test case :

RGBY
RGBY

answer :
equal

According to the segement discussion, re-paintings include [a,b] and we have to output (a,b),
So how the output for TEST1 is (47,634) ?
According to me the output should be (47,635) as 634 will be painted due to 3rd repainting.

My accepted program gives me :
test 1 :
5
2 6854726
5664 2
2 34
1 5646532
40 1100000
answer :
2
0
2
5
8

#include <iostream>
using namespace std;

int main()
{
    int N, M;  int S[15001];
 int S1[1000001];

    cin >> N;
    for (int i1 = 1; i1 <= N; i1++)
        cin >> S[i1];


    int f = 0;

    cin >> M;
    for (int i = 1; i <= M; i++)
        cin >> S1[i];

    for (int i1 = 1; i1 <= N; i1++)
    for (int i = 1; i <= M; i++)
    if (S[i1] == S1[i]) f++;

    cout << f;
}

the product is : n*(n-k)*(n-2*k)... till---> k if divisible by k, i.e. the last non zero  value of n OR n(mod)k which is again the last non zero value of n.
You don't have to multiply it with k or n mod k again.

May be your program is O(n^2)
you can use an array.
f[i] means there are f[i] '1' after s[i].

(I can't speak English well.)

I know how to identify end of console in Pascal, C/C++, but in Java it's not working. Anybody knows?

What is the bug in my code?

import java.io.*;
public class ACM1601 {

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


        boolean isNewSntc = true;
        String line="";
        int ch;
        while ((ch = System.in.read()) != -1) {
            if (isNewSntc){
                line += Character.toString((char)ch);
                isNewSntc = false;
            }
            else {
                line += Character.toString(lowerCase((char)ch));
            }
            if (ch == 46 || ch == 63 || ch == 33)
                isNewSntc = true;
        }

        System.out.println(line);
    }

    public static char lowerCase(char in) {
        if ((int)in > 64 && (int)in < 91)
            return (char) ((int)in + 32);
        return in;
    }
}

Edited by author 31.12.2013 22:26

Edited by author 31.12.2013 22:26

please send me the test data at least for test 1,
i have trouble understanding what is wrong with the code just going blindly,
after running all the tests i could find and succeeding still not able to pass first one.

tx

//1639. Chocolate 2
#include<stdio.h>
int main()
{
    int  n;
    printf((scanf("%d",&n)*n*scanf("%d",&n)*n)%2==0?"[:=[first]\n":"[second]=:]\n");

}

it works fine in my compiler..but don't know why it shows wrong answer here while submission...

Edited by author 31.12.2013 01:59

Deleted

Edited by author 30.12.2013 20:05

I had some huge problems getting from WA9 to AC. Here is a test that might help you out:
Input:
3
)))((
))))
(((((
Output:
14 3
3 1 2

I am using greedy strategy
coeff = (unhappiness - possible reduced unhappiness)*length of line
x is position to cut the line for min unhappiness
:::::
add main list to priority queue with initial unhappiness
while(cuts<K){
remove List with max coeff;
cut in 2 parts for min unhappiness, calc coeff (for each list);
add the 2 parts of list to queue;
cuts++;

}
then poll all queue members(i.e. all cut lines) and add their unhappiness

I'm getting all posted tests right.. I tried all variations I could, and I'm getting right answers, but I am getting WA#2, please help!!!!


Edited by author 30.12.2013 21:16

My question is not particularly about this problem,
but about printf output with precision 10^-6 or better.

In this problem when I use float and output with printf("%.6f", ... ) I get WA8.
When I use double and printf("%.6lf", ... ) I get WA1.
What is the correct approach please?

Input:
2
1 1
Output 1:
1
-1 3
Output 2:
2
-1 1
1 3
Какой из ответов правильный?

If you work in C, just use int64_t for N and it will work! I spend like 2 hours solving bugs just because of datatypes (facepalm)

#include <stdio.h>
#include <math.h>

int foo () {
long long int a;
while(scanf ("%lld",&a)!=EOF)
{ foo ();
  printf ("%.4f\n",sqrt ((double)a));
  return 0; }
}

int main () {
foo ();
return 0;
}

Well, I guess there is some sneaky thing. Someone can give me test 4 please. Thanks

import java.util.Scanner;
public class problem1197 {

    public static void main(String args[]) {
        Scanner in = new Scanner(System.in);
        int I = in.nextInt();
        int J = in.nextInt();

        if (I == 1) {
            System.out.println(1);
            return;
        }
        if (I == J) {
            System.out.println(I);
            return;
        }
        int deyer[] = new int[100001];
        int idx = 0;
        int min = 10000;
        int mineded = 0;

        for (int i = J; i >= I; i--) {

            for (int j = i / 2; j >= 1; j--) {

                if (i % j == 0) {
                    if (j < min) {
                        min = j;
                        mineded = i;
                        if (j == 1) {
                            System.out.println(i);
                            return;
                        }
                    }

                    break;
                }
            }
        }
        System.out.println(mineded);

    }
}