wa#4 for countless times, seek for kind help!
thanks

var a,b,c,i,k:real;
begin
read(a,b,c);
while a>b do
begin
a:=a-(a*c/100);
I:=i+1;
end;
write(i);
end.

Here is the code (python 3.6). Can anyone explain, why it fails on test 9?


import math
string = input().split()
start = int(string[0])
end = int(string[1])
commission = int(string[2])
if start <= end:
    n = 0
else:
    q = 1 - commission / 100
    n = math.log((end / start), q)
    n = round(n)
print(n)

1) Find the prime numbers up to 10,000,100 (Sieve of Eratosthenes)
2) if the sum is equal to 19 degrees, then the residue is prime to check
What can you advise?

In Idea my code works, but here is a runtime error:

import java.util.Scanner;

public class MathematiciansAndBerries {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        byte a1 = scanner.nextByte();
        byte b1 = scanner.nextByte();
        byte a2 = scanner.nextByte();
        byte b2 = scanner.nextByte();
        byte a3 = scanner.nextByte();
        byte b3 = scanner.nextByte();

        System.out.println((a1 - a3) + " " + (b1 - b2));
    }
}

Help pls, why runtime error occurs?

Edited by author 06.12.2018 21:47

I think I can not get AC because of "Enter"  key on Windows  is "\r\n"  and C++ scanf()  reads it as "\n"  and printf()  writes "\n" as "\r\n".
So,  if in input is some "\n",  then I would read it as "\n" and then print it as "\r\n"!!!  I don't know how to fix that.
Also there were problems with EOF-character.

The test is empty. There is nothing there.

I really don't understand why my solution is wrong but here it is:

letters = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ"
start = 0
s = input()

for i in range(0, len(s)):
    if not(s[i] in letters):
        word = s[start:i]
        other = s[i:i+1]

        if (word != ""):
            print(word[::-1], end="")

        print(other, end="")
        start = i+1

if (start != len(s)):
    print(s[start:len(s)][::-1], end="")

What's wrong with this solution?
WA (Test 2)

Edited by author 30.03.2013 23:10

верно ли,что при n=48,ответ  793566522023068987143383410279408208612410000
а при n=100 3834688188199309756675466732967588185862422492337212276545580039517258085033387305291001857619600

Hello.

I have received strange 'Compilation Error' verdict for my solution on the 1314 task under Microsoft Visual C++ 2017 compiler.

Compilation log is as follows:
/////////////////////////////////////////
lg1vr1-w2nnli(24): error C2062: type 'float' unexpected
lg1vr1-w2nnli(84): warning C4244: 'argument': conversion from 'float' to 'const int', possible loss of data
lg1vr1-w2nnli(94): warning C4018: '<': signed/unsigned mismatch
/////////////////////////////////////////

Lines 23 and 24 of my code are:
/////////////////////////////////////////
const int MAX_STATION_ID = 32767L;
const int INFINITY = 10L * MAX_STATION_ID;
/////////////////////////////////////////


Commits with CE are 8162255, 8162250, 8162248.

But there is successful (Accepted) commit 8162257 (the same code as 8162255) under G++7.1 commit.

Could you check if the CE message was correct or not?

Best regards.

Edited by author 04.12.2018 02:01

Please give me test 12.

if you have TLE #20. Use BufferedReader and Printwriter in Java. Good luck...

