#include<iostream>
#include<cmath>
#include<iomanip>
#define pi 3.1415926
using namespace std;

int main(){
    float side,len;
    cin>>side>>len;
    if(len>(side/2)*sqrt(2)){
        cout<<fixed<<setprecision(3)<<side*side;
        return 0;
    }
    if((side/2)>=len){
        cout<<fixed<<setprecision(3)<<len*len*pi;
        return 0;
    }
    float cosx=(side/2)/len;
    float sinx=sqrt(1-cosx*cosx);
    float cosA=2*sinx*cosx;
    float theta=acos(cosA);
    cout<<fixed<<setprecision(3)<<len*len*0.5*theta*4+sqrt(len*len-(side/2)*(side/2))*(side/2)*4;

return 0;
}

qalesan

GOOD peoples Please help me! I have got WA#5! I used DSU!! My program gives right answer for all my tests! But again WA#5 and I can not found my mistake! I shall wait advices or tests! Thank you advance!!!

just consider this condiction:
2
A\A
B\A
the correct output is
A
 A
B
 A
but if you are mistaken, you will output
A
 A

Please help me. TLE #39

I think that most people trying to solve this problem will try to deal with two problems, they are:
1. How to calculate a[n] fast?
2. How to find those n that a[n] is the maximum in [1...n]?
Here is my method.

1. Calculating a[n] (without calculating a[1...n-1]).
Directly calculation will lead to a long time.
But, after some observation, I found that:
a[4n] = a[n]
a[4n + 1] = a[2n] + a[2n + 1] = 2a[n] + a[n + 1]
a[4n + 2] = a[n] + a[n + 1]
a[4n + 3] = a[n] + 2a[n + 1]
So, a[4n] to a[4n + 3] could be presented as the linear form of a[n] and a[n + 1].
We will soon realise that, a[2^k n] to a[2^k n + (2^k - 1)]  could also be presented as linear form of a[n] and a[n + 1].
So, pre-calculation is done to calculate all the linear forms of a[65536n] to a[65536n + 65535]. This will take about O(65536 * 2) time.
After this pre-calc, calculation for any number K will go very fast.

2. Finding the maximum.
After listing some of the 'maximum index', we found that:
1
3
5
9
11
19
21
35
37
43
......
Maybe my math is not so good, so I could not find so useful properties in the array, but I noticed one thing:
------------------------------------------------------------------------------------------
Every 'maximunm index' could be presented as its max 2-power and some former 'maximum index'.
------------------------------------------------------------------------------------------
Sorry for my poor English, let's take a look at the examples:
1
3 = 2 + 1
5 = 4 + 1
9 = 8 + 1
11 = 8 + 3
19 = 16 + 3
21 = 16 + 5
35 = 32 + 3
37 = 32 + 5
43 = 32 + 11
......
So, we just need to use every 2^k number to ITERATE the list of 'maximum index'. As the list is of size O(Log N), the algo runs quite fast.

With above methods, I got AC in 0.046 sec.

Well, if my algo is too stupid for you, don't hesitate to post your algo here.

Have fun and good luck.

 #include <iostream>
using namespace std;

char paper[4][4], pass[4][4],rPass[16],TempPaper[4][4];
int v, i, j;

void check(char paper[4][4], char pass[4][4])
{
    for (int j = 0; j < 4; j++)// X/.
    {
        for (int i = 0; i < 4; i++)
        {
            if (paper[j][i] == 'X')
            {
                rPass[v] = pass[j][i];
                v++;
            }

        }
    }
}

int main()
{
    for (int j = 0; j < 4; j++) //paper cin
    {
        for (int i = 0; i < 4; i++)
        {
            cin >> paper[j][i];
        }
    }

    for (int j = 0; j < 4; j++)//зашифрованый пароль
    {
        for (int i = 0; i < 4; i++)
        {
            cin >> pass[j][i];
        }
    }
    for (int j = 0; j < 4; j++)//temp
    {
        for (int i = 0; i < 4; i++)
        {
            TempPaper[j][i] = paper[j][i];
        }
    }
    check(paper, pass);
    for (i = 0; i < 4; i++)
    {
        for (j = 0; j < 4; j++)
        {
            int t=0, y;
            if (j == 0)
            {
                int u = 3;
                for (y = 0; y < 4; y++)
                {
                    paper[t][u] = TempPaper[0][y];
                    t++;
                }
            }
            if (j == 1)
            {
                int u = 2;
                for (y = 0; y < 4; y++)
                {
                    paper[t][u] = TempPaper[1][y];
                    t++;
                }
            }
            if (j == 2)
            {
                int u = 1;
                for (y = 0; y < 4; y++)
                {
                    paper[t][u] = TempPaper[2][y];
                    t++;
                }
            }
            if (j == 3)
            {
                int u = 0;
                for (y = 0; y < 4; y++)
                {
                    paper[t][u] = TempPaper[3][y];
                    t++;
                }
            }
        }
        check(paper, pass);
        for (int j = 0; j < 4; j++)
        {
            for (int i = 0; i < 4; i++)
            {
                TempPaper[j][i] = paper[j][i];
            }
        }
        }
    return 0;
}

It contains more than 20 zeroes.

Totally discouraged with it. Please give some test.

