see https://github.com/yosupo06/library-checker-problems/issues/610

my AC does not pass this short test

#include<algorithm>
#include<cstdio>
using namespace std;
#define MAX_N 250000

int n, i, t1, t2;
int *heap;

int main(){
    scanf("%d", &n);
    heap = new int[n/2+2];
    for(i = 0; i < n/2+1; ++i){
        scanf("%d", heap+i);
        push_heap(heap, heap+i+1);
    }

    for(i = 0; i < n/2-(n+1)%2; ++i){
        scanf("%d", heap+n/2+1);
        push_heap(heap, heap+n/2+2);
        pop_heap(heap, heap+n/2+2);
    }

    if(n%2){
        printf("%d.0\n", heap[0]);
    }else{
        t1 = heap[0];
        pop_heap(heap, heap+n/2+1);
        t2 = heap[0];
        if((t1+t2)%2){
            printf("%d.5\n", t1/2+t2/2);
        }else{
            printf("%d.0\n", t1/2+t2/2+t1%2);
        }
    }
    delete[] heap;

    return 0;
}

#include <iostream>
#include <algorithm>

using namespace std;

int main()
{
    int i,j,k,l,ii;
    int m,n;
    int girls[101];
    int boys[101];
    int count = 0;
    int upsets = 0, girlUpsets = 0, boyUpsets = 0, currentUpsets = 0;
    int happyGirls = 0;
    int chooseGirl = 0, chooseBoy = 0;

    for (i = 0; i <= 100; i ++)
    {
        girls[i] = 0;
        boys[i] = 0;
    }

    cin >> n >> m; //n girls m boys
    for (i = 1; i <= n; i ++)
    {
        cin >> girls[i];
    }
    for (j = 1; j <= m; j ++)
    {
        cin >> boys[j];
    }

    count = min (m, n);
    //计算无配对upset值
    for (i = 1; i <= n; i ++)
        upsets += girls[i];


    happyGirls = 0;


    for (i = 1; i <= count; i ++)

    {
        for (j = 1; j <= n; j ++)
            for ( k = 1; k <= m; k ++)
            {
                girlUpsets = 0;
                boyUpsets = 0;
                //计算girl upset值
                for (l = 1; l <= n; l ++)
                {
                    if (l != j)
                    {
                        girlUpsets += girls[l];
                    }
                }


                //计算boy upset值
                for (l = 1; l <= m; l ++)
                {
                    if (l != k)
                        boyUpsets += boys[l] * (happyGirls + 1);
                    //cout << "werqe: " << boys[l] << '\t' << boyUpsets << endl;
                }


                currentUpsets = girlUpsets + boyUpsets;
                if (currentUpsets < upsets)
                {
                    upsets = currentUpsets;
                    chooseGirl = j;
                    chooseBoy = k;
                }

            }

        if (chooseBoy != 0  && chooseGirl != 0)
        {
            happyGirls ++;

            //删除第chooseGirl个girl
            for (ii = chooseGirl; ii <= n - 1; ii ++)
               girls[ii] = girls[ii + 1];
            n --;

            //删除第chooseBoy个boy
            for (ii = chooseBoy; ii <= m - 1; ii ++)
                boys[ii] = boys[ii + 1];
            m --;
        }

    }
    cout << upsets << endl;
    return 0;
}

Find a robust feature to distinguish different bones with same total dots.
This really is an easy problem once you find that feature.

Try this tests

INPUT
tram
OUTPUT
Tram driver

INPUT
tram.tram
trolleybus
OUTPUT
Tram driver

i need some test... i guess the mistake is about "tram", i don't know which pattern of "tram*" should be counted

My solution is fast! valery@uni.lg.ua

Why is here the runtime error on the 6 test?
What test is 6?