Hints for Java'ers who also hates problems like this and doesn't want to write automation-based solution.

new BigDecimal(line1) - it's almost cool, but you need some fixes before use constructor
1. Check for empty fraction case, e.g. "1.", "1.e2", "1.E2" - these are NANs according P1074 grammar, but OK according BigDecimal grammar (this is the only difference in grammars).
2. Fix big exponent to something smaller, for BigDecimal to be able to parse it
e.g. e1000000000000 => e1000
     e-1000000000000 => e-1000

Edited by author 09.02.2013 06:24

<input (replace 'S' with spaces!):>
S
1

34

10

-
1
-.
1
-.e0
1
-.1e
1
#123
1
--123
1
-1.e6
1
S4
4
12345678e-100000000000000000000000000000
10
0.0e10000000000000000000000
1
#



<output:>
Not a floating point number
34.0000000000
Not a floating point number
Not a floating point number
Not a floating point number
Not a floating point number
Not a floating point number
Not a floating point number
Not a floating point number
Not a floating point number
0.0000000000
0.0

I use segment tree to solve this problems(O(n^2logn)).
And I have checked my program with an solution(already accepted) for many times.
All my answer is correct.
But still wa#2.
Sorry for my poor English.
Please help me,Thank you.

Что нужно поправить,чтобы эта программа реализовывала тест '60 100 300 1'?вообще она ломается на 10 тесте при сдаче, но после того как я посмотрела обсуждения и прогнала её по найденным тестам  поняла что не могу реализовать только один выше сказанный ответ кстати (как я думаю) 298.
вот тесты на которых пробовала.
к стати я очень невнимательная, поэтому сильно на меня не ругаться )))
150 50 1000 100
ответ - 450

1 4 2 2
ответ - 2

1 3 5 2
ответ - 4

2 1 1 1
ответ - 2

1 3 4 3
ответ - 4

1000 2 2000 3
ответ - 140

11 1 10 2
ответ - 11

4 3 6 1
ответ - 6

1 2 12 3
ответ - 6

var a,b,c,d:integer;
begin
read(a,b,c,d);
if a>c then write(a)
else
begin
while a<>c do
begin
a:=a+b;
if (a>c-d)and(c>a) then c:=a
         else if a+b>c then a:=c
                     else if a=c then break
                                 else c:=c-d;
end;
write(c);
end;
end.

Заранее спасибо)♥_♥

Edited by author 28.12.2017 21:55

Edited by author 28.12.2017 21:56

I had WA4 because of this: AnD <=> and, NOt <=> not, etc.
I had WA9 because of this: empty strings (only spaces) - this is not a command))
I had TLE 13 because of this: we should ignore as a command lines has only a label)
Cool problem))
Good Luck)

Edited by author 07.04.2008 12:41

#include <iostream>
#include <math.h>
#include <iomanip>

using namespace std;

int main(int argc, char const *argv[])
{

    long number;
    double sqrts[256];

    int k = 0;
    while ( cin >> number )
    {
        sqrts[k] = sqrt(number);
        k++;
    }

    for ( int n = k - 1; n > -1; n-- ) {
        cout << fixed << setprecision(4) << sqrts[n] << endl;
    }

    return 0;
}

Edited by author 29.11.2018 20:32

скорее всего ошибка с функцией Eof() ибо впервые ей пользуюсь и смутно представлю принцип работы. Если её нельзя использовать, то как найти конец вводимого файла?

var s,sk:string;
var i,fl,j,code:integer;
var n:real;
var a:array [1 ..256] of real;
var f:array [1 ..256] of integer;
begin
while (Eof()<>True ) do
 begin
 readln(s);
 for i:=1 to length(s) do
 begin
 if (s[i]<>' ') and (fl=0) then begin
                               fl:=1;
                               sk:=sk+s[i];
                               end
else  if (s[i]<>' ') and (fl=1) then sk:=sk+s[i];

 if (((s[i]=' ') and (fl=1)))or ((i=length(s))and(fl=1)) then begin
                                 fl:=0;
                                 val(sk,n,code);
                                 for j:=1 to 256 do
                                 if f[j]<>1 then begin
                                                 a[j]:=sqrt(n);
                                                 f[j]:=1;
                                                 sk:='';
                                                 n:=0;
                                                 break;
                                                 end;

                                 end;
 end;end;
for i:=1 to 256 do
if (f[i]=1) then writeln(a[i]:0:4);
end.