some Test Please !!!

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace _1489
{
    class Point
    {
        public double x;
        public double y;
        public double z;
        static void Main(string[] args)
        {
            int A, B, C;
            string S = Console.ReadLine();
            string[] SS = S.Split();
            A = Convert.ToInt32(SS[0]);
            B = Convert.ToInt32(SS[1]);
            C = Convert.ToInt32(SS[2]);

            double x1, y1, x2, y2;
            double[] mas = Console.ReadLine().Split(' ').Select(double.Parse).ToArray();
            x1 = Convert.ToDouble(mas[0]);
            y1 = Convert.ToDouble(mas[1]);
            double[] mas2 = Console.ReadLine().Split(' ').Select(double.Parse).ToArray();
            x2 = Convert.ToDouble(mas2[0]);
            y2 = Convert.ToDouble(mas2[1]);

            Point a = new Point();

                // A - x B - y C - z
                // x
                if (x1 <= C) a.x = 0;
                else if (x1 <= C + A) a.x = x1 - C;
                else a.x = A;

                // y

                if (y1 <= B) a.y = B - y1;
                else if (y1 <= B + C) a.y = 0;
                else if (y1 <= 2 * B + C) a.y = y1 - B - C;
                else a.y = B;

                // z
                if (y1 <= B) a.z = 0;
                else if (y1 <= B + C && x1 <= C + A && x1 >= C) a.z = y1 - B;
                else if (x1 <= C && y1 <= 2 * B + C) a.z = x1;
                else if (x1 >= C + A && y1 <= 2 * B + C) a.z = A + C + C - x1;
                else if (y1 >= B + C + B) a.z = 2 * (B + C) - y1;
                else a.z = C;

            Point b = new Point();

                // x
                if (x2 <= C) b.x = 0;
                else if (x2 <= C + A) b.x = x2 - C;
                else b.x = A;

                // y
                if (y2 <= B) b.y = B - y2;
                else if (y2 <= B + C) b.y = 0;
                else if (y2 <= 2 * B + C) b.y = y2 - B - C;
                else b.y = B;

                // z
                if (y2 <= B) b.z = 0;
                else if (y2 <= B + C && x2 <= C + A && x2 >= C) b.z = y2 - B;
                else if (x2 <= C && y2 <= 2 * B + C) b.z = x2;
                else if (x2 >= C + A && y2 <= 2 * B + C) b.z = A + C + C - x2;
                else if (y2 >= B + C + B) b.z = 2 * (B + C) - y2;
                else b.x = C;

            double difx = (a.x - b.x);
            double dify = (a.y - b.y);
            double difz = (a.z - b.z);
            //if (difx * difx + dify * dify + difz * difz > 0)
                Console.WriteLine((Math.Sqrt(difx * difx + dify * dify + difz * difz)));
            //else Console.Write(0);
        }
    }
}

TEST PLEASE

#include <stdio.h>
#include <math.h>
main(void)
{
long long i;
    while(scanf("%lli", &i) != EOF)
           printf("%.4f\n", (double)sqrt(i));
    }

[code deleted]

Edited by moderator 19.11.2019 23:31

public class Problem {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int r = sc.nextInt();
        double l = 0;
        if (n != 1) {
        double x1 = sc.nextDouble();
        double y1 = sc.nextDouble();
        double memoryX = x1, memoryY = y1;
        for (int i = 1; i <= n; i++) {
            if (i == n) {
                l += Math.sqrt((memoryX - x1) * (memoryX - x1) + (memoryY - y1) * (memoryY - y1));
                continue;
            }
            double x = sc.nextDouble();
            double y = sc.nextDouble();
            l += Math.sqrt((x - memoryX) * (x - memoryX) + (y - memoryY) * (y - memoryY));
            memoryX = x;
            memoryY = y;
        }
        }
        l += 2*Math.PI*r;
        System.out.printf("%.2f", l);
        sc.close();
    }

}

Radius and coords can be float
N can be 1

Below test cases helped me to check the program:

2 3
.**|_|**.
**.|_|***
ans: [good test]
POBEDA
1A
2C
1F

5 6
...|_|***
...|_|***
...|_|***
...|_|***
...|_|***
ans:
POBEDA
1A
1C
3A
3C
5A
5C

5 12
...|_|...
...|_|...
...|_|...
...|_|...
...|_|...
ans:
POBEDA
1A
1C
3A
3C
5A
5C
1D
1F
3D
3F
5D
5F


6 6
*..|_|***
...|_|***
...|_|***
...|_|***
*..|_|***
..*|_|***
ans:
POBEDA
1C
2A
3C
4A
5C
6A


6 6
*..|_|***
..*|_|***
...|_|***
...|_|***
*..|_|***
...|_|***
ans:
POBEDA
1C
2A
3C
4A
5C
6A


6 6
*..|_|***
*.*|_|***
...|_|***
...|_|***
*..|_|***
...|_|***
ans:
PORAZHENIE

I didn't get this task.
Is 1001 unlucky number or 1010?

There a[i][j] in [-127..127]  and  n <= 100 -- May be there bitmask tricks, or other tricks????

Can anyone please explain the sample solution given in the problem

Edited by author 09.11.2019 20:00

I have tried everything.

Here are some testcases from my algorithm:
0: 10
1: 1
2: 2
3: 3
4: 4
5: 5
6: 6
7: 7
8: 8
9: 9
10: 25
12: 26
14: 27
15: 35
16: 28
18: 29
20: 45
21: 37
24: 38
25: 55
27: 39
28: 47
30: 56

Some big numbers:
16859136: 267778888
16875000: 3555555589
16934400: 355778889
16941456: 277777789
17006112: 48999999
17010000: 555567899
17146080: 256778999
17150000: 1555557778
17203200: 155788888
17210368: 177777888
17222625: 355579999
17280000: 555568889
17287200: 455777789
17294403: 177777777
17360406: 67799999
17364375: 555577799
17418240: 256788899
17496000: 355588999
17500000: 4555555578
17503290: 257777999
17561600: 455777888
17578125: 5555555559
17635968: 67889999
17640000: 555577889
17647350: 556777777
17694720: 256888889
17714700: 255699999
17718750: 2555555799

long long int minimoentero3(long int num)
{
     long long int nf=result;
     return nf;
}


int main()
{
    long int n;
    cin >> n;

    cout<<minimoentero3(n); <- Here the result is sent.
}

Edited by author 26.10.2017 18:28

#include <bits/stdc++.h>

using namespace std;

int main()
{
    map<string, int> d;
    string s;
    cin >> s;
    string c = "";
    int n = s.size();
    for (int i = 0; i < n; i++)
    {
        c = s[i];
        d[c]++;
        for (int j = i + 1; j < n; j++)
        {
            c += s[j];
            d[c]++;
        }
    }
    cout << d.size();
}

Player with name "HIT"

The answer is "N*(M+1)"

I got time limit in test#5 and I can't find out why

I get TLE on 22 test. What can you advise?

Each number maps to several letters of the alphabet. However, notice that several letters map to a single digit. If you can't go to the mountain, bring the mountain to you. Convert the words found in the dictionary for each case into digits, using the given map. Looking up which digit a letter maps to can be done in constant time if you use the right data structure. Then all that is left is to compare against the phone number given.

I got AC! It's nice problem!
My algo works at O(3*N)=O(N).
1)BFS from any vertex and find the most remote vertex (let V);
2)BFS from V and find the most remote vertex (let U);
3)Create path from U and V; vertex(vertexes) in the middle is answer.

Edited by author 04.08.2015 12:56