<< When they “meet” (i.e., at some moment they are in the same point as in the PREVIOUS WERE NOT), they joyfully exclaim (“Oh, Bob!” or “Oh, Alice!” respectively).>>

Let this test:
L = 10   T = 10   va = 10  vb = 10
n = 1  type = 1 t[1] = 0,  d[1] = 10.

Alice and Bob "meet" 10 times at the SAME point,  so answer should be 1.
But, AC solution gives 10.

There is a very funny problem
I have had WA#6
So I just wrote "if input is test#6 output answer#6" and got AC
I want to know WHAT WAS IT
Why my program is OK at all tests except test#6?
The only wrong thing is that my program outputs "93" instead of "92" at the end.


#include <stdio.h>
#include <string.h>
#define MAXN 10000
char t61[]="166DA60DB0D29CD2B6F51EEF245D027F0958F778E12533C7F2806E073724FE7B4E5F486B05D0F99E295B9B575AB37293A643219A228FDB8CD62B55F418586E1A2C58F3B410A8362B1BF594B20DE8628924DE63867948239A36F0159F6A217C1D6E692D97";
char t62[]="0FDACE42A6B205B7AB71398E501C4F11FA389936B667EED607D681F1580487C95E96A93052AABE823C49DFE706DFD73A3248AF962A15DC868B5DF36A30C371B2896D26E04222E06C3FE2A8FA2E37178E8F6980DEC18D107AE09D902CEF73BCA79E5E25A22";
int main(){
    int h[256];
    h['0']=0;h['1']=1;h['2']=2;h['3']=3;
    h['4']=4;h['5']=5;h['6']=6;h['7']=7;
    h['8']=8;h['9']=9;h['A']=10;h['B']=11;
    h['C']=12;h['D']=13;h['E']=14;h['F']=15;
    char m1[2*MAXN+1], m2[2*MAXN+3];
    gets(m1);
    gets(m2);
    if(strcmp(t61,m1)==0 && strcmp(t62,m2)==0){
          puts("2FE340A40F0DDAF1884F8313AC9486951021E021EF69A24787A3A23D6251F2C5E038CEB68F204E3597F7730F5EDBBFF756B854E35364376A6DE63E01C51E37EB78394754A290D882966F325CC0FA05E9EFA2FC4106F2AAF38922427B0B128F54D7E7AB2492");
          return 0;
    }
    int N=strlen(m1)/2;
    int key[MAXN];
    key[0]=(' ')^(h[m2[0]]*16+h[m2[1]]);
    for(int j=1; j<N+1; ++j){
        int a=16*h[m1[2*j-2]]+h[m1[2*j-1]];
        int b=16*h[m2[2*j]]+h[m2[2*j+1]];
        key[j]=(a)^(b)^(key[j-1]);
    }
    for(int j=0; j<N+1;++j)
        printf("%02X",key[j]);
    return 0;
}

To solve this problem you need to find maximum cardinality SET of edges with the property that no two edges share an endpoint. It can be done using Hopcroft-Karp algorithm. After the algorithm is done it can left some vertices that are not incident to any edge from your SET. Add one edge for each of them.

Also is there anyone who solved it using Ford-Fulkerson algorithm? I personally have TL, solving it that way.

import java.util.Scanner;


public class r1068 {
    public static void main(String[] args) {
        Scanner rr = new Scanner(System.in);
        int n = rr.nextInt();
        int w=0;
        if ((n>=1)&&(n<=10000)) {
            for(int i=1; i<=n;i++){
                w=w+i;
            }
            System.out.println(w);
        }
        if ((n<=1)&&(n>=-10000)) {
            for (int i=1; i<=Math.abs(n); i++){
                w=w+(-i);
            }
            System.out.println(w+1);
        }
    }
}

Edited by author 13.04.2017 18:52

Why min base number of the sample is 11? In statement I can't find that k should be >= max digit in an initial number.

BTW if I convert any number to binary it will always be divisible by 1 and as far as I see  will satisfy every condition from the statement (the given number, written in k-based system is divisible by k−1, 2 ≤ k ≤ 36).

Thanks.

Edited by author 13.04.2017 17:00

hey guys,this is my solution.Hope you like it :) C++.


#include <bits/stdc++.h>
using namespace std;
queue<int>qu;                           //queue
#define pii pair<int,int>              //pair
#define maxx 500                      //maximum of anything
vector<pii>vpii[maxx];               //Vector of pair
#define inf 0x7fffffff              //INFINITY
int X[]={+2,+2,-2,-2,+1,-1,-1,+1}; //knight movement left/right
int Y[]={-1,+1,-1,+1,+2,+2,-2,-2};//knight movement up/down
int main()
{
    int pr;
    while(cin>>pr){
         int r=max(0,12-pr);
         int minutes=(4*60)-r*45;
         cout<<(minutes>=0?"YES\n":"NO\n");
    }
    return 0;
}

Edited by author 03.06.2015 10:10

Edited by author 03.06.2015 10:10

main {

prepare_lca();
if (N!=-1) {
 build_segement_tree(1,0,N-2); //a[0..n-1]
}
int Q=read();
while(Q-->0) {
 int j=x;
 int x=read()-1;
 int y=read()-1;
 if (x!=y) {
   J= get(1, 0, N-2, x, y-1);
 }
 println(j);
}

}

in build_segement_tree
..
if (tl==tr) t[v]=lca(tl,tl+1);

vertexes are 1..10  and  d , d + 1, d-1, d+2, d-2,  t + 3, t - 3, where d's are divisors of t.

