暑期ACM集训

2019-07-17

08:42:11

这是总结昨天的做题情况

总体来说，好久的没做题了，实力下降了许多，这一个月假又学习吧！！！！

A - Ropewalkers

Polycarp decided to relax on his weekend and visited to the performance of famous ropewalkers: Agafon, Boniface and Konrad.

The rope is straight and infinite in both directions. At the beginning of the performance, Agafon, Boniface and Konrad are located in positions $a$

Ropewalkers can walk on the rope. In one second, only one ropewalker can change his position. Every ropewalker can change his position exactly by $1$

You should find the minimum duration (in seconds) of the performance. In other words, find the minimum number of seconds needed so that the distance between each pair of ropewalkers can be greater or equal to $d$

Ropewalkers can walk to negative coordinates, due to the rope is infinite to both

Input

The only line of the input contains four integers $a$

Output

Output one integer — the minimum duration (in seconds) of the performance.

Examples

Input

5 2 6 3

Output

Input

3 1 5 6

Output

Input

8 3 3 2

Output

Input

2 3 10 4

Output

Note

In the first example: in the first two seconds Konrad moves for 2 positions to the right (to the position $8$

$8$

#include <bits/stdc++.h>
using namespace std;
int main()
{
    long long a, b, c, d;
    scanf("%lld %lld %lld %lld", &a, &b, &c, &d);
    long long s[3] = {a, b, c};
    sort(s, s + 3);
    long long int s1 = s[1] - s[0];
    long long int s2 = s[2] - s[1];
    int j = 2;
    if (s1 > d)
    {
        s1 = 0;
        j--;
    }
    if (s2 > d)
    {
        s2 = 0;
        j--;
    }
    cout << d * j - s1 - s2;
    return 0;
}

B - Email from Polycarp

Methodius received an email from his friend Polycarp. However, Polycarp's keyboard is broken, so pressing a key on it once may cause the corresponding symbol to appear more than once (if you press a key on a regular keyboard, it prints exactly one symbol).

For example, as a result of typing the word "hello", the following words could be printed: "hello", "hhhhello", "hheeeellllooo", but the following could not be printed: "hell", "helo", "hhllllooo".

Note, that when you press a key, the corresponding symbol must appear (possibly, more than once). The keyboard is broken in a random manner, it means that pressing the same key you can get the different number of letters in the result.

For each word in the letter, Methodius has guessed what word Polycarp actually wanted to write, but he is not sure about it, so he asks you to help him.

You are given a list of pairs of words. For each pair, determine if the second word could be printed by typing the first one on Polycarp's keyboard.

Input

The first line of the input contains one integer $n$

The first line of each description contains a single non-empty word $s$

It is guaranteed that the total length of all words $s$

Output

Output $n$

Examples

Input

4
hello
hello
hello
helloo
hello
hlllloo
hello
helo

Output

YES
YES
NO
NO

Input

5
aa
bb
codeforces
codeforce
polycarp
poolycarpp
aaaa
aaaab
abcdefghijklmnopqrstuvwxyz
zabcdefghijklmnopqrstuvwxyz

Output

NO
NO
YES
NO
NO

题解:我这道题开始想的时用#include <unordered_set>这个做,但是编译器不许通过😭
我有换了另一种方法
这个问题的难点就是hello中出现了两个ll
解决方法就是让两个hello helllloo同步

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int main()
{
    ll n;
    scanf("%lld", &n);
    while (n--)
    {
        string a;
        string b;
        cin >> a >> b;
        ll lenth1 = a.length();
        ll lenth2 = b.length();
        if (lenth1 > lenth2)
        {
            cout << "NO" << endl;
            continue;
        }
        if (a[0] != b[0])
        {
            cout << "NO" << endl;
            continue;
        }

        int i, j;
        int flag = 0;
        for (i = 0, j = 0; i < lenth1; i++)
        {
            if (a[i] == a[i + 1] && a[i] == b[j])
            {
                j++;
                continue;
            }

            if (a[i] != b[j])
            {
                flag = 1;
                break;
            }
            else
            {
                while (a[i] == b[j])
                {
                    j++;
                }
            }
        }
        if (i != lenth1 || j != lenth2 || flag)
        {
            cout << "NO" << endl;
        }
        else
        {
            cout << "YES" << endl;
        }
    }
    return 0;
}

C - Exam in BerSU (easy version)

The only difference between easy and hard versions is constraints.

