Codeforces Educational Codeforces Round 45 (Rated for Div. 2)题解

A. Commentary Boxes

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Berland Football Cup starts really soon! Commentators from all over the world come to the event.

Organizers have already built $n$ commentary boxes. $m$ regional delegations will come to the Cup. Every delegation should get the same number of the commentary boxes. If any box is left unoccupied then the delegations will be upset. So each box should be occupied by exactly one delegation.

If $n$ is not divisible by $m$, it is impossible to distribute the boxes to the delegations at the moment.

Organizers can build a new commentary box paying $a$ burles and demolish a commentary box paying $b$ burles. They can both build and demolish boxes arbitrary number of times (each time paying a corresponding fee). It is allowed to demolish all the existing boxes.

What is the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$)?

Input

The only line contains four integer numbers $n$, $m$, $a$ and $b$ ($1\le n,m\le {10}^{12}$, $1\le a,b\le 100$), where $n$ is the initial number of the commentary boxes, $m$ is the number of delegations to come, $a$ is the fee to build a box and $b$ is the fee to demolish a box.

Output

Output the minimal amount of burles organizers should pay to satisfy all the delegations (i.e. to make the number of the boxes be divisible by $m$). It is allowed that the final number of the boxes is equal to $0$.

Examples

input

Copy

9 7 3 8

output

Copy

15

input

Copy

2 7 3 7

output

Copy

14

input

Copy

30 6 17 19

output

Copy

0

Note

In the first example organizers can build $5$ boxes to make the total of $14$ paying $3$ burles for the each of them.

In the second example organizers can demolish $2$ boxes to make the total of $0$ paying $7$ burles for the each of them.

In the third example organizers are already able to distribute all the boxes equally among the delegations, each one get $5$

 boxes.

【分析】比较一下 n % m * b和 (m - n % m) * a的大小即可

【代码】

#include
using namespace std;

long long min(long long a, long long b)
{
    if (a > b) return b;
    return a;
}

int main()
{
    long long n, m, ans, c;
    int a, b;
    scanf("%I64d%I64d%d%d", &n, &m, &a, &b);
    c = n % m;
    printf("%I64d", min(c * b, (m - c) * a));
    return 0;
}

B. Micro-World

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

You have a Petri dish with bacteria and you are preparing to dive into the harsh micro-world. But, unfortunately, you don't have any microscope nearby, so you can't watch them.

You know that you have $n$ bacteria in the Petri dish and size of the $i$-th bacteria is $a_i$. Also you know intergalactic positive integer constant $K$.

The $i$-th bacteria can swallow the $j$-th bacteria if and only if $a_i>a_j$ and $a_i\le a_j+K$. The $j$-th bacteria disappear, but the $i$-th bacteria doesn't change its size. The bacteria can perform multiple swallows. On each swallow operation any bacteria $i$ can swallow any bacteria $j$if $a_i>a_j$ and $a_i\le a_j+K$. The swallow operations go one after another.

For example, the sequence of bacteria sizes $a=[101,53,42,102,101,55,54]$ and $K=1$. The one of possible sequences of swallows is: $[101,53,42,102,\underset{–}{101},55,54]$ $\to$ $[101,\underset{–}{53},42,102,55,54]$ $\to$ $[\underset{–}{101},42,102,55,54]$ $\to$ $[42,102,55,\underset{–}{54}]$ $\to$ $[42,102,55]$. In total there are $3$ bacteria remained in the Petri dish.

Since you don't have a microscope, you can only guess, what the minimal possible number of bacteria can remain in your Petri dish when you finally will find any microscope.

Input

The first line contains two space separated positive integers $n$ and $K$ ($1\le n\le 2\cdot {10}^5$, $1\le K\le {10}^6$) — number of bacteria and intergalactic constant $K$.

The second line contains $n$ space separated integers $a_1,a_2,\dots ,a_n$ ($1\le a_i\le {10}^6$) — sizes of bacteria you have.

Output

Print the only integer — minimal possible number of bacteria can remain.

