#include <iostream>

int main()
{
    int a=1+5;
    std::cout<<a;
}

I killed a lot of time writing the solution in C++, Python, Java. Java gives AC with BigInteger -_-
It is NOT solvable with int64

What answer on this test?

0 3 1 1 1 0 3 0

Edited by author 04.01.2009 19:53

#include <iostream>
#include <stdio.h>
#include <iomanip>
#include <math.h>
using namespace std;

void main()
        {
            int n=0;
            int i=1,j;
            double number;
             float a[140000];


            while (cin)
            {


                cin>>number;
                a[i] = sqrt(number);
                n+=1;
                i++;

            }
            for (j = n-1; j>=1; j--)
            {
            printf("%f \n",a[j]);
            }
    system("pause");
        }

Edited by author 13.05.2015 15:57

For some reason I get answers that are very close but not equal to the ones in example:

14.495347194
0.000000000
56.489368526

Maybe I misunderstood something in the model?

Really dont know what the mistake???


int main() {
    char src[1405][1405];

    int n;
    cin >> n;

    for(int i = 0; i < n; i++) {
        for(int z = 0; z < n; z++) {
            cin >> src[i][z];
        }
    }

    int s = 0, S = 0;

    for(int i = 0; i < n; i++) {
        int tmps = 0, tmpS = 0;
        for(int z = 0; z < n; z++) {
            if(src[i][z] == 's') {
                tmps++;
                s = max(s, tmps);
                tmpS = 0;
            } else {
                tmpS++;
                S = max(S, tmpS);
                tmps = 0;
            }
        }
    }

    for(int i = 0; i < n; i++) {
        int tmps = 0, tmpS = 0;
        for(int z = 0; z < n; z++) {
            if(src[z][i] == 's') {
                tmps++;
                s = max(s, tmps);
                tmpS = 0;
            } else {
                tmpS++;
                S = max(S, tmpS);
                tmps = 0;
            }
        }
    }

    for(int i = 0; i < n; i++) {
        int tmps = 0, tmpS = 0;
        for(int z = 0; z <= i; z++) {
            if(src[i - z][z] == 's') {
                tmps++;
                s = max(s, tmps);
                tmpS = 0;
            } else {
                tmpS++;
                S = max(S, tmpS);
                tmps = 0;
            }
        }
    }

    for(int i = 0; i < n; i++) {
        int tmps = 0, tmpS = 0;
        for(int z = i; z < n; z++) {
            if(src[z][z - i] == 's') {
                tmps++;
                s = max(s, tmps);
                tmpS = 0;
            } else {
                tmpS++;
                S = max(S, tmpS);
                tmps = 0;
            }
        }
    }

    if(s > S) {
        cout << "s\n" << s;
    } else if(s < S) {
        cout << "S\n" << S;
    } else {
        cout << "?\n" << s;
    }
}

// ConsoleApplication4.cpp : Defines the entry point for the console application.
//
#include <cstdlib>
#include <iostream>
#include <stdio.h>
#include <math.h>
#include <string.h>

using namespace std;


int main()
{
    int count = 0;
    int maxCount = 0;

    long long MyArray[255];

    for(int i=0; i<=255; i++){
        MyArray[i] = -1;
    }

    while (scanf("%llu", &MyArray[count]) != EOF){

        count++;
        maxCount = count;
        if (count > 255){
            break;
        }
    }


    for (int i = maxCount-1; i >= 0; i--){

        if(MyArray[i] !=   -1){
            long double result = 0;
            result = sqrt((long double)(MyArray[i]));
            printf("%.4Lf\n", result);
        }
    }

    system("PAUSE");
    return EXIT_SUCCESS;
}

I'm afraid that i'm getting WA on 7-th test because of hash collisions. How to avoid it? Please, give me some recipes) I tried to change prime base - 31,43,57 - anyway WA#7.

Edited by author 07.11.2010 10:15

Sorry for my english.
First of all check the value of k: 0 < k < n. If False, then the result is n, otherwise -- use formulas.

My solution only outputs alpha=180.
Is there any case that it's possible and alpha can't be 180?

У меня на первый тест (n=3) ответ 1 6 2. Вроде ответ правильный: (1+6)=7 (2 делителя: 1,7), (1+6+2)=9 (3 делителя: 1,3,9).

А система выдаёт мне WA на етом тесте. Пожалуйста проверьте чекер :)

The test has only one number:
1
2

I am not allocating any memory in heap, everything is defined globally, why do I get Memory Limit Exceeded error on test 5? Why this error is not given for test 1? I'm not even using STL.

It also happened to me solving another problem(1220. Stacks), I got memory limit exceeded on test 10, same situation as here. Does IO operations use memory? I assume I get this error as test-cases get larger in size... (I also tried disabling IO buffering)

4
2 1 => 2 0 1 1 1
4 3 => 2 0 2 0 1
3 1 => 4 0 0 0 1
1 5 => 4 в 1 => нестабильная смесь => улетает чистый кислород.

I had found O(n) solution but I need to out ~2.7 Mb. It's takes ~0.12 sec.

So, how this problem can be solved in 0.062?

My first solution outputs points P and Q with coordinates more than 2000 on test

0 1000 0
1000 999 0
0 0 0
1 0 0

but passes jury tests.

Edited by author 03.05.2015 19:00

Edited by author 03.05.2015 19:08

Edited by author 03.05.2015 20:28

System.out.println((Math.min(N - 1L, M - 1L) << 1) + (N > M ? 1 : 0));

First of all make sure that you are not reading newlines and carriage return symbols. Input can be read like:

    char c;
    int scanfRez;
    do {
        scanfRez = scanf("%c", &c);
        if (scanfRez != EOF && c != '\n' && c != '\r') {
            chs[chCount++] = c;
        }
    }while(scanfRez != EOF);

I was getting WA7 with this input reading:

    char c;
    int scanfRez;
    do {
        scanfRez = scanf("%c", &c);
        if (c != '\n' && c != '\r') {
            chs[chCount++] = c;
        }
    }while(scanfRez != EOF);

If you get WA3 make sure that you are processing sequence correctly with given sequence length > N. For example:
let N = 11, input sequence = "abcdefghijkl" (without quotes) then letters should be removed in this order:
k 1
j 2
l 3
b 4
e 5
i 6
g 7
h 8
d 9
a 10
c 11
f 12

It does not matter if you put newline at the end of output.

Boost up your brains, coders.
Length of the thread is just sum of two simple components. One is the polygon perimeter, the other is [think on yourself, mate. The other component contains PI number]. Hello to your school teacher of elementary geometry.

4 5
1 2
2 3
3 4
4 5

7 6
1 2
2 3
3 4
4 1
4 5
5 6
6 1

2 5
1 2
2 3

2 4
1 2
3 4

3 4
3 1
1 2
2 4

4 4
1 2
3 1
2 4
3 4

3 3
1 2
2 3
1 3