A session has begun at Beland State University. Many students are taking exams.

Polygraph Poligrafovich is going to examine a group of $n$

The $i$
if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
if the student finds the ticket easy, he spends exactly $t_{i}$

Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.

The duration of the whole exam for all students is $M$

For each student $i$

Input

The first line of the input contains two integers $n$

The second line of the input contains $n$

It's guaranteed that all values of $t_{i}$

Output

Print $n$

Examples

Input

7 15
1 2 3 4 5 6 7

Output

0 0 0 0 0 2 3

Input

5 100
80 40 40 40 60

Output

0 1 1 2 3

Note

The explanation for the example 1.

Please note that the sum of the first five exam times does not exceed $M = 15$

In order for the $6$

In order for the $7$

$7$

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int m;
    cin >>m;
    int a[150]={0};
    for (int i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    int b[150]={0};
    b[0] = 0;
    for (int i = 1; i < n; i++)
    {
        int q = m;
        q -= a[i];
        int c[150] ={0};
        for (int z = 0; z < i; z++)
        {
            c[z] = a[z];
        }
        sort(c, c + i);
        for (int j = 0; j < i; j++)
        {
            if (q - c[j] >= 0)
            {
                b[i]++;
                q -= c[j];
            }
            else
            {
                break;
            }
        }
    }
    for (int i = 0; i < n; i++)
    {
        if (i != n - 1)
            cout <<  i - b[i] << " ";
        else
        {
            cout << i - b[i];
        }
    }
    return 0;
}

D - Exam in BerSU (hard version)

The only difference between easy and hard versions is constraints.

If you write a solution in Python, then prefer to send it in PyPy to speed up execution time.

A session has begun at Beland State University. Many students are taking exams.

Polygraph Poligrafovich is going to examine a group of $n$

The $i$
if this ticket is too hard to the student, he doesn't answer and goes home immediately (this process is so fast that it's considered no time elapses). This student fails the exam.
if the student finds the ticket easy, he spends exactly $t_{i}$

Students take the exam in the fixed order, one-by-one, without any interruption. At any moment of time, Polygraph Poligrafovich takes the answer from one student.

The duration of the whole exam for all students is $M$

For each student $i$

Input

The first line of the input contains two integers $n$

The second line of the input contains $n$

It's guaranteed that all values of $t_{i}$

Output

Print $n$

Examples

Input

7 15
1 2 3 4 5 6 7

Output

0 0 0 0 0 2 3

Input

5 100
80 40 40 40 60

Output

0 1 1 2 3

Note

The explanation for the example 1.

Please note that the sum of the first five exam times does not exceed $M = 15$

In order for the $6$

In order for the $7$

$7$

$7$

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
ll a[200000];
ll b[200000];
ll c[200000];
int main()
{
    ll n;
    cin >> n;
    ll m;
    cin >>m;
    for (ll i = 0; i < n; i++)
    {
        cin >> a[i];
    }
    ll sum = 0;
    ll count[200]={0};
    for(int i = 0; i < n; i++)
    {
        sum += a[i];
        if(sum > m)
        {
            ll ant = 0;
            ll summ = 0;
            for(int j = 100; j >= 1; j--)
            {
                summ += j * count[j];
                ant += count[j];
                if(sum - summ <= m )
                {
                    int x = sum - m;
                    int y = summ - x;
                    ant -= (y/j);
                    cout << ant << " ";
                    break;
                }
            }
            count[a[i]]++;
        }
        else
        {
            cout << 0 << " ";
            count[a[i]]++;
        }
        
    }
    return 0;
}

E - Vus the Cossack and a Contest

Vus the Cossack holds a programming competition, in which $n$ people participate. He decided to award them all with pens and notebooks. It is known that Vus has exactly $m$ pens and $k$ notebooks.

Determine whether the Cossack can reward all participants, giving each of them at least one pen and at least one notebook.

题解:简单

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n,m,k;
    cin >> n >> m >> k;
    if(n > m || n > k)
    {
        cout << "NO";    
    } 
    else
    {
        cout << "YES";
    }
    return 0;
}

E - Vus the Cossack and a Contest

posted @ 2019-07-17 09:02 Artimislyy 阅读(252) 评论(0) 编辑收藏举报

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