if you get WA1 or WA2 try to test:

3
0 0 1 1
0 3 1 4
4 2 4 4
3 1

right answer = 3 1 2
._
|_|
.
...../\
.....\/
._
|_|
.

This is square but not rectangle! ;)
I used simple geometric from

http://hardfire.jino-net.ru/index.php?id=77&cp=40563b52a55df6f5dfc9732f78ae175b

Good luks! :)

Edited by author 29.06.2008 14:30

/******************************************************************************

                              Online C++ Compiler.
               Code, Compile, Run and Debug C++ program online.
Write your code in this editor and press "Run" button to compile and execute it.

*******************************************************************************/

#include <bits/stdc++.h>

using namespace std;

int main()
{
    int N,a[55],b[55],i,c[55],j,d[55],e[55],m,n;
    unsigned long long x,y,arr[55];
    cin>>N;
    for(i=0;i<N;i++){
        cin>>a[i]>>b[i]>>c[i]>>d[i];
        e[i]=i;
    }
    cin>>m>>n;
    for(i=0;i<N;i++){
        x=max(a[i]-m,m-c[i]);
        if(x<0){
            x=0;
        }
        y=max(b[i]-n,n-d[i]);
        if(y<0){
            y=0;
        }
        arr[i]=x*x+y*y;
    }
    for(i=0;i<N;i++){
        for(j=i+1;j<N;j++){
            if(arr[i]>a[j]){
                arr[i]=arr[j];
                e[i]=e[j];
            }
        }
    }
        for(i=0;i<N;i++){
            cout<<e[i]+1<<endl;
        }

    return 0;
}

#include<stdio.h>

int main(){
    int N, n, k, i, a, b, x, res;
    scanf("%d\n", &N);
    for(i=0; i<N; i++){
        scanf("%d %d", &n, &k);
        if(n%k==0){
            x=n/k;
            res=n*(n-x)/2;
            printf("%d\n", res);
        }
        else {
            x=n/k;
            b=n-k*x;
            a=k-b;
            res=((n-x)*a*x+(n-x-1)*(x+1)*b)/2;
            printf("%d\n", res);
        }
    }
    return 0;
}

#include <bits/stdc++.h>
string c[4] = {"88", "68", "06", "16"};
int main() {
    cin >> n;
    if (n > 4) {
        cout << "Glupenky Pierre" << endl;
        return 0;
    }
    for (int i = n - 1; i >= 0; i--)
        cout << c[i] << " ";
}

"68 88" is good, isn't?

n, m = map(int,input().split(' '))
voices = []
p = ['p']
h = 0
for i in range(m):
    voices.append(int(input()))

for i in range(n):
    p.append(0)

for i in voices:
    for c in range(n+1):
        if i == c:
            p[c] += 1

for i in range(1, len(p)):
    h = p[i] / m * 100
    p[i] = str(round(h, 2))
    if len(p[i]) == 4:
        p[i] += '0'
    print(p[i] + '%')
Help. what's wrong

You should add some test.
So my AC programm give wrong answer on this test
20 0 0
-20 0 0
0 0 0
10
---
Write answer is 51.42;
and my answer is 45.11
Please do something!!!

what is there in test 6 ?? Can anyone please provide some test cases. I have tried all test cases provided in earlier discussions, all of these pass. Still getting WA6. Want to jump off some cliff  now, if that might solve this WA6.

just use DSU to check teams, and then build answer

My programm write this Answers:

Test 1
1999
2002

Test 2
1250025
1251521

Test 3
100003
101101
Test 4

1000003
1001001

Test 4
11111
11111
Give me please some tests?



Edited by author 04.12.2006 03:35

DELETED

Edited by author 15.08.2019 10:11

in my accepted program I checked if i <= 10000 but printed i + k, so for test like 6 0 2 it printed 10001 but I think correct answer should be 0

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

typedef long long                   ll;
typedef pair<int,int>               pii;
typedef pair<ll,ll>                 pll;
typedef pair<string , string>       pss;

#define all(x)                      (x).begin(),(x).end()
#define Sort(x)                     sort(all((x)))
#define F                           first
#define S                           second
#define sep                         ' '
#define file_io                     freopen("input.txt" , "r+" , stdin) ; freopen("output.txt" , "w+" , stdout);
#define fast_io                     ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define error(x)                    cerr << "$$$ " << x << '\n';
#define error2(x,y)                 cerr << "#F : " << x << "      #S : " << y << '\n';
#define pb                          push_back

const ll MX5 = 3*1e5+3;
const ll MX6 = 2*1e6+3;
const ll INF = 8e18;
const ll MOD = 1e9 + 7;
const ll MOD2= 998244353;

ll power(ll a, ll b, ll md) {
    return (!b ? 1 : (b & 1 ? a * power(a * a % md, b / 2, md) % md : power(a * a % md, b / 2, md) % md));
}
ll bmm(ll a,ll b){
    return (a%b==0 ? b : bmm(b,a%b));
}

ll A[MX5];


string num(ll n){
    string ans="";
    while(n){
        ans=(char)(n%10+'0')+ans;
        n/=10;
    }
    return ans;
}

string An(ll t,ll n){
    string a=num(t);
    if(t==n){
        return "sin("+a+")";
    }
    if(t%2==1){
        return "sin("+a+"-"+An(t+1,n)+")";
    }
    return "sin("+a+"+"+An(t+1,n)+")";
}
string Sn(ll t,ll n){
    string a=num(n-t+1);
    if(t==1){
        return An(1,1)+"+"+a;
    }
    return "("+Sn(t-1,n)+")"+An(1,t)+"+"+a;
}
int main(){
    fast_io
    //file_io
    ll n;
    cin >> n;
    string answer=Sn(n,n);
    cout << answer;
    return 0;
}