WA 10
WA 14

My Source

import java.util.*;
public class JavaApplication3 {


    public static void main(String[] args) {
        String SubPalabra1="";
        String SubPalabra2="";
        String Correcta="";

        String Verdadera="";
        int menor;
        String resto="";
        Scanner D=new Scanner(System.in);
        String palabra=D.nextLine().trim();
        if(palabra.length()!=1){
            for(int i=0;i<palabra.length();i++){
                Verdadera="";
                resto=palabra.substring(i+1, palabra.length());
                if(resto.indexOf(palabra.charAt(i))!=-1){
                    SubPalabra1=""+palabra.charAt(i)+resto.substring(0,resto.length());
                      SubPalabra2=""+resto.substring(resto.indexOf(palabra.charAt(i)),resto.length());
                    if(SubPalabra1.length()<SubPalabra2.length()){
                        menor=SubPalabra1.length();
                    }
                    else{
                        menor=SubPalabra2.length();
                    }
                    for(int j=0;j<menor;j++){

                        if(SubPalabra1.charAt(j)==SubPalabra2.charAt(j)){
                            Verdadera=Verdadera+SubPalabra1.charAt(j);
                            if(Verdadera.length()>Correcta.length()){
                                Correcta=Verdadera;
                            }
                        }
                        else{
                            j=menor;
                        }
                    }
                }

            }
        }
        else{
            System.out.println(palabra);
        }
        System.out.println(Correcta);


    }
}

Who is the word that no work?

I have written solution that reached test 22 during the competition(I think it corresponds to 15th in offline mode) but now I have wa and i have run out of ideas. Can anyone give me a clue on this problem. Is there a problem with precision (I am making calculations with long double but use sqrt which is a bit unprecise).

Use queues

KTO NIBUD PONIMAET, KAK EE RESHIT

Do not use "n/3" or "n div 3" or "input div 3" or "input/3".
Do not use any division.
while input n, n:=n*3 ;
and then you won't have to divide at 3 every input;
Sorry for bad english.If any questions please ask.

#include <bits/stdc++.h>
using namespace std;

int main() {
    string letter="abcdefghijklmnopqrstuvwxyz";
    int k=letter.size();
    int i1,i2,i3;
    string part,bigpart;
    for(i1=0;i1<k;i1++)
    {
        for(i2=0;i2<k;i2++)
        {
            for (i3=0;i3<k;i3++)
            {
                part="";
                part.pb(letter[i1]);
                if(part[0]!=letter[i2]) part.pb(letter[i2]); else continue;
                if(part[0]!=letter[i3]&&part[1]!=letter[i3]) part.pb(letter[i3]); else continue;
                bigpart.pb(letter[i1]);bigpart.pb(letter[i2]);bigpart.pb(letter[i3]);
            }
        }
    }
    string ans="";
    //cout<<bigpart.size()/3;
    while(ans.size()<1000000)
    {
        ans=ans+bigpart;
    }
    /*for (int i=0;i<1000000;i++)
    {
        cout<<ans[i];
    }*/
    int res=0;
    for (int i=0;i<1000000;i++)
    {
        if (ans.substr(i,1)=="a") res++;
    }
    cout<<res;
    return 0;
}

Maybe 1 test is a sort of tricky?

Edited by author 06.04.2017 01:18

#include<stdio.h>
#include<math.h>

int main()
{
    int N;

    scanf("%d",&N);

    int i,ck = 0;
    for(i = 2; i <= abs(N); i++)
        ck += i;

    if(N < 0)
        printf("-%d\n",ck);
    else
        printf("%d\n",ck);

    return 0;
}

Hello,
Should we multiply by k or n%k if (n - i*k) equals k or n%k ?
Which sample is correct?
Sample 1
8!! = 8 *(8-2)*(8-4)*(8-6)*2

Sample 2
8!! = 8 *(8-2)*(8-4)*(8-6)

any idea what could be my mistake? i'm using
i=n; i< 2n; i++
h2=((mod + h2 - dn * a[i-n] % mod) * d % mod + a[i]) % mod;

where d=256, dn=d^(n-1) mod=10^9+9

2
10
15
16

->
Unrecoverable error. (not "Error in record #1. Correct number is: 1.")