#include <iostream>
#include <cmath>
using namespace std;

void ghjkl(long double p)
{cout << fixed;
cout.precision(4);
cout << p << endl;}

int main() {
long double a[1000]={-1.0};
long long i=0;
while(cin)
{cin >> a[i];
i++;}
i=i-2;
long double b[1000];
for(int k=0; k<1000; k++)
{b[k]=sqrt(a[k]);}
for(int m=i; m>=0; m--)
{ghjkl(b[m]);}
return 0;
}

Edited by author 24.03.2021 16:53

Edited by author 24.03.2021 16:54

Hi,

Well it took me a while to figure out as to how to approach this problem but as it turned out it is pretty straightforward. I divided the problem into two parts. First I printed out the elements up until the middle diagonal. And then the elements after that.

Here is the format of the numbers for 4 X 4 array

0,0  0,1  0,2  0,3
1,0  1,1  1,2  1,3
2,0  2,1  2,2  2,3
3,0  3,1  3,2  3,3

We can see a pattern here. First print the element 0,0.

Then -> 1,0 to 0,1    (i = 1 to i = 0 and j = 0 to j = 1)
Then -> 2,0 to 0,2    (i = 2 to i = 0 and j = 0 to j = 2)
Then -> 3,0 to 0,3    (i = 3 to i = 0 and j = 0 to j = 3)

So, basically as we are heading towards the main diagonal, we are increasing the value of i and at the same time running j from 0 to i.

Implementing this will get you the first 10 out of 16. For the rest of them, we have the same pattern again but this time its reverse

Then -> 3,1 to 1,3    (i = 3 to i = 1 and j = 1 to j = 3)
Then -> 3,2 to 2,3    (i = 3 to i = 2 and j = 2 to j = 3)
Then the last element 3,3

So, this just requires some clever manipulation of the for loops or while loops.

gives me WA 1

In order to achieve absolute safety providing electricity to the camps, besides an electric supplying system, the host organization set up a path from a reserved electricity generator (which is placed in one of the camps) to every camp once,
this means that, thare is only and only one path from Pi camp to Pj camp.

Можете дать подсказку?
Я понял что количество всех пар всегда три
И число дней как посчитать
Но не могу подобрать алгоритм чтобы вывести в каком порядке будут солдаты выходить на дежурство

Edited by author 02.06.2021 13:24

Hint: Binary search!!

Solution: This is a pretty easy binary search problem. All we need is to check for every number of the second list(as first list is sorted in ascending order) let's denote as x, if we can find y = 10000-x in the first list. If we find y for any x(from the 2nd list) then we will print "YES". Because we already got a pair x, y for which x + y = 10000.

So, the complexity is NlogN(as we are doing binary search for every element of a list in the worst case).

[Code was deleted]

Edited by moderator 06.06.2021 03:06

WA on test 2 don't know why  :(

Edited by author 01.07.2016 14:09

Edited by author 01.07.2016 14:09

Edited by author 01.07.2016 14:09

Edited by author 01.07.2016 14:10

We want to use segment tree, but we need to know how express some segment in order to preserve validity of multiplication and order calculation.
I noticed any segment can be expressed by just 2 integers. I cannot proof this, my math background is not good here.
Multiplication is trivial.
Now we need ability to merge 2 segments (expressed by 1-2 integers) and get 2 another integers. I spent tens of hours thinking on this, but cannot come up with correct merge algorithm.

Tried this:
merge(x1, x2, x3, x4):
    if exists such i,j,k: where x[i]*x[j]%p=x[k], then we can remove x[k] if we preserve order
    else try to set x[i] = x[i]*x[j]%p for some random i,j
    repeat

My merge procedure is not correct. Am I heading in the right direction?

My accepted program 9380769 fails on the test:
1 3 100 1.1 3

why wrong answer?


#include"iostream"

int main()
{

         int banki=10;
    int Garry,Larry;
    std::cin>>Garry>>Larry;
    std::cout<<banki-Garry<<" "<<banki-Larry;
    return 0;
}

I spent 3 hours on it.But I find it is expected to use "short" at last!!!