Examples

input

Copy

7 1
101 53 42 102 101 55 54

output

Copy

3

input

Copy

6 5
20 15 10 15 20 25

output

Copy

1

input

Copy

7 1000000
1 1 1 1 1 1 1

output

Copy

7

Note

The first example is clarified in the problem statement.

In the second example an optimal possible sequence of swallows is: $[20,15,10,15,\underset{–}{20},25]$ $\to$ $[20,15,10,\underset{–}{15},25]$ $\to$ $[20,15,\underset{–}{10},25]$ $\to$$[20,\underset{–}{15},25]$ $\to$ $[\underset{–}{20},25]$ $\to$ $\left[25\right]$.

In the third example no bacteria can swallow any other bacteria.

【分析】排序一下，a[i] < a[i + 1] && a[i] + k >= a[i + 1] 就吃，注意判断一下a[i] == a[i + 1] 的情况。

【代码】

#include
#include

using namespace std;

priority_queue, greater> q;

int main()
{
    int n, k, i, x, a1, a2, cnt = 1;
    scanf("%d%d", &n, &k);
    for (i = 1; i <= n; ++i)
    {
        scanf("%d", &x);
        q.push(x);
    }
    a1 = q.top();
    q.pop();
    if (q.empty())
    {
        printf("1");
    }
    else
    {
        while (!q.empty())
        {
            a2 = q.top();
            q.pop();
            if (a2 > a1 && a1 + k >= a2)
                n -= cnt, cnt = 1;
            else if (a2 > a1) cnt = 1;
            else if (a2 == a1) ++cnt;
            a1 = a2;
        }
        printf("%d", n);
    }
    return 0;
}

C. Bracket Sequences Concatenation Problem

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A bracket sequence is a string containing only characters "(" and ")".

A regular bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters "1" and "+" between the original characters of the sequence. For example, bracket sequences "()()", "(())" are regular (the resulting expressions are: "(1)+(1)", "((1+1)+1)"), and ")(" and "(" are not.

You are given $n$ bracket sequences $s_1,s_2,\dots ,s_n$. Calculate the number of pairs $i,j(1\le i,j\le n)$ such that the bracket sequence $s_i+s_j$ is a regular bracket sequence. Operation $+$ means concatenation i.e. "()(" + ")()" = "()()()".

If $s_i+s_j$ and $s_j+s_i$ are regular bracket sequences and $i\ne j$, then both pairs $(i,j)$ and $(j,i)$ must be counted in the answer. Also, if $s_i+s_i$ is a regular bracket sequence, the pair $(i,i)$ must be counted in the answer.

Input

The first line contains one integer $n(1\le n\le 3\cdot {10}^5)$ — the number of bracket sequences. The following $n$ lines contain bracket sequences — non-empty strings consisting only of characters "(" and ")". The sum of lengths of all bracket sequences does not exceed $3\cdot {10}^5$.

Output

In the single line print a single integer — the number of pairs $i,j(1\le i,j\le n)$ such that the bracket sequence $s_i+s_j$ is a regular bracket sequence.

Examples

input

Copy

3
)
()
(

output

Copy

2

input

Copy

2
()
()

output

Copy

4

Note

In the first example, suitable pairs are $(3,1)$ and $(2,2)$.

In the second example, any pair is suitable, namely $(1,1),(1,2),(2,1),(2,2)$.

【分析】用cnt统计每一串括号的情况，遇到'(' +1 遇到 ')' -1，那么cnt为相反数的括号是相互匹配的。但是要注意类似')('这种括号是不能与任何括号匹配的，其特征为 前缀cnt  < 整串cnt && 前缀cnt < 0 ，枚举这个前缀即可。比赛的时候ans用longlong了，但是中间爆了。。debug了一个小时，不知道在玩什么。

【代码】

#include
#include
#define ZJ 310000
using namespace std;

long long d[710000];
char s[310000];
int js[310000];
int main()
{
    int i, j, n, cnt;
    bool a, b;
    long long ans = 0;
    scanf("%d", &n);
    for (i = 1; i <= n; ++i)
    {
        a = 0;
        cnt = 0;
        scanf("%s", s);
        
        for (j = 0; j < strlen(s); ++j)
        {
            if (s[j] == '(') ++cnt;
            else --cnt;
            js[j + 1] = cnt;
        }
        for (j = 1; j <= strlen(s); ++j)
        {
            if (js[j] < 0 && js[strlen(s)] > js[j]) a = 1;
            js[j] = 0;
        }
        if (a) continue;
        else d[ZJ + cnt]++;
    }
    for (i = 0; i <= ZJ; ++i) ans += d[ZJ + i] * d[ZJ - i];
    printf("%I64d", ans);
    return 0;
}

