Sorry for my bad English.
I use

    int ch;
    while ((ch = System.in.read()) != -1) {
        // my solution
        System.out.print(<some char>);
    }
    System.out.println();

But I get WA #1 with or without last code line. I also used System.out.flush (without last line), but it do not help me.

Edited by author 24.06.2019 13:55

hEdited by author 19.07.2019 12:15


Edited by author 19.07.2019 12:16

import math

nm = input().split()
n = int(nm[0])
m = int(nm[1])
res = set()
result = []
for i in range(m):
    x = int(input())
    if x not in res:
        res.add(x)
    result.append(x)
answer = []
for i in res:
    answer.append(result.count(i))
while n - len(answer) > 0:
    answer.append(0)
for i in answer:
    v = str(round(i*100/m, 2))
    if v[-2] + v[-1] == '.0':
        print(str(round(i*100/m, 2)) + '0%')
    else:
        print(str(round(i*100/m, 2)) + '%')














Edited by author 24.02.2019 01:39

Edited by author 09.03.2019 03:19

input:
10 5 4 4
2
1 1 4 0 4
3 0 10 4 10

output:
1 4

#include <iostream>
#include <cstdlib>
#include <deque>

using namespace std;

int main()
{
    int h;
    char s[200001];
    scanf("%s", &s);
    int n=strlen(s);
    deque<char> v (s, s + n  );

    if(v.size()==2  && v[0]==v[1] ) return 0;
    b: h=2;
    for(int i=0; i<v.size()-1 ;i++)
          if(v[i] == v[i+1])
        {
            v.erase(v.begin() + i);
            v.erase(v.begin() + i);
            h=1;
        }

    if(h==1) goto b;

    for(int i=0; i<v.size(); i++)
        cout << v[i];

        return 0;
}

very similar to the NP-complete problem 3-dimensional matching (3-сочетания)

You can solve the question very fast with appropriate precomputed values.
The sequence is known as the Stern-Brocot sequence, I was not aware of it before!

Edited by author 12.07.2019 04:29

#include<iostream>
using namespace std;

int main(){
    int petr,a,taxi,b;
    int temp;

    while(cin>>petr>>a>>taxi>>b){
        if(petr>=taxi)cout<<petr<<endl;

        else{
            temp=taxi-petr;
            if(a<=b){
                while(temp>b){
                    petr+=a,taxi-=b;
                    temp=taxi-petr;
                }
                cout<<petr+a<<endl;
            }
            else{
                while(temp>a){
                    petr+=a,taxi-=b;
                    temp=taxi-petr;
                }
                cout<<taxi<<endl;
            }
        }
    }
}

i pass all the test that the discuss give
but i still get wa at #4
could u give me some new test?

First and foremost, is seems to me that there are only two test cases:
(1) the sample case
(2) an 8MB test case

For WA this test helped me:
input:
2
bbaadbdcc
aadbdbcbadb
aadbdbcbadb
bbaadbdcc

output:
Case #1: 3 2 5
Case #2: 5 2 3

For TLE: parse the input
I optimized all sorts of things but eventually I ran out of ideas. Then I remembered someone else was mentioning he submitted the same code again and passed got rid of TLE. Also, from my own experience, I noticed there was a sort of "randomness" in the time execution from one submission to another. (I was able to detect it by killing my program after processing 6MB of the input file.)

At this point I was seriously considering speeding up the input reading part of my program using custom code instead of stdin.h. I got 760ms for processing 6MB of the input and I said to myself I'm really close: my code should run in 1.040s.

And so I decided to parse the input: I got AC in 560ms! A 480ms boost of speed! So... long story short: parse the input!

5
1 9
14 15
1 11
12 13
10 15

Right answer is 3, but some algorithmes may give answer 2 for this test.
P.S. I had this mistake:(
P.P.S. Sorry, for my poor English...

WA14((
No ideas...
I tried any tests my brain could imagine.
Can anybody help, plz?

I have try many of my tests, but I don't know what is wrong in my code. What is test #2, please?

Why does this give runtime error #8 .
n = int(raw_input())
k = int(raw_input())
dp = [[-1 for x in xrange(15)]for y in xrange(1910)]
def solve(index,prev) :
    if(index > n ) :
        return 1
    if(dp[index][prev] != -1) :
        return dp[index][prev]
    res = 0
    start = 0
    if(index == 1) :
        start = 1
    for i in xrange(start,k) :
        if(prev == 0 and i == 0) :
            continue
        res = res + solve(index+1,i)
    dp[index][prev] = res ;
    return  res
print solve(1,0)

var str: string;
i,kol,n,pol: integer;
begin
readln(n);
pol:=1;
kol:=0;
for i:=1 to n do
begin
readln(str);
if (str='Alice')or(str='Ariel')or(str='Aurora')or(str='Phil')or(str='Peter')or(str='Olaf')or(str='Phoebus')or(str='Ralph')or(str='Robin') then
begin
kol:=kol+abs(pol-1);pol:=1;
end
else
if (str='Bembe')or(str='Belle')or(str='Bolt')or(str='Mulan')or(str='Mowgli')or(str='Mickey')or(str='Silver')or(str='Simba')or(str='Stitch') then
begin
kol:=kol+abs(pol-2);pol:=2;
end
else
if (str='Dumbo')or(str='Genie')or(str='Jiminy')or(str='Kuzko')or(str='Kida')or(str='Kenai')or(str='Tarzan')or(str='Tiana')or(str='Winnie') then
begin
kol:=kol+abs(pol-3);pol:=3;
end;
end;
write(kol);
end.

Edited by author 17.11.2016 23:36

Edited by author 17.11.2016 23:36

input:
6 5 6 4
2
1 0 6 1 5
3 0 6 0 6

output:
2 6

Why the second difference is 1.937 and no 2.000?

input:
1
'a-z' like 'a-z'

output:
YES

Edited by author 03.07.2019 23:01

Edited by author 03.07.2019 23:01

Hi,
I have tried like million times different combinations, and it always gives wrong answer, always on 6. case.

Does anyone have 6. case? Does any anyone have any idea what kind of strings/patterns are in 6. case?

Thanks in advance