journ = []
x = 0
k = 0
sold = 0
cmd = ',,'
while cmd[0] != 'QUIT':
    cmd = input().split()
    oper_name = cmd[0]
    if cmd[0] == 'BID':
        x = cmd[1]
        journ.append(x)
    elif cmd[0] == 'DEL':
        x = cmd[1]
        journ.remove(x)
    elif cmd[0] == 'SALE':
        x = cmd[1]
        k = int(cmd[2])
        for i in journ:
            if float(i) >= float(x):
                k -= 1
                sold += 1
            else:
                continue
print(sold * 0.01)

What is wrong ?

Can I use CompletePack to solve it?
Give me a test about wa15.

Input:

_123

My first solution output:

_<span class=number>123</span>

My second solution output:

_123

But all passed. So which answer are correct?

.I'm already tired of solving this number..Mods please give me 19th test case...

Please, if you know, give an advice, how to pass this problem

O(n^3) to calculate dp values.
o(1) to answer query.

dp[floors][eggs] -- answer.

ограничение по времени для Питона не выполняется...

A test like this is possible:
15
0 35
12000000
12000000
-1

First time it occurs in Test 9.
This means that you should pay the same tax for any amount of salary. It is ok to output 0.00 for all such tests, despite it is not correct always because of interim roundings of regional coefficient.
Everything else is well-described in previous comments, basically correct numbers round and correct searching of initial salary (you should search for max possible value with 2 decimal places in it) are the only things that can make the initial problem description being unclear.

Test case 2:
9
-1 1
-2 2
-3 3
-10 10
3 6
4 5
2 7
1 8
0 9
Answer:
6
6 5 7 8 9 4

Test case 3:
100
272 422
352 367
102 435
274 469
532 554
88 468
28 761
334 381
62 236
858 955
104 930
4 892
467 749
39 695
189 709
217 790
571 999
126 355
138 779
570 645
31 222
372 511
345 888
535 731
435 953
87 869
172 863
442 936
450 704
452 697
862 896
887 977
510 844
135 225
95 932
632 888
30 704
165 523
47 306
107 645
22 34
237 712
779 796
505 964
419 495
171 751
449 616
62 642
355 954
101 266
71 585
284 723
69 900
781 998
21 338
619 950
173 239
394 502
183 605
113 999
30 748
122 833
541 829
65 453
650 741
141 562
355 928
360 852
83 997
563 590
482 668
318 980
84 170
233 858
9 484
637 799
104 611
483 856
29 889
16 463
530 709
774 979
414 654
332 527
406 476
257 735
57 257
201 508
259 806
62 955
410 790
37 720
490 893
573 779
241 400
315 640
242 940
165 929
54 814
356 501
Answer:
13
2 8 95 88 38 66 77 48 14 92 61 79 12

It says we have to start at the lowest left tile. Does that mean that the euclidean distance of the tile from origin should be minimum?? or does it mean that we have to look for left most layer and lowest tile in that??

try this test:

label: a = 1
goto label

I try to solve this problem as "offline LCA problem" with recursive DFS, but I have TLE? Where is mistake?

P.S. Нашёл баг, из-за которого был WA: если вы в Pascal пишите z:=n*(n-1), где z-int64, n-longint, то результат z будет типом longint; правильно так: z:=n; z:=z*(n-1).
И ещё: лучше выводить не 4 знака в ответе, а все.

Edited by author 03.08.2016 11:44

KMP + DP, O(n ^ 2)

2
87 4547
87 2945 1516
64 913 9864
-----------
9904.472185221

A solution that doesn't make use of the max number being 10,000 is using inclusion-exclusion principle. Because when we take the sets of K that have a gcd divisible by 2 and the sets of K that have a gcd divisible by 3, you count twice the ones with a gcd divisible by 6, so you have to subtract those. It's similar to a solution to this problem: https://open.kattis.com/problems/coprimeintegers

import java.util.*;
public class j {
    public static void main(String[] args){
        Scanner input = new Scanner(System.in);
        int bin,sum = 0;
        int n=input.nextInt();
        int m=input.nextInt();
        int [] mas = new int [n];
        for(int i = 0;i<m;i++){
            bin = input.nextInt();
            sum += 1;
            mas[bin-1] = mas[bin-1] + 1;

        }
        for(int i = 0;i<n;i++){
            System.out.println(String.format("%.2f",mas[i]/(sum*0.01))+"%");
        }

    }
}

import java.util.*;
public class Main {

    public static void main(String[] args) {
        ArrayList<Integer> mas = new ArrayList<>();
        Scanner input = new Scanner(System.in);
        int n,bin,delta = 0;

        n = input.nextInt();
        for (int i = 0; i <n;i++){
            bin = input.nextInt();
            mas.add(bin);
        }

        n = input.nextInt();
        for (int i = 0; i <n;i++){
            bin = input.nextInt();
            mas.add(bin);
        }

        n = input.nextInt();
        for (int i = 0; i <n;i++){
            bin = input.nextInt();
            mas.add(bin);
        }

        Collections.sort(mas);
        delta = mas.size();
        for (int i = 0;i<mas.size()-1;i++){
            bin = mas.get(i);
            delta = delta - (mas.lastIndexOf(bin)-mas.indexOf(bin)+1);


        }
        delta = -1*delta-mas.size();
        System.out.print(delta);
    }
}

:P

import java.util.*;

public class ReverseRoot {
    public static void main(String[] args) {
        ArrayList<Long> list = new ArrayList<>();
        Scanner scn = new Scanner(System.in);
        while (scn.hasNextLong()) {
            long p = scn.nextLong();
            list.add(p);
        }
        scn.close();
        for (int i = list.size() - 1; i >= 0; i--) {
            System.out.printf("%.4f%n", Math.sqrt((double) list.get(i)));
        }
    }
}