D. Graph And Its Complement

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Given three numbers $n,a,b$. You need to find an adjacency matrix of such an undirected graph that the number of components in it is equal to $a$, and the number of components in its complement is $b$. The matrix must be symmetric, and all digits on the main diagonal must be zeroes.

In an undirected graph loops (edges from a vertex to itself) are not allowed. It can be at most one edge between a pair of vertices.

The adjacency matrix of an undirected graph is a square matrix of size $n$ consisting only of "0" and "1", where $n$ is the number of vertices of the graph and the $i$-th row and the $i$-th column correspond to the $i$-th vertex of the graph. The cell $(i,j)$ of the adjacency matrix contains $1$ if and only if the $i$-th and $j$-th vertices in the graph are connected by an edge.

A connected component is a set of vertices $X$ such that for every two vertices from this set there exists at least one path in the graph connecting this pair of vertices, but adding any other vertex to $X$ violates this rule.

The complement or inverse of a graph $G$ is a graph $H$ on the same vertices such that two distinct vertices of $H$ are adjacent if and only if they are not adjacent in $G$.

Input

In a single line, three numbers are given $n,a,b(1\le n\le 1000,1\le a,b\le n)$: is the number of vertexes of the graph, the required number of connectivity components in it, and the required amount of the connectivity component in it's complement.

Output

If there is no graph that satisfies these constraints on a single line, print "NO" (without quotes).

Otherwise, on the first line, print "YES"(without quotes). In each of the next $n$ lines, output $n$ digits such that $j$-th digit of $i$-th line must be $1$ if and only if there is an edge between vertices $i$ and $j$ in $G$ (and $0$ otherwise). Note that the matrix must be symmetric, and all digits on the main diagonal must be zeroes.

If there are several matrices that satisfy the conditions — output any of them.

Examples

input

Copy

3 1 2

output

Copy

YES
001
001
110

input

Copy

3 3 3

output

Copy

NO

【分析】画下图可以发现，a和b至少有一个为1才能符合题意。针对b == 1 随意分成a块，补图一定只有一块，针对

a == 1 ，可以swap(a, b) ， 然后输出其补图即可。a == 1 && b == 1 的几种特殊情况要特判。

【代码】

#include
using namespace std;

int d[1100][1100];
int main()
{
    int i, j, k, a, b, n;
    scanf("%d%d%d", &n, &a, &b);
    if (a != 1 && b != 1)
    {
        printf("NO");
    }
    else
    {
        if (a == 1 && b == 1)
        {
            if (n == 1) printf("YES\n0");
            else if (n == 2 || n == 3) printf("NO");
            else
            {
                printf("YES\n");
                for (i = 1; i <= n; ++i)
                    for (j = 1; j <= n; ++j)
                        if (j - 1 == i) d[i][j] = d[j][i] = 1;
                for (i = 1; i <= n; ++i)
                {
                    for (j = 1; j <= n; ++j) printf("%d", d[i][j]);
                    printf("\n");
                }
            }
        }
        else if (b == 1)
        {
            printf("YES\n");
            for (i = a; i < n; ++i) d[i][i + 1] = d[i + 1][i] = 1;
            for (i = 1; i <= n; ++i)
            {
                for (j = 1; j <= n; ++j) printf("%d", d[i][j]);
                printf("\n");
            }
        }
        else
        {
            printf("YES\n");
            for (i = b; i < n; ++i) d[i][i + 1] = d[i + 1][i] = 1;
            for (i = 1; i <= n; ++i)
            {
                for (j = 1; j <= n; ++j)
                    if (i != j) printf("%d", !d[i][j]);
                    else printf("%d", d[i][j]);
                printf("\n");
            }
        }
    }
    return 0;
}