WA6?

Edited by author 21.01.2019 19:40

If it is a real tree, why after N lines, describing conviviality ratings of employees,
we have not simply (N - 1) lines, describing supervisor relations?

Does such input format mean that some employees may have more than one supervisors?

Edited by author 21.06.2012 18:08

Why do I memory limit exceeded even if I have use pointer?
Here is my code:

const mn=6000;
      inf=-100000000;
type rec=record
           father:integer;
           num:integer;
           child:array[1..mn]of ^integer;
         end;
var value:array[1..mn]of ^integer;
    f:array[1..mn,1..2]of ^longint;
    t:array[1..mn]of ^rec;
    n,i,a,b,root:integer;
    ans:longint;

function max(a,b:longint):longint;
  begin
    if a>b
      then exit(a)
      else exit(b);
  end;

function dp(root,lab:integer):longint;
  var i:integer;
  begin
    if root=0 then exit(0);
    if f[root,lab]^>inf then exit(f[root,lab]^);
    if lab=1
      then begin
        f[root,lab]^:=value[root]^;
        for i:=1 to t[root]^.num do
          inc(f[root,1]^,dp(t[root]^.child[i]^,2));
      end
      else begin
        f[root,lab]^:=0;
        for i:=1 to t[root]^.num do
          inc(f[root,2]^,max(dp(t[root]^.child[i]^,1),dp(t[root]^.child[i]^,2)));
      end;
    exit(f[root,lab]^);
  end;

begin
  readln(n);
  for i:=1 to n do
    begin
      new(value[i]);
      new(t[i]);
      new(f[i,1]);
      new(f[i,2]);
      readln(value[i]^);
    end;
  while not eof do
    begin
      readln(a,b);
      if(a=0)and(b=0)then break;
      t[a]^.father:=b;
      inc(t[b]^.num);
      new(t[b]^.child[t[b]^.num]);
      t[b]^.child[t[b]^.num]^:=a;
    end;
  for i:=1 to n do
    if t[i]^.father=0
      then begin
        root:=i;
        break;
      end;
  for i:=1 to n do
    begin
      f[i,1]^:=inf;
      f[i,2]^:=inf;
    end;
  ans:=max(dp(root,1),dp(root,2));
  writeln(ans);
end.

Who can HELP me?

Edited by author 05.04.2010 16:57

Some test cases that helped me:

6 4
001000 100001
000100 000010
010000 100100
100000 111000
100000
ans:
111111


6 4
010000 100000
001000 100001
100000 111000
000100 000001
100000
ans:
111001


6 4
100000 111000
010000 100000
001000 100001
000100 000001
000000
ans:
000000


6 4
111100 111000
111110 100000
000000 100001
111111 000001
010100
ans:
110101


6 4
111000 101000
111100 111000
111110 100000
111111 000001
011100
ans:
011100


6 0
010100
ans:
010100

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

WHY answer is:
2
1 5
5 9
I think [2;6] and [3;7] have common interior point such as 3, 4, 5, 6.

What's the meaning of "Land fragments that are adjacent to the map's
border are not considered as islands." ???

I can hardly understand it ! in the Example input, There is two
island adjacent to the map's border(the north-east corner and the
south-west corner) . but the answer is 3 !!! can anyone help me ?

/\/\/
\/\/\
/\*\/
\/\/\
/\/\/
ans: FAIL

\/\/\
\/\/*
//\//
\\\//
\/\\/
ans: U2

/\///
\/\\/
/\\*\
\/\//
/\/\\
ans: WIN

\\///
//\//
*\\\\
\\\/\
//\//
ans: U1

\/\//
\/\\\
/////
*\\\\
\//\/
ans: WIN

so this is a NP complete problem ,how can it be solved with 1000 range.. I don't know

Can anybody tell me test 5,i am getting WA.I have tried my code on many random cases and it works fine.

