mxw can be 1, and all w is about 10^9 and a can be 10^9, then in the worst case we need to withdraw the sum of 10^18 about 10^5 times, so long double should not fit, but it fits

What is in test7?


Edited by author 23.02.2023 13:59

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0

I have WA in test 13

Edited by author 21.02.2023 01:48

Anybody else?
I guess it's the epsilon. But I failed all trial.

I got a question about the problem description.

In other games, the losers hope to last the game longer for as many turns/rounds/moves as possible.

In this game, is the losers' optimal move to
(1) to minimize the difference,
(2) to maximize theirselves' scores, or
(3) to minimize the opponents' scores?

Or are the 3 goals lead to the same move?

Thanks.

"Your task is to make the largest number of the Great Discoveries and maximally to delay the doomsday."

Notice, that we are not only maximazing the period, but also the k^period. So we have to find strictly maximal "generator"/"primitive"/"первообразная"/"образующая".
There can be problem with understanding the way of minimizing these values. But as period it divisor of (mod - 1), and there are many generator, and they are quite uniformly distributed, probably k^(mod - 1) will always be better, than some p^((mod - 1) / q), where p > k.

#include<iostream>
#include<iomanip>
#include<bits/stdc++.h>
#include<string>
using namespace std;
int main(){
    float n, k;
    int mark, a;
    a=0;
    k=0;

    cin >> n;
    for(int i=0; i<n; i++){
        cin >> mark;
        a+=mark;
    }
    cout << fixed;
    cout << setprecision(1);
    k+=a/n;

        if(a%5==0 && mark>=5){
            cout <<"Named" << endl;
        }
        else if(mark>3 && k>=4.5){
            cout <<"High"<<endl;
        }
        else if(mark<=3){
            cout <<"None" << endl;
        }
        else{
            cout << "Common" << endl;
        }

return 0;
}

For those who have WA on test 7, please try following one:
4/2*3+

For n = 10 and k = 20, what should be the output? Well, obviously it's not 0!

I created every stack using pointers, but nodes in this stack saved data of ten last pushed elements, not of one last. So this is a stack of arrays and not the stack of single elements. Also I had an additional array of int[1000], that showed how many positions in last node of each stack is already used.

Conclusion: you need to do such a struct of that has an array of ten elements and a pointer to previous node. On every push-call you should create another node(if previous is full) and add an element on the first open space of node, then increase the number of used positions. On every pop-call you should go to previous node and delete current node(if current node is empty) and print the last element of node, then decrease the number of used positions.

Hope it'll help you. And I apologize for my poor english)

a = input()
row = []
for i in a:
    row.append(i)
for i in range(len(row)):
    row[i] = int(row[i])
suml = row[0] * 100 + row[1] * 10 + row[2]
sumr = row[3] * 100 + row[4] * 10 + row[5]
if suml - sumr == 1 or sumr - suml == 1:
    print('yes')
else: print('no')

n,k = map(int, input().split())
kol = 0
u = 1
for i in range(n):
    a, b = map(int, input().split())
    if a - b == 2:
        kol += u
    else: u = 0
if kol == 0:
    print('Big Bang!')
else: print(kol)

const STACK_SIZE: usize = 128 * 1024 * 1024;

fn run() {
    //...
}

fn main() {
    let child = std::thread::Builder::new()
        .stack_size(STACK_SIZE)
        .spawn(run)
        .unwrap();
    child.join().unwrap();
}

There were mentioned, that it is required 64-bit integer type. I also add, that it has to be unsigned, otherwise you get WA 8.

Can somebody give me anti-greedy tests so that a greedy solution will not give correct answer? A test like #39? I'm sure that greedy is the right solution, i cannot find counter-test for my solution.

Guess, why I had wasted 18 attempts?
The exponent can have + sign: 1.3e+32

All bad wished to author.

#include <iostream>
#include<vector>
#include<algorithm>
using namespace std;
int main()
{
    int n, a, razn, k1 = 0, k2 = 0;
    cin >> n;
    vector<int>num;
    for (int i = 0; i < n; i++) {
        cin >> a;
        num.push_back(a);
    }
    sort(num.begin(), num.end());
    int kol = 0;
    for (int i = 0; i < n - 1; i++) {//все камни кроме максимума
        kol += num[i];
    }
    if (kol < num.back()) {//если все камни меньше самого максимального
        cout << num.back() - kol;
    }
    else if (kol == num.back()) {//если все камни pавны максимальному
        cout << 0;
    }
    else if (kol > num.back()) {//если все камни больше максимального
        k2 = num.back();
        for(int i =n-2;i>=0;i--){
            if (k1 < k2) {
                k1 += num[i];
            }
            else if (k1 > k2) {
                k2 += num[i];
            }
            else if (k1 == k2) {
                k1 = num[i];
                k2 = 0;
            }
        }
        cout << abs(k1 - k2);
    }
}

//my program is fine solves any of my values. what is in test 5?

n, m = map(int, input().split())
a = []
b = [1]
c = []
for i in range(m):
    f = int(input())
    a.append(f)
a = sorted(a)
f = 0
for i in range(1, len(a)):
    if a[i] == a[i - 1]:
       b[f] += 1
    elif a[i] != a[i - 1]:
        b.append(1)
        f += 1
for i in range(len(b)):
    b[i] = float(b[i] / m * 100)
for i in range(len(b)):
    c.append('%.2f' % b[i] + '%')
for i in range(n):
    try:
        print(c[i])
    except:
        print('0.00%')