Edited by author 12.08.2019 13:01

#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

const ll MAX = 1e5+10;
const ll ZERO = 0;

#define endl '\n'
#define dokme(x) cout << x ;  return(0);
#define migmig ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

stack < char > stk;

int main(){
    migmig;
    stk.push('.');
    char c[200001];
    cin >> c;
    int n=strlen(c);
    for (int i =0; i < n ; i ++ ){
        if (stk.top()!=c[i]){
            stk.push(c[i]);
        }
        else{
            stk.pop();
        }
    }
    n=stk.size()-1;
    memset(c,0,sizeof(c));
    for (int i = 0 ; i<n ; i ++ ){
        c[n-i-1]=stk.top();
        stk.pop();
    }
    cout << c;
}

Im struggling with it for few days, but absolutely have no idea how to do it!
Any hints??

import java.util.Scanner;

public class Main {

  private static void checking(double a){
        if(Math.sqrt(a)%1==0){
            System.out.printf("%.0f",Math.sqrt(a));
        }else{
            System.out.printf("%.4f",Math.sqrt(a));
        }
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        System.out.print("Сколько чисел хотите ввести: ");
        int a = scanner.nextInt();

        double[] b = new double[a];

        for (int i = 0; i < b.length; i++) {
            System.out.print("Число " + (i+1) + ": ");
            b[i] = scanner.nextDouble();
        }

        for(double c : b){
            System.out.print("Корень " + c + " = ");
            checking(c);
            System.out.println();
        }
    }
}

import math
import re

input_string =  ''

while True:
    line = input()
    if line:
        input_string = input_string + ' ' + line
    else:
        break


input_digits = [int(x) for x in input_string.replace('\n',' ').replace('\t', ' ').split()]
input_digits = input_digits[::-1]

for int in input_digits:
        sqr_root = math.sqrt(int)
        print(format(sqr_root, '.4f'))



#Thanks for the help guys!

2^19 or another where factorized number = 19

IF you have TL, try this test:
18
aasda0sd basdasdasd casdasd
asdasd33a 0b12asdasd ca45sdasd
cdasdas3das dasd0asda2ds e1dasqew
sdsdad4asdsaf dasda0sdasd 0gasdasd
asdasda5sa asdasdasd3b asda0sdasdc
aqwewq6e bqweqwe cqweq0we
cqweqw7e dqweqwe eqw3ewqees
fsadasd8asd dasdasdas0d gasdasd
asdasds9adasa abasdas3dads casdasd12
qweqwew123qea bqweqw3ewqe cweqqwee
csadas12d dasdas0d ed0dsdadsa
fdasdasd33 dasdasds30ad gasasdsad
adasdasd123d badas3ddasda cdasds0adas
asdads455as basdasda2sd casdasdd0sa
cqwewq123e deeqw12e eqw1easda8d
fasd123asd dasdasds3a1dqw gqwe9wqewes
jeqweqw123e eqweq123wewee hwewe9wqq
ldszczx123c ad123sdasdaq qweweq09f

ans: 18

Edited by author 10.08.2019 22:28

Edited by author 10.08.2019 22:28

Edited by author 10.08.2019 22:28

#include<bits/stdc++.h>
using namespace std;
int main()
{
    double a,b,c,d,m,n,l,p,t,s,k;
    cin>>a>>b>>c>>d>>m>>n>>l;
    p=sqrt((a-m)*(a-m)+(b-n)*(b-n));
    t=sqrt((c-m)*(c-m)+(d-n)*(d-n));
    s=max(p,t);
    if(s>l){s-=l;
    }
    else{ s=0;}
    k=min(p,t);
    if(k<=l){
        k=0;
    }
    else if(p==t){
        a=(a+c)/2,b=(b+d)/2;
        k=sqrt((a-m)*(a-m)+(b-n)*(b-n));
        if(k<=l){
        k=0;
    }
    else{
        k-=l;
    }
    }
    else{
    while(abs(p-t)>.001){
        if(p<t){
            c=(a+c)/2,d=(b+d)/2;
            t=sqrt((c-m)*(c-m)+(d-n)*(d-n));
        }
        else{
            a=(a+c)/2,b=(b+d)/2;
            p=sqrt((a-m)*(a-m)+(b-n)*(b-n));
        }
    }
    k=min(p,t);
        if(k<=l){
        k=0;
    }
    else{
        k-=l;
    }
    }
    cout<<fixed<<setprecision(2)<<k<<endl;
    cout<<fixed<<setprecision(2)<<s<<endl;
}

There seems to be some junk at the end of test 27. My solution with "while (read()) solve();" got WA but when I removed it I got OK.