize(int value) {
    if (127 < value) {
        return 255;
    }
    return 0;
}

I'm using this function:

int quantize(int value, int randomness) {
    if (127 - randomness + rng.nextInt(2 * randomness + 1) < value) {
        return 255;
    }
    return 0;
}

This provides me with magical parameter "randomness" varying which and playing tambourine I got my AC. 8-] I think I should add constraints checking and reprocessing in case the check was a fail, but I'm too lazy, I guess in case of possible future rejudging everything will be fine. :)

I express my sincerest thanks to authors of this beautiful problem. It has bugged me a lot in era of matrix printers how can they do this funny dotted images. Finally I have my answer and I can even do it myself!!!

Edited by author 17.01.2018 02:43

#include <iostream>
using namespace std;

int main()
{
    int k, n, res=0;
    cin >> k >> n;
    int *mas = new int[n];

    for (int i=0; i!=n; i++)
    {
        cin >> mas[i];

        if (mas[i] > k)
        {
            res = res + (mas[i] - k);
        }
        else
        {
            if (res!=0 )
            {
                if ((k - mas[i]) > res)
                {
                    res = 0;
                }
                else
                {
                    res = res - (k - mas[i]);
                }
            }
        }

    }
    cout << res;

    return 0;
}