Code:-

#include<bits/stdc++.h>
#define PB push_back
#define MP make_pair
#define F first
#define S second
#define SZ(a) (int)(a.size())
#define SET(a,b) memset(a,b,sizeof(a))
#define LET(x,a) __typeof(a) x(a)
#define TR(v,it) for( LET(it,v.begin()) ; it != v.end() ; it++)
#define loop(i,a,b) for(int i=a;i<b;i++)
#define si(n) scanf("%d",&n)
#define sll(n) scanf("%lld",&n)
#define sortv(a) sort(a.begin(),a.end())
#define all(a) a.begin(),a.end()
#define bitcount(n) __builtin_popcount(n)
#define DRT()  int t; cin>>t; while(t--)
#define TRACE
#ifdef TRACE
#define trace1(x)                cerr << #x << ": " << x << endl;
#define trace2(x, y)             cerr << #x << ": " << x << " | " << #y << ": " << y << endl;
#define trace3(x, y, z)          cerr << #x << ": " << x << " | " << #y << ": " << y << " | " << #z << ": " << z << endl;
#define trace4(a, b, c, d)       cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << endl;
#define trace5(a, b, c, d, e)    cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << endl;
#define trace6(a, b, c, d, e, f) cerr << #a << ": " << a << " | " << #b << ": " << b << " | " << #c << ": " << c << " | " << #d << ": " << d << " | " << #e << ": " << e << " | " << #f << ": " << f << endl;
#else
#define trace1(x)
#define trace2(x, y)
#define trace3(x, y, z)
#define trace4(a, b, c, d)
#define trace5(a, b, c, d, e)
#define trace6(a, b, c, d, e, f)
#endif
using namespace std;
typedef long long int lli;
typedef pair<int,int> ii;
typedef vector<int> vi;
typedef vector< vi > vvi;
typedef vector< ii > vii;
lli modpow(lli a,lli n,lli temp){lli res=1,y=a;while(n>0){if(n&1)res=(res*y)%temp;y=(y*y)%temp;n/=2;}return res%temp;}
//***********************************END OF TEMPLATE*********************************************************************
const int MAX = 50005,MAX2=log2(MAX);
int d[MAX],h[MAX],P[MAX][MAX2],n;
vector<vii> graph(MAX);
void dfs(int u,int p){
    P[u][0]=p;
    h[u]=h[p]+1;
    for(auto k:graph[u]){
        if(k.F==p)d[u]=d[p]+k.S;
        else dfs(k.F,u);
    }
}
void init(){
    for(int i=0;i<n;++i){
        for(int j=0;(1<<j)<n;++j)P[i][j]=-1;
    }
    d[0]=0;
    h[0]=-1;
    dfs(0,0);
    for(int j=1;(1<<j)<n;++j){
        for(int i=0;i<n;++i){
            if(P[i][j-1]!=-1)P[i][j]=P[P[i][j-1]][j-1];
        }
    }
}
int lca(int u,int v){
    if(h[u]>h[v])swap(u,v);
    int l=log2(h[v]);
    for(int i=l;i>=0;--i){
        if(h[v]-(1<<i)>=h[u]&&P[v][i]!=-1)v=P[v][i];
    }
    //u and v at same level
    if(u==v)return u;
    for(int i=l;i>=0;--i){
        if(P[u][i]!=P[v][i]&&P[u][i]!=-1){
            u=P[u][i];
            v=P[v][i];
        }
    }
    return P[u][0];
}
int query(int u,int v){
    int lc = lca(u,v);
    return d[u]+d[v]-2*d[lc];
}
int main(){
    int u,v,w,q;
    si(n);
    loop(i,1,n){
        si(u);si(v);si(w);
        graph[u].PB(MP(v,w));
        graph[v].PB(MP(u,w));
    }
    init();
    si(q);
    while(q--){
        si(u);si(v);
        printf("%d\n",query(u,v));
    }
    return 0;
}

Edited by author 17.01.2016 01:13

wa#4 for countless times, seek for kind help!
thanks

var a,b,c,i,k:real;
begin
read(a,b,c);
while a>b do
begin
a:=a-(a*c/100);
I:=i+1;
end;
write(i);
end.

Here is the code (python 3.6). Can anyone explain, why it fails on test 9?


import math
string = input().split()
start = int(string[0])
end = int(string[1])
commission = int(string[2])
if start <= end:
    n = 0
else:
    q = 1 - commission / 100
    n = math.log((end / start), q)
    n = round(n)
print(n)

1) Find the prime numbers up to 10,000,100 (Sieve of Eratosthenes)
2) if the sum is equal to 19 degrees, then the residue is prime to check
What can you advise?

In Idea my code works, but here is a runtime error:

import java.util.Scanner;

public class MathematiciansAndBerries {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        byte a1 = scanner.nextByte();
        byte b1 = scanner.nextByte();
        byte a2 = scanner.nextByte();
        byte b2 = scanner.nextByte();
        byte a3 = scanner.nextByte();
        byte b3 = scanner.nextByte();

        System.out.println((a1 - a3) + " " + (b1 - b2));
    }
}

Help pls, why runtime error occurs?

Edited by author 06.12.2018 21:47

I think I can not get AC because of "Enter"  key on Windows  is "\r\n"  and C++ scanf()  reads it as "\n"  and printf()  writes "\n" as "\r\n".
So,  if in input is some "\n",  then I would read it as "\n" and then print it as "\r\n"!!!  I don't know how to fix that.
Also there were problems with EOF-character.

The test is empty. There is nothing there.

I really don't understand why my solution is wrong but here it is:

letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
start = 0
s = input()

for i in range(0, len(s)):
    if not(s[i] in letters):
        word = s[start:i]
        other = s[i:i+1]

        if (word != ""):
            print(word[::-1], end="")

        print(other, end="")
        start = i+1

if (start != len(s)):
    print(s[start:len(s)][::-1], end="")

What's wrong with this solution?
WA (Test 2)

Edited by author 30.03.2013 23:10

верно ли,что при n=48,ответ  793566522023068987143383410279408208612410000
а при n=100 3834688188199309756675466732967588185862422492337212276545580039517258085033387305291001857619600

Hello.

I have received strange 'Compilation Error' verdict for my solution on the 1314 task under Microsoft Visual C++ 2017 compiler.

Compilation log is as follows:
/////////////////////////////////////////
lg1vr1-w2nnli(24): error C2062: type 'float' unexpected
lg1vr1-w2nnli(84): warning C4244: 'argument': conversion from 'float' to 'const int', possible loss of data
lg1vr1-w2nnli(94): warning C4018: '<': signed/unsigned mismatch
/////////////////////////////////////////

Lines 23 and 24 of my code are:
/////////////////////////////////////////
const int MAX_STATION_ID = 32767L;
const int INFINITY = 10L * MAX_STATION_ID;
/////////////////////////////////////////


Commits with CE are 8162255, 8162250, 8162248.

But there is successful (Accepted) commit 8162257 (the same code as 8162255) under G++7.1 commit.

Could you check if the CE message was correct or not?

Best regards.

Edited by author 04.12.2018 02:01