#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    pair<int ,int > a[150000];
    for(int i=0;i<n;i++)
    {
        cin >> a[i].first >> a[i].second;
    }
    for(int i=100;i>=0;--i)
    {
        for(int j=0;j<n;j++)
        {
            if(i==a[j].second) cout << a[j].first << " " << a[j].second << endl;
        }
    }

}

WA #2

a = input().split()
s = input().split()
n, k = int(a[0]), int(a[1])
out = ["" for x in range(n//k+1)]
for i in range(k):
    if i != k - 1:
        for j in range(n//k+1):
            if len(s[0]) == 1:
                out[j] += "   " + s[0]
                s.pop(0)
            elif len(s[0]) == 2:
                out[j] += "  " + s[0]
                s.pop(0)
            else:
                out[j] += " " + s[0]
                s.pop(0)
    else:
        for j in range(n%k):
            if len(s[0]) == 1:
                out[j] += "   " + s[0]
                s.pop(0)
            elif len(s[0]) == 2:
                out[j] += "  " + s[0]
                s.pop(0)
            else:
                out[j] += " " + s[0]
                s.pop(0)
for i in out:
    print(i)

Edited by author 09.03.2018 19:59

It seems that the main problem is precision.
If it is so, how could we compute the answer more accurately?
e.g., set what Epsilon threshold for determining a value being 0 ?

output limit exceeded!
what's the case 4 like?
3Q

#include<stdio.h>
int main()
{
    int n , m , sum=0 ;
    double ava;
    scanf("%d",&n);
    int mark[11];
    for(m=0;m<n;m++){
        scanf("%d",&mark[m]);    }
    for(m=0;m<n;m++){
        sum=sum+mark[m];
    }
    ava=(double)sum/n;
    if(ava<=3)printf("None\n");
    else if (ava>=5)printf("Named\n");
    else if(ava>=4.5)printf("High\n");
    else printf("Common\n");
}

RT

My solution is crushed on 11 test.
please help me...

Any hints?

Hello all. I am doing task #1701, but ,actually, my question is also related to other equal tasks: how can you not exceed the time limit when the task supposes large array as the answer? For instance, in mentioned task the time is limited to 2 seconds. In case when N == 50 000, if possible set of emplyee's salaries exists, you need write 50 000 strings to console. But, as i have found out by measurig, only outputting array to the console takes over 3-5 seconds. Test it using C# and C++ with array of int values set to "0".

Edited by author 05.03.2018 09:04

If you use set<A*> or std::sort/std::stable_sort vector/array of pointers, your comparator won't work, because you need special comparator, like this

struct APtrCmp
{
  bool operator()(const A* lhs, const A* rhs) const  { /* implement logic here */ }
};
set<A*, APtrCmp> yourSet;

Good luck! ;)

#include <stdio.h>
#include <math.h>

int plus(int number){
    return number*(number+1)/2;
}
int minus(int number){
    return (-(number*(number+1)/2))+number;
}
int main(){
    int N;

    scanf("%d",&N);
    if(abs(N)>=1  && abs(N)<=10000){

        if(N<0)
            printf("%d",minus(N));
        else
            printf("%d",plus(N));
    }


}

this task is the same as 1075, but with a different number of dimensions(it doesn't really matter)

3 different solutions gets WA24. Is this test correct or I don't understand the statement?

If we run GCC 7.1, we can as well use -std=gnu++17

Before sending your solution try this test on your pc and make sure
that pc can calculate it faster than 10 hours :))

10000
10000
10000
10000
10000
10000
10000
10000
10000
10000
1

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace РџРѕРІРѕСЂРѕС‚
{
    class Program
    {
        static void Main(string[] args)
        {
            string[] mas = Console.ReadLine().Split(' ');
            int k = int.Parse(mas[0]);
            int n = int.Parse(mas[1]);
            int[] A = Console.ReadLine().Split(new char[] { ' ' }, StringSplitOptions.RemoveEmptyEntries).Select(i => int.Parse(i)).ToArray();
            int M = 0;
            for (int i = 0; i < n; i++)
            {
                M += A[i];
                if (M - k >= 0)
                {M -= k;}
                else
                {M = 0;}
            }
            Console.WriteLine(M);
        }
    }
}
Надеюсь понятно всё.

Have solved this task in Python many years ago.
How test my code on Python and have Accept
This code on Ruby have WA on test №3