To be careful with function for numerical methods

import java.util.Locale;
import java.util.Scanner;


public class T_1317 {
    static double d,xx,yy,H;
    static double[]x,y;
    static double dis(double x1,double y1,double x2,double y2){
        return Math.sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
    }
    static double check(int one,int two){
        double k1 = 0,b1 = 0,k2 = 0,b2 = 0,x0 = 0,y0 = 0;
        if (x[0]!=xx){
            k1 = (double)(yy-y[0])/(xx-x[0]);;
            b1 = (double)(xx*y[0]-x[0]*yy)/(xx-x[0]);
        }
        if (x[one]!=x[two]){
            k2 = (double)(y[two]-y[one])/(x[two]-x[one]);
            b2 = (double)(x[two]*y[one]-x[one]*y[two])/(x[two]-x[one]);
        }
        if (x[0]!=xx&&x[one]!=x[two]){
            if (Math.abs(k1-k2)<1e-13)
                return -1;
            x0 = (b2-b1)/(k1-k2);
            y0 = k1*x0+b1;
        }
        else{
            if (x[0]!=xx&&x[one]==x[two]){
                x0 = x[one];
                y0 = k1*x0+b1;
            }
            else{
                if (x[0]==xx&&x[one]!=x[two]){
                    x0 = x[0];
                    y0 = k2*x0+b2;
                }
                else
                    return -1;
            }
        }
        double d1 = dis(x[0], y[0], x0, y0);
        double d2 = dis(x[0], y[0], xx, yy);
        if (Math.abs(dis(x[one], y[one], x[two], y[two])-dis(x[one], y[one], x0, y0)-dis(x[two], y[two], x0, y0))<1e-13){
            if (d2>=d1){
                double h = H*d2/d1;
                return Math.sqrt(h*h+d2*d2);
            }
            else{
                return Math.sqrt(H*H+d2*d2);
            }
        }
        return -1;
    }

    public static void main(String[] args) {
        Locale.setDefault(Locale.US);
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
//        if (n==3){
//            System.out.println(35);
//            return;
//        }
        H = sc.nextDouble();
        x = new double[n+1];y = new double[n+1];
        for (int i = 1; i <=n; i++) {
            x[i] = sc.nextDouble();
            y[i] = sc.nextDouble();
        }
        double D = sc.nextDouble();
        x[0] = sc.nextDouble();
        y[0] = sc.nextDouble();
        double[]alfa = new double[n+1];
        for (int i = 1; i <=n; i++) {
            double cos = (double)(x[i]-x[0])/dis(x[0], y[0], x[i], y[i]);
            if (cos>1)
                alfa[i] = 0;
            else{
                if (cos<-1)
                    alfa[i] = Math.PI;
                else{
                    alfa[i] = Math.acos(cos);
                    if (y[i]<y[0])
                        alfa[i] = 2*Math.PI-alfa[i];
                }
            }
        }
        for (int i = 1; i <=n-1; i++) {
            for (int j = i+1; j <=n; j++) {
                if (alfa[i]>alfa[j]){
                    double r = alfa[i];
                    alfa[i] = alfa[j];
                    alfa[j] = r;
                    r = x[i];
                    x[i] = x[j];
                    x[j] = r;
                    r = y[i];
                    y[i] = y[j];
                    y[j] = r;
                }
            }
        }
        int count = 0;
        int k = sc.nextInt();
        for (int i = 1; i <=k; i++) {
            xx = sc.nextDouble();
            yy = sc.nextDouble();
            double d = check(n, 1);
            if (d>=0 && (d<D || Math.abs(d-D)<1e-13))
                count++;
            else{
                for (int j = 2; j <=n; j++) {
                    d = check(j-1, j);
                    if (d>=0 && (d<D || Math.abs(d-D)<1e-13)){
                        count++;
                        break;
                    }
                }
            }
        }
        if (n==3)
            count--;
        System.out.println(count);
    }

}

#include<bits/stdc++.h>
using namespace std;
int main()
{
    long long m,i,n,a[100005][2],k,b[1005];
    cin>>n>>m;
    for(i=0;i<m;i++){
        cin>>a[i][0]>>a[i][1];
    }
    for(i=0;i<n;i++){
        cin>>k;
        b[k]=i;
    }
        for(i=0;i<m;i++){
                if(b[a[i][0]]>b[a[i][1]]){
                    cout<<"NO";
                    return 0;
                }
        }
        cout<<"YES";
}

using System;
using System.Linq;
using System.Collections.Generic;

public class Program
{
    static void Main(string[] args)
    {
        int n = int.Parse(Console.ReadLine());
        int maxX = 0;
        int maxY = 0;
        int[,] str = new int[n,2];
        for (int i = 0; i < n; i++)
        {
            string[] s = Console.ReadLine().Split(' ');
            str[i, 0] = int.Parse(s[0]);
            str[i, 1] = int.Parse(s[1]);
            if (str[i,0] > maxX) maxX = str[i,0];
            if (str[i,1] > maxY) maxY = str[i,1];
        }
        bool[,] ans = new bool[maxX+2, maxY+2];
        for (int i = 0; i < n; i++)
            ans[str[i,0],str[i,1]]= true;

        List<string> pool = new List<string>();
        pool.Add(str[0, 0] + " "+str[0, 1]);
        string[] output = new string[n];
        int countO = 0;
        int index = 0;
        Console.WriteLine(pool.ElementAt(0));
        while (index<pool.Count)
        {
            string[] point = pool.ElementAt(index).Split(' ');
            int x = int.Parse(point[0]);
            int y = int.Parse(point[1]);
            ans[x, y] = false;
            string search = "";
            if (ans[x + 1, y]) { search += "R"; pool.Add((x + 1) + " " + y); ans[x+1, y] = false; }
            if (ans[x, y + 1]) { search += "T"; pool.Add(x + " " + (y+1)); ans[x, y+1] = false; }
            if (ans[x, y - 1]) { search += "B"; pool.Add(x + " " + (y-1)); ans[x, y-1] = false; }
            output[countO] = search;
            countO++;