I used a simple algo... I am just checking 3 consecutive numbers... If any 3 consecutive numbers from the given numbers are like this, n1>n3>n2 (mid one is the smallest, first one is the largest and the last one lies in between these two values) , then the sequence is invalid (Cheater) ... else , I print "Not a proof" ...

Is my algo ok? If so, then please help me with some test cases...
I have tried all the test cases found in the discussions and my program passed all of them...

#include<bits/stdc++.h>

using namespace std;

int main(void)   {

    int t_h,s_h,pb;


    while(cin>>pb)   {
     s_h=0,t_h=0;

    if(pb>=1 && pb<=11)    {

    s_h=12-pb;
    t_h=s_h*45;

   if(t_h<=240)                    cout<<"Yes"<<endl;
   else                            cout<<"No"<<endl;

    }

 }



return 0;

}

#include <stdio.h>
int main() {
unsigned long long int a;
scanf("%llu", &a);
if (a/7!=0)
printf("%llu\n", a%7);
else if (a < 7)
printf("%d\n", a); }

I`m sure that written algo is correct, but I might make stupid mistake...
My solutions can pass all my tests and my friends` also.

Please, give me some tricky tests!

Thanks a lot!

I keep getting non-zero return from WA #9 (meaning some of my scanf() fails)
Could you please forward WA #9 problem so that I can check parsing is proper?
Thank you,

import java.io.*;
import java.util.*;

public class task {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);
        int n = in.nextInt();
        int k = in.nextInt();
        int t;

        if(n>=1 && k<=1000) {
            if((int) (n * 2.0 / k) == (n * 2.0 / k)) t = (int) (n * 2.0 / k);
            else t = 1 + (int) (n * 2.0 / k);
            out.print(t);

        }
        out.flush();
   }
}

#include <iostream>
#include <stdio.h>
#include <vector>

using namespace std;

class Place {
public:
    char x;
    int y;

    bool isOnboard() {
        if (x < 'a' || x > 'h' || y < 1 || y > 8) return false;
        return true;
    }



    Place(char x, int y) {
        this->x = x;
        this->y = y;
    }

};



int main(int argc, const char * argv[])
{
    int N;
    cin >> N;


    vector<Place> places;

    for (int i = 0; i < N; i++) {
        char x;
        int y;
        cin.get(x);

        if (x == '\n') {
            i--;
            continue;
        }

        cin >> y;

        Place place (x, y);
        places.push_back(place);

    }


    for (int i = 0; i < N; i++) {
        Place place = places[i];
        int moves = 0;


        //left 2 up 1
        place.x -= 2;
        place.y -= 1;
        moves += place.isOnboard();

        //left 2 down 1
        place.y += 2;
        moves += place.isOnboard();

        //right 2 down 1
        place.x += 4;
        moves += place.isOnboard();
        //right 2 up 1
        place.y -= 2;
        moves += place.isOnboard();

        //right 1 up 2
        place.x -= 1;
        place.y -= 1;
        moves += place.isOnboard();

        //left 1 up 2
        place.x -= 2;
        moves += place.isOnboard();

        //left 1 down 2
        place.y += 4;
        moves += place.isOnboard();

        //right 1 down 2
        place.x += 2;
        moves += place.isOnboard();

        cout << moves << endl;

    }



}

I get, Runtime error (non-zero exit code), even if I send a program such:

fn main() {
    println!("{}",-1);
}

Когда использую циклы такого вида выдает ошибку Runtime error (non-zero exit code)

i=n;
while(i>0) {
    name[i]=calloc(1,30);
    scanf(" %s",name[i]);
    i--;
}
i=n;
while(i>0) {
    char *buf=name[i];
    while(*buf!='\0') {
        if(*buf=='+') {
            kol++;
            break;
        } else {
            buf++;
        }
    }
    i--;
}

А здесь AC
for(i=0;i<n;i++) {
    name[i]=calloc(1,30);
    scanf(" %s",name[i]);
}
for(i=0;i<n;i++) {
    char *buf=name[i];
    while(*buf!='\0') {
        if(*buf=='+') {
            kol++;
            break;
        } else {
            buf++;
        }
    }
}

try
6
6 0
3 0
5 0
5 6 0
0
0

and

7
2 0
5 0
6 4 0
5 7 0
0
0
0

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;

public class Lesson {

    public static void main(String[] args) throws Exception {

        BufferedReader bufferedReader = new BufferedReader(new InputStreamReader(System.in));
        int countPrepod = Integer.parseInt(bufferedReader.readLine());
        TreeSet<Long> prepod = new TreeSet();
        for (int i = 0; i < countPrepod; ++i) {
            prepod.add(Long.parseLong(bufferedReader.readLine()));
        }

        int countStudent = Integer.parseInt(bufferedReader.readLine());
        ArrayList<Long> student = new ArrayList();
        for (int i = 0; i < countStudent; ++i) {
            student.add(Long.parseLong(bufferedReader.readLine()));
        }

        student.sort(((o1, o2) -> Long.compare(o1,o2)));
        int result=0;
        for (long find:prepod)
        {
                while (Collections.binarySearch(student,find)>=0)
                {
                    ++result;
                    student.remove(find);
                }
        }
        System.out.println(result);
    }

}