tovars,prices,recomendations= [],[],[]
for i in range(6):
    name = input()
    tovar = input()
    price = int(input())
    if tovar in tovars:
        if prices[tovars.index(tovar)] > price:
            prices[tovars.index(tovar)] = price
        recomendations[tovars.index(tovar)] +=1
    else:
        tovars.append(tovar)
        prices.append(price)
        recomendations.append(1)
if recomendations.count(max(recomendations)) == 1:
    print(tovars[recomendations.index(max(recomendations))])
else:
    ans = []
    ans1 = []
    l = len(tovars)
    mm = max(recomendations)
    for i in range(l):
        if recomendations[i]  == mm:
            ans.append(tovars[i])
            ans1.append(prices[i])
    print(ans[ans1.index(min(ans1))])

Could you please add Common Lisp as a language? SBCL is the most popular, and most performant, implementation of Common Lisp, and has been for some time (forked from CMUCL nearly 20 years ago)! It is also mostly maintained by Stas Boukarev. It has static typing, and it would be a beautiful language to solve Timus problems in. Common Lisp has been around for a very long time and its ANSI Standard has been the same for quite a number of years, so it is possible that there would not need to be many updates after it is added once.

Adding the SBCL implementation as a supported language would be greatly appreciated by me and the Common Lisp community to be able to solve the amazing problems on this website
with this very performant meta programming language. While attempting to get SBCL added, this repo could be used for reference -- https://github.com/optimisticlisper

I really hope you will consider it.

#include<stdio.h>
#include<math.h>
int main()
{
    int  n;

    printf("%d\n",scanf("%d",&n)*n*scanf("%d",&n)*n*scanf("%d",&n)*n*2);
}

At first,I got wa on test1,like many other authors.But later,I found it!
All in one sentence,test the data below:


1
inputon

Create struct:

struct Animal which has:
- ID // ID of hero
- name // name of animal (char *) (cin.getline) (cin.ignore())
- strength(int) // power of animal
- frequency // number of hits in one iteration
- HP // amount of life

Your task is to show animal which won most rounds in the game.

Input:
n (how many animals) [2 <= n <= 10]
n-times:
ID
name
strength
frequency goals
HP goals
Output:
winner with amount of victory.

Example:

Input:
10
1 Lion John 30 2 100 // (0 1 0 0 1 0 1 0 0) 3 victory
2 Hen Rebecca 60 1 200 // (1 1 1 0 1 0 1 0 1) 6 victory
3 Tiger Alex 40 2 100 // (1 0 0 0 1 0 1 0 1) 4 victory
4 Buffalo James 40 3 100 // (1 1 1 0 1 0 1 0 1) 6 victory
5 Elephant Jack 50 1 300 // (1 1 1 1 1 0 1 0 1) 7 victory
6 Rhino Michael 30 1 200 // (1 0 0 0 0 0 1 0 0) 2 victory
7 Snake Oz 50 2 400 // (1 1 1 1 1 1 1 1 1) 9 victory
8 Cow Bettie 40 1 100 // (0 0 0 0 0 0 0 0 0) 0 victory
9 Crocodile Nick 20 3 500 // (1 1 1 1 1 1 0 1 1) 8 victory
10 Bear Drake 40 2 150 // (1 1 1 1 0 1 0 1 0) 6 victory

Output :
7 Snake Oz 50 2 400 9 victory

use "long double"
and to avoid rounding when you print the result, use
cout << fixed <<setprecision(0)<< result << endl;

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?