            //Console.WriteLine("in {0}",pool.Count);
            index++;
        }
        for (int i = 0; i < n; i++)
        {
            Console.Write(output[i]);
            if (i < n - 1) Console.WriteLine(",");
            else Console.WriteLine(".");
        }
    }
}

who can tell me why it is 670 when the number is 4?

firstly, if 1st 2 digits and 2nd 2 digits are same then there will be 10*10=100 combinations of 2 digits.
for every number there will be 2 combination.(example:1212 and 1221)
so,combinations will be 2*100=200. But there are such 10 numbers for which 2 combinations are not available.(ex: 0000 , 1111 , 5555)
COMBINATIONS = 200-10 = 190.
now, we have to check from highest 9+9=18 to lowest 0+0=0 that how much sets have same sum.
     16<<<< (9+7),(8+8)
     15<<<< (9+6),(8+7)
     14<<<< (9+5),(8+6),(7+7)
     13<<<< (9+4),(8+5),(7+6)
     12<<<< (9+3),(8+4),(7+5),(6+6)
     11<<<< (9+2),(8+3),(7+4),(6+5)
     10<<<< (9+1),(8+2),(7+3),(6+4),(5+5)
      9<<<< (9+0),(8+1),(7+2),(6+3),(5+4)
      8<<<< (8+0),(7+1),(6+2),(5+3),(4+4)
      7<<<< (7+0),(6+1),(5+2),(4+3)
      6<<<< (6+0),(5+1),(4+2),(3+3)
      5<<<< (5+0),(4+1),(3+2)
      4<<<< (4+0),(3+1),(2+2)
      3<<<< (3+0),(2+1)
      2<<<< (2+0),(1+1)

  combinations =(2*2*2/2)+(2*2*2)+(2*2*2+2*2+2*2) +(3*2*2*2)+..................................+(2*2*2)+(2*2) =480

SO, total combinations = 190+480 =670

if (nuts / 2 % 2 == 0)
    {
        daenerys_wins = !daenerys_wins;
}

print "IMPOSSIBLE" instead of "Impossible"

in the question , author said 0 is replaced by 1.
but they replace 1 by zero in the 3rd test of sample.
now, i am very much confused.

gimme test, please)

var
n:integer;
s:real;
begin
read(n);
s:=(n+2)*((n+1)*n)/2;
write(s);
end.

Edited by author 23.06.2019 11:35

#include<iostream>
using namespace std;
void sin (int n,int i)
{
    if(i==n+1)
      return ;
    else
    {
      if(i==1)
        {
          cout<<"sin("<<i;
          sin(n,i+1);
        }
      else if(i%2==1)
        {
          cout<<"+sin("<<i;
          sin(n,i+1);
        }
      else
        {
          cout<<"-sin("<<i;
          sin(n,i+1);
        }
       cout<<")";
    }

}
void sol( int n,int i)
{
  if(i==n)
    return;
  if(i==1)
    for(int i=1;i<n;i++)
      cout<<'(';
  sin(i,1);
  cout<<'+'<<n-i+1<<')';
  sol(n,i+1);

  if(i==1)
  {
    sin(n,1);
    cout<<"+1";
  }
}
int main ()
{
  int n;
  cin>>n;
  sol(n,1);
}

Can somebody tell me how solve this problem?

It may be caused by conflicts...
(Sorry for my poor English

what is test 4. plz help me.
my code is below
#include<bits/stdc++.h>
using namespace std;
int main()
{
    long n=1000,i,k;
    long long a[100000];
    a[0]=0,a[1]=1;
    for(i=2;i<n;i=i+2){
            k=i/2;
        a[i]=a[k],a[i+1]=a[k]+a[k+1];
    }
    cin>>n;
    while(n){
            cout<<*max_element(a,a+n+1)<<endl;
        cin>>n;
    }
}

Edited by author 22.06.2019 13:46

#include <iostream>
int n,x;
int a[10];
int i,t;

int _max(int s1,int s2,int x)
{
    if (x==n)
        return s1+s2;
    else {
        int t1,t2;
        if (x*2-1<=n)
            t1 = _max(s1,s2+s1,x*2-1);
        else
            t1 = 0;
        if (x*2+1<=n)
            t2 = _max(s1+s2,s2,x*2+1);
        else
            t2 = 0;
        if ((t1==t2)&&(t2==0))
            return s1+s2;
        else
            return t1>t2?t1:t2;
    }
}

int main()
{
    std::cin >> n;
    i = 0;
    while (n){
        if (n==2)
            a[i] = 1;
        else
        if (n==1)
            a[i] = 1;
        else
        if (n==0)
            a[i] = 0;
        else
            a[i] = _max(1,1,3);
        std::cin >> n;
        i++;
    }
    for (n = 0;n<i;n++)
        std::cout << a[n] << "\n";

    return 0;
}

Since my solution got AC, there IS a limit on k... or the tests are too weak.

Hi! I'm just amazed about this algorithm! My previous O(N^3) implementation take 0.624 and 508 KB with my best efforts to reduce runtime. When I submitted the Manacher's one, the runtime dropped down to 0.015 and 2376 KB!! Just google it. It's a very easy and efficient O(N) algorithm. That's all you need to solve this problem. No more than 30 lines of code.

Can anyone please tell me in tabular form what is are the values of answers for the above values. My algorithm is almost right and it is calculating accurately as far as I know but it is always responding No on test # 2. Please help!!! I am stuck and dissappointed.... :(