Can anyone give me some test cases because I really don't know where my program is wrong ?

Is there a way to know more precisely what is going on on the test machine?
I've tested my python solution on my machine on large data (N=10^6) with a lot of copying operations (geometric progression) and got answer with memory used < 64Mb.

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    string s;
    int a=1;
    int counter=0;
    for(int i=0; i<n; i++)
    {
        cin>>s;
        if((s[0]=='A' || s[0]=='P' || s[0]=='R' || s[0]=='O' ) && a==2) { a=1; counter++;}
        else
        if((s[0]=='A' || s[0]=='P' || s[0]=='R' || s[0]=='O' ) && a==3) { a=1; counter+=2;}
        else
        if((s[0]=='B' || s[0]=='M' || s[0]=='S') && a==1){ a=2; counter++;}
        else
        if((s[0]=='B' || s[0]=='M' || s[0]=='S') && a==3){ a=2; counter++;}
        else
        if((s[0]=='D' || s[0]=='G' || s[0]=='J' || s[0]=='K' || s[0]=='T' || s[0]=='W' ) && a==1) { a=3; counter+=2;}
        else
        if((s[0]=='D' || s[0]=='G' || s[0]=='J' || s[0]=='K' || s[0]=='T' || s[0]=='W' ) && a==2) { a=3; counter++;}

    }
    cout<<counter;
    system("pause");
    return 0;
}

my code is wrong in the test 8 but i dont know the test, can you help me?

#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
int main(){
    string st;
    int W,n,i=0,A[1000];
    cin >> n;
    do{
        cin >> st;
        if(((int)st[0]==65)||((int)st[0]==80)||((int)st[0]==79)||((int)st[0]==82)) A[i]=1;
        if(((int)st[0]==66)||((int)st[0]==77)||((int)st[0]==83)) A[i]=2;
        if(((int)st[0]==68)||((int)st[0]==71)||((int)st[0]==74)||((int)st[0]==75)||((int)st[0]==84)||((int)st[0]==87)) A[i]=3;
        i++;
    }while(i<n);
    for(i=0;i<n-1;i++) {
        int S;
    S=abs(A[i]-A[i+1]);
    W+=S;
    }
    cout<<W;
    return 0;
}
Please,help me!!!

import sys

n = input()

arr = bytearray(n)
i = 0

while i < n:
    line = raw_input().split()
    arr[i] = (int(line[0]) + int(line[1]))
    i += 1

result = ""

i = n-1
while i > 0:
    if arr[i] > 9:
        arr[i] -= 10
        arr[i-1] += 1
    result = str(arr[i]) + result
    i -= 1

arr[0] = arr[0]%10
result = str(arr[0]) + result

print result

f(x) = k * (x^2 -2(a1)x + (a1)^2 + x^2 -2(a2)x + (a2)^2 + ... + x^2 -2(an)x + (an)^2) / n
      = k (x^2 - (2/n)(a1 + a2 + ... + an)x + ((a1)^2 + (a2)^2 + ... + (an)^2)/n

so we have
f'(x) = k(2x - (2/n)(a1 + a2 + ... + an))

question: what is k?
(explaining from previus post)

Edited by author 05.05.2016 14:53

Edited by author 05.05.2016 14:54

Earlier I had got a WA 9 but now I'm getting a WA 5 with the same code. What's the logic?

4686044    16:50:13   21 Dec 2012    Marian Darius    1297. Palindrome    C++    Accepted      0.031    108 KB

Just try iterating the middle of the palindrome and extend to both ends, until you find two elements that are different.

3 days overall. I'm your fan forever.

Edited by author 03.05.2016 14:58

Can someone help to post test #2? Thanks

1. Sort it by endTime, if endTime equals another endTime then sort them by startTime.
2. Open one iteration from 2 to n, init just check that,  is endTime<starTime, if so then ans++;
3. print ans; endTime and startTimes is one array with unchanged same index...

Sorry for poor English

I found hah)

Edited by author 01.05.2016 17:16

What is the 12 test?


Edited by author 29.11.2014 12:19

import math
values = raw_input().split(' ')
a = float(values[0])
b = float(values[1])
c = float(values[2])
d = float(values[3])
if (math.fabs(a*b-c*d)<0.001):
    print "Impossible."

j = 0.0
x = a**2 + b**2 #DC**2(1)
y = c**2 + d**2 #DC**2(2)
x1 = - 2 * a * b
y1 = - 2 * c * d
l = math.fabs(x - y)
r = math.fabs(x1 - y1)
if r == 0:
    False
else:
    j = l / r #cos
    res = a**2.0 + b**2.0 - 2.0 * a * b *j #DC**2
    res = math.sqrt(res)*1000 #DC
    print "Distance is %(res)d km." %{"res": res}


Edited by author 29.04.2016 15:30

Edited by author 29.04.2016 16:01

new one but still WA9
package banky;
import java.util.*;
public class test1 {

    private static Scanner scan;

    public static void main(String [] args){

        scan = new Scanner(System.in);
        int k=0;
        int r=0;
        int pr=0;
        while (scan.hasNext()){
            String s = scan.nextLine();
            //s  = s.toUpperCase();
            s  = s.toLowerCase();
            char c [] = s.toCharArray();
            if (k<=0||r>0)
                c[0] -= 32;
            int i ;
            for(i=1;i<s.length();i++){
                if  ((c[i] == 46) || (c[i] == 33) || (c[i] == 63))
                    {pr= 1;
                    c[i]=c[i];}
                else if  (pr==1&&c[i]!=32)
                { c[i] -= 'a' - 'A';
                pr = 0;}}
            for(i=0;i<s.length();i++){
                if  ((c[i] == 0)|| (c[i] == 14) || (c[i] == 1) || (c[i] == 31))
                    c[i] += 'a' - 'A';}
            for(i=1;i<s.length();i++){
                if  (((c[i-1]< 'a')&&(c[i-1]>='A'))&&((c[i]< 'a')&&(c[i]>='A')))
                      {c[i] += 'a' - 'A';}}

            if  ((c[s.length()-1] == 46) || (c[s.length()-1] == 33) || (c[s.length()-1] == 63))
            {r=r+1;}
            else
            {r=0;}
            //System.out.println(k);
            //System.out.println(r);
            System.out.println(c);
            k=k+1;


        }
    }
}

Edited by author 30.04.2016 13:09

I've got quadratic equation. It can be calculated if you know just what is arithmetical progression. What I should to know in maths to find max P (I found P from the equation).

Edited by author 29.04.2016 01:05

#include <iostream>

using namespace std;

int main() {
    int mashines, minutes;
    cin >> mashines >> minutes;

    int arr[minutes];

    for(int i = 0; i < minutes; ++i) {cin >> arr[i];}

    int m = 0;
    for(int i = 0; i < minutes; ++i) {
        m += arr[i];

        if(m - mashines >= 0) {m -= mashines;}
        else {m = 0;}
    }

    cout << m;
}

Hah
I decided to check all variants, but due to misprint my program become greedy)