1
0 0 1
2

2
0 -1 1
0 3 3
3

2
-1 0 1
1 0 2
4

2
0 0 5
3 0 2
3

3
0 0 6
-1 0 5
5 0 1
5

4
0 0 5
10 10 5
10 0 5
0 10 5
6

8
0 0 1
2 0 1
4 0 1
4 2 1
4 4 1
2 4 1
0 4 1
0 2 1
10

2
1 1 5
1 1 5
2

3
0 -5 6
0 0 2
0 0 4
6

i couldn't understand, why my programm got Runtime error 15. When i increased MAXN from 1e4 + 10 to 1e5 + 10, i got AC. So the vertices number are in the range [1, 1e5], but not in [1, 1e4], as written in the statement

Как понять когда она завершила ввод сообщения?

don't round up. or you will get wa8.



Edited by author 28.05.2023 19:05

I'll just combine two previous hints.
Use bubble sort and THEN use random.

Hi, I'm still getting WA-1, can anybody give me answers to these tests:

80 60
60 80
60 60

Thanks.

a-b and c-d are close, but not touching each other.

-2 0 0
-1 0 0
0 1 0
0 1000000 1

In this test all numbers are big (~1e9), so it can be precision errors/overflow. You should use NTT with big modulo to avoid doubles

Use long long

The width of the last column of ranklist, "Last AC", is a little narrower than needed. The ranklist table always displayes abnormal. Please can you take a look?

And so does the bottom ranklist table in the author page.

Solved the problem using segment tree.
Im using VC++ 12 and it's giving the correct answer for the sample.
Still while sending (VC++ 2010) im getting WA 1. Changing it to G++ C++ 11 gives AC. What's the difference and why am i getting WA 1 when i choose VC++?

Code:
#include <iostream>
#include <vector>
using namespace std;
const int MAXN=100000;
pair<int, int> t[4*MAXN];
int z=1;


void build_tree(int v, int tl, int tr)
{
    if (tl==tr)
    {
        t[v]=make_pair(1,z++);
        return;
    }
    int tm=(tl+tr)/2;
    build_tree(2*v, tl, tm); build_tree(2*v+1, tm+1, tr);
    t[v].first=t[2*v].first+t[2*v+1].first;
    t[v].second=-1;
}


int req(int v, int tl, int tr, int n)
{
    if (tl==tr)
    {
        --t[v].first;
        return t[v].second;
    }
    int tm=(tl+tr)/2;
    t[v].first--;
    if (t[2*v].first>=n)
        req(2*v, tl, tm, n);
    else
        req(2*v+1, tm+1,tr,n-t[2*v].first);




}


int main()
{
    int n, k; cin>>n>>k;
    build_tree(1,1,100000);
    int cur=k;
    for (int i=0; i<n; ++i)
    {
        int h=req(1,1,100000, cur);
        cout<<h<<" ";
        if (i==n-1)
            break;
        cur=(cur-1+k)%(n-1-i);
        if (cur==0)
            cur+=n-1-i;
    }
}

Edited by author 10.03.2013 07:20

He donated all the money to the poor and went to bed with a clear conscience...

WpKRwvgKenn
output shold be
"WpKRwvgKenneKgvwRKpW"
not
"WpKRwvgKennneKgvwRKpW"

Этот код работает везде, кроме как на этом сайте
Runtime error test 1. Почему???

import sys
array4 = dict()
result = ""
for i in sys.stdin.read().split("\n")[1:]:
    array4[str(i.split(" ")[0])] = str(i.split(" ")[1])
def QSort(arr3):
    if len(arr3) < 2:
        return arr3
    else:
        privot, do3, sered, posle = int(arr3[list(arr3.keys())[0]]), dict(), dict(), dict()
        for x, i in arr3.items():
            if int(i) > privot:
                do3[x] = i
            elif int(i) < privot:
                posle[x] = i
            else:
                sered[x] = i
    return {**QSort(do3), **sered, **QSort(posle)}
for t, y in QSort(array4).items():
    result += f"{t} {y}\n"
print(result.strip('\n'))

# include <bits/stdc++.h>

using namespace std;
using ll = long long;

#define vec vector
#define int long long
#define ld long double

#define f first
#define s second

#define pb push_back

#define fe(x, a) for (auto& x : a)

#define pw(x) (1ll << x)
#define sz(x) (int)x.size()
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()

using pii = pair<int, int>;

const int mod = 1e9 + 7;
const ll OO = 1e16;
const int N = 1e5 + 2;
const ld eps = 1e-3;

template<typename T>
bool umn(T &a, T b) { return a > b ? (a = b, 1) : 0; }

template<typename T>
bool umx(T &a, T b) { return a < b ? (a = b, 1) : 0; }

void solve() {
    int n, best = -1, ans = 0;
    cin >> n;
    set<pii> setik;
    for (int i = 0, x; i < n + 1; ++i) {
        x = -OO;
        if (i < n)
            cin >> x;
        if (i && x < (*setik.begin()).f)
            if (umx(best, i - (*setik.begin()).s - !(*setik.begin()).s))
                ans = (*setik.begin()).s;

        setik.insert({x, i});
    }
    cout << ans + 1;
}

int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int t = 1;
//    cin >> t;
    while (t--)
        solve();
}

Input:
9
16372 24788
000010111
001111101
010010001
010011001
111100101
010100100
110011000
100000000
111110000

Possible answer (a's should be the same, d's can differ in you answer):
163720
000000000
000000d0d
0000d000d
0000dd00d
00dd00d0d
000d00d00
0d00dd000
000000000
0dddd0000

Matrix for the graph in the answer:
000010111
001111000
010000000
010000000
110000000
010000000
100000000
100000000
100000000

Edited by author 12.05.2023 13:49

If you have a mistake, most likely you delete connections of the input graph in a wrong way.
And most likely, because you search the connected components incorrectly.

If you are solving using formula, there are some corner cases.

IN
0 0
0 0
OUT
0

IN
17 0
0 17
OUT
17

And just some random test.

IN
7 9
6 8
OUT
15.165177459575631

Who can tell me…………
Why got WA#12?

#include <iostream>

int main()
{
    long long int y1;
    int y2, y3, y4;

    std::cin;
    std::cin >> y1;
    y4 = y1 / 16777216;
    y1 = y1 % 16777216;
    y3 = y1 / 65536;
    y1 = y1 % 65536;
    y2 = y1 / 256;
    y1 = y1 % 256;
    y1 = y1 * 16777216 + y2 * 65536 + y3 * 256 + y4;
    std::cout << y1;
    return 0;
}
в данном коде есть проблема: в данном компиляторе первый cin ничего не делает и второй получает на вход строку.