I thought my solution is quite "brute", are there any other solutions?
And I have to say it is a really good problem for sufficient samples and a clear statement.

I cant ivent any tests that my solution will fail, but I have WA1!
Can any body help...
Give me some test!
Plz...

8
AA
BB
CC
DD
EE
FF
GG
HH
6
AA HIT BB IN HEAD
AA HIT CC IN HEAD
AA HIT DD IN HEAD
AA REVIVE BB
AA REVIVE CC
EE REVIVE DD

CORRECT
AA BB CC FF
DD EE GG HH

100000000
10
1 5 even
6 10 even
11 18 even
19 25 even
1 8 even
16 25 even
16 40 even
9 40 odd
1000 5000 odd
1000 6000 odd
-1

answer: 7

I had runtime error 25, but when i surrounded sorting with try catch it accepted
Any ideas why?

#solution 1068
x=int(input("enter your number:  "))
def factorial(x):
    if x>0:
        return x + factorial(x - 1)
    elif x<0:
        return x + factorial(x+1)
    if x==0:
        return 0
y=factorial(x)
if x<0:
    print(y+1)
else:
    print(y)

n = int(input())
if n > 0:
    print(int(n*(n+1)/2))
elif n == 0:
    print('1')
else:
    print(int(n*(n-1)/-2 + 1))

delete

Edited by author 04.10.2020 16:38

#include<bits/stdc++.h>
using namespace std;
map<string,int>mp;
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    string s;
    string tr="tram";
    string tro="trolleybus";
    string last="";
    while(getline(cin,s)) {
        string subject=s;
        regex re("\\w+");
        sregex_iterator next(subject.begin(), subject.end(), re);
        sregex_iterator end;
        while (next != end) {
            smatch match = *next;
            last=match.str();
            if(last==tr)
                mp[tr]++;
            else if(last==tro)
                mp[tro]++;
            next++;
        }


    }
    if(mp[tro]>mp[tr])
        cout<<"Trolleybus driver"<<endl;
    else if(mp[tr]>mp[tro])
        cout<<"Tram driver"<<endl;
    else
        cout<<"Bus driver"<<endl;
}

What's wrong with test #38? I got AC with sorting but failed on majority search?

input :
15 4
1 2 30 994 50 600 700 999 900 990 991 992 993 40 800
output :
   1  50 900 993
   2 600 990  40
  30 700 991 800
 994 999 992

If you have WA4, try a 100x100 matrix with all 1 or with all 0.

Edited by author 12.05.2019 20:53

wa18 ...please

import java.util.Scanner;


public class task9 {


    public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int n = sc.nextInt();
          int m = sc.nextInt();
          int x = sc.nextInt();
          int y = sc.nextInt();
          int cl = 0;
          int stroka =1;
          int[][] mas = new int[n][m];
         for(int i=0;i<n;i++){
             if(stroka%2==1){
                 for(int j=0;j<m;j++){
                 mas[i][j]=cl;
                 cl++;

                 }
                 stroka++;
             }
             else{
                 for(int j=m-1;j>=0;j--){
                 mas[i][j]=cl;
                 cl++;
                 }
                 stroka++;
             }
         }
        System.out.println(mas[x-1][y-1]);

            }


}

The problem is to find how many l*1 or 1*l while rectangles that do not *included* by each others.

My solution is O(NlogN),but I thought there may be some solutions use only O(N) time, thank you for your sharing!

Question requirements are ordered without changing order

#include <iostream>
#include <cstdio>
#include <algorithm>
using namespace std;

const int maxn = 150005;
struct Node
{
    int index;
    int ID;
    int mount;
}node[maxn];

int N;

bool cmp(Node &a, Node &b)
{
    if (a.mount != b.mount)
    {
        return a.mount > b.mount;
    }
    else
    {
        return a.index < b.index;
    }
}

int main()
{
    scanf("%d", &N);
    for (int i=0; i<N; i++)
    {
        scanf("%d%d", &node[i].ID, &node[i].mount);
        node[i].index = i;
    }

    sort(node, node + N, cmp);

    for (int i=0; i<N; i++)
    {
        printf("%d %d\n", node[i].ID, node[i].mount);
    }
    return 0;
}

I tried so many time until I change `#include<bits/stdc++.h>` to `#include<stdio.h>`.

And then I did a small experience: for the same code,
Visual C++ 2017 -- 612 KB
Visual C 2017 -- 620 KB
GCC 7.1 / Clang++ 4.0.1 -- 652 KB
G++ 7.1 -- 700 KB

Hope it helps~

#include <iostream>
int main(void){

char nameOfKid[8];

int noOfLetters;
std::cin >> noOfLetters;
std::cin.ignore( 256, '\n') ;
int cur, temp, steps, newcur;
newcur = 0; cur = 0; steps= 0 ;
for (int i = 0 ; i < noOfLetters; ++i){

        std::cin.getline(nameOfKid, 8);

        temp = static_cast<int>(nameOfKid[0]);

        switch(temp){
            case 65:
            case 80:
            case 79:
            case 82: newcur = 1;
                   break;
            case 66:
            case 77:
            case 83: newcur = 2;
                   break;
            case 68:
            case 71:
            case 74:
            case 75:
            case 84: newcur = 3;
                   break;

        }
        if ((newcur!=cur) && (i!=0)){
        steps = steps + ((newcur>cur) ? (newcur-cur): (cur-newcur));

        }
        cur = newcur;

    }

std::cout << steps;
return 0;
}

Solving the problem using the merge sort approach (counting inversions while merging subsequences - plus one inversion in a case of element from the left subseq greater than the element from the right) looks feasible (and maybe evident) to implement. But there is one caveat: it is crucial to count not only the one inversion in a case *cur_left > *cur_right, but count elements in a range [cur_left + 1, right_begin) as "inverted" too. That follows from the fact that merged subarrays are sorted in ascending order, so if we encountered the (*cur_left > *cur_right) case, then elements in a range [cur_left + 1, right_begin) satisfies that too and could be counted as inversions.
Hope I stated that clearly.

Edited by author 04.05.2019 18:59

Edited by author 06.05.2019 06:44