есть ли способы на питоне обойти прожорливость памяти?

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?