So, when user starts looking for words, my program immediately writes the answer. Maybe that's why it isn't correct. For example:
user -> 2
user -> ab
program -> abc
user -> xz
program -> xzy

We can qsort(n) and enumeration to solve it.
Notice that n is a prime.There are 2500 primes under 32767.
Enumeration can solve this problem.
And every ask have mostly 2 answers.最多2个解

I wrote the solution and had the correct answers on test cases, but i have wa1. Can someone help me or give advice?

This is my solution:

#include <iostream>
#include <iomanip>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <set>
#include <map>
#include <vector>
#include <string>
#include <cstring>
#include <cstdlib>

using namespace std;
#define task


void print(int array[], int n) {
     for (int i = 0; i < n; i++) {
          cout << array[i] << ' ';
     }
}

void optimization() {
     cin.tie(nullptr);
     ios_base::sync_with_stdio(false);
}

int factorial(int x) {
     int a = 1;
     for (int i = 2; i < x + 1; i++) {
          a *= i;
     }
     return a;
}


#ifdef task

int main() {
     optimization();

     int n, m;
     cin >> n >> m;
     int **field = new int *[n];
     for (int i = 0; i < n; i++) {
          field[i] = new int[m];
     }

     for (int i = 0; i < n; i++) {
          for (int j = 0; j < m; j++) {
               cin >> field[i][j];
          }
     }
     int q;
     int k;
     int x = 0; int y = 0;
     int max_drop = -1;
     int part_sum = 0;
     int move_to_right = -1;
     cin >> q;
     for (int quest = 0; quest < q; quest++) {
          cin >> k;
          for (int i = 0; i < n; i++) {
               for (int j = 0; j < m; j++) {
                    move_to_right = k + 1;
                    part_sum = 0;
                    for (int down = i; down < i + k + 1; down++) {
                         for (int right = j; right < j + move_to_right; right++) {
                              if (down < n && right < m) {
                                   part_sum += field[down][right];
                              }
                         }
                         move_to_right--;
                    }

                    if (part_sum > max_drop) {
                         max_drop = part_sum;
                         x = i + 1;
                         y = j + 1;
                    }
               }
          }

          cout << max_drop << ' ' << x << ' ' << y << endl;
          max_drop = -1;

     }

     return 0;
}

#endif

Edited by author 14.11.2019 22:47

In the example 2 2 3, z==5. Why do everybody say that z==2?

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