Codeforces Round #487 (Div. 2) ABC题解

A. A Blend of Springtime

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

When the curtains are opened, a canvas unfolds outside. Kanno marvels at all the blonde colours along the riverside — not tangerines, but blossoms instead.

"What a pity it's already late spring," sighs Mino with regret, "one more drizzling night and they'd be gone."

"But these blends are at their best, aren't they?" Absorbed in the landscape, Kanno remains optimistic.

The landscape can be expressed as a row of consecutive cells, each of which either contains a flower of colour amber or buff or canary yellow, or is empty.

When a flower withers, it disappears from the cell that it originally belonged to, and it spreads petals of its colour in its two neighbouring cells (or outside the field if the cell is on the side of the landscape). In case petals fall outside the given cells, they simply become invisible.

You are to help Kanno determine whether it's possible that after some (possibly none or all) flowers shed their petals, at least one of the cells contains all three colours, considering both petals and flowers. Note that flowers can wither in arbitrary order.

Input

The first and only line of input contains a non-empty string $s$ consisting of uppercase English letters 'A', 'B', 'C' and characters '.' (dots) only ($\left|s\right|\le 100$) — denoting cells containing an amber flower, a buff one, a canary yellow one, and no flowers, respectively.

Output

Output "Yes" if it's possible that all three colours appear in some cell, and "No" otherwise.

You can print each letter in any case (upper or lower).

Examples

input

Copy

.BAC.

output

Copy

Yes

input

Copy

AA..CB

output

Copy

No

Note

In the first example, the buff and canary yellow flowers can leave their petals in the central cell, blending all three colours in it.

In the second example, it's impossible to satisfy the requirement because there is no way that amber and buff meet in any cell.

题意：问串中是否存在连续的‘ABC’的全排列.

题解：模拟下即可。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long double ll;
ll a[10005],b[10005];
int main()
{
	string s;
	cin>>s;
	for(int i=2;i<s.size();i++)
	{
		if(s[i]!='.'&&s[i-1]!='.'&&s[i-2]!='.'&&s[i]!=s[i-1]&&s[i-1]!=s[i-2]&&s[i]!=s[i-2])
		{
			printf("Yes\n");return 0;
		}
	}
	cout<<"No\n";
	return 0;
}

B. A Tide of Riverscape

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Walking along a riverside, Mino silently takes a note of something.

"Time," Mino thinks aloud.

"What?"

"Time and tide wait for no man," explains Mino. "My name, taken from the river, always reminds me of this."

"And what are you recording?"

"You see it, tide. Everything has its own period, and I think I've figured out this one," says Mino with confidence.

Doubtfully, Kanno peeks at Mino's records.

The records are expressed as a string $s$ of characters '0', '1' and '.', where '0' denotes a low tide, '1' denotes a high tide, and '.' denotes an unknown one (either high or low).

You are to help Mino determine whether it's possible that after replacing each '.' independently with '0' or '1', a given integer $p$ is not a period of the resulting string. In case the answer is yes, please also show such a replacement to Mino.

In this problem, a positive integer $p$ is considered a period of string $s$, if for all $1\le i\le \left|s\right|-p$, the $i$-th and $(i+p)$-th characters of $s$ are the same. Here $\left|s\right|$ is the length of $s$.

Input

The first line contains two space-separated integers $n$ and $p$ ($1\le p\le n\le 2000$) — the length of the given string and the supposed period, respectively.

The second line contains a string $s$ of $n$ characters — Mino's records. $s$ only contains characters '0', '1' and '.', and contains at least one '.' character.

Output

Output one line — if it's possible that $p$ is not a period of the resulting string, output any one of such strings; otherwise output "No" (without quotes, you can print letters in any case (upper or lower)).

Examples

input

Copy

10 7
1.0.1.0.1.

output

Copy

1000100010

input

Copy

10 6
1.0.1.1000

output

Copy

1001101000

input

Copy

10 9
1........1

output

Copy

No

Note

In the first example, $7$ is not a period of the resulting string because the $1$-st and $8$-th characters of it are different.

In the second example, $6$ is not a period of the resulting string because the $4$-th and $10$-th characters of it are different.

In the third example, $9$ is always a period because the only constraint that the first and last characters are the same is already satisfied.

Note that there are multiple acceptable answers for the first two examples, you can print any of them.

题意：可以把原串中的‘.’变为0或1，问是否存在一个串使的串中某个位置s[i]!=s[i+p].

题解:直接模拟下即可，记得最后把所有‘.’都变成0或1。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
typedef long double ll;

int main()
{
	int n;cin>>n;
	string s;int p;cin>>p>>s;
	int flag=0;
	for(int i=0;i<n;i++)
	{
		if(s[i]=='.')
		{
			if(i+p<n)
			{
				flag=1;
				if(s[i+p]=='1')s[i]='0';
				else if(s[i+p]=='0') s[i]='1';
				else 
				{
					s[i]='1';s[i+p]='0';
				}
				break;
			}
		}
		if(s[i]=='.')
		{
			if(i-p>=0)
			{
				flag=1;
				if(s[i-p]=='1')s[i]='0';
				else if(s[i-p]=='0')s[i]='1';
				else if(s[i-p]=='.')
				{
					s[i]='1';s[i-p]='0';
				}
				break;
			}
		}
	}
	for(int i=0;i<s.size();i++)
	{
		if(i+p<n) if(s[i]!=s[i+p]) flag=1;
	}
    if(flag==1)
    {
    	for(int i=0;i<n;i++) if(s[i]=='.')s[i]='1';
	}
	if(flag)
	cout<<s<<endl;
	else cout<<"No\n";
	return 0;
}

C. A Mist of Florescence

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

As the boat drifts down the river, a wood full of blossoms shows up on the riverfront.

"I've been here once," Mino exclaims with delight, "it's breathtakingly amazing."

"What is it like?"

"Look, Kanno, you've got your paintbrush, and I've got my words. Have a try, shall we?"

There are four kinds of flowers in the wood, Amaranths, Begonias, Centaureas and Dianthuses.

The wood can be represented by a rectangular grid of $n$ rows and $m$ columns. In each cell of the grid, there is exactly one type of flowers.

According to Mino, the numbers of connected components formed by each kind of flowers are $a$, $b$, $c$ and $d$ respectively. Two cells are considered in the same connected component if and only if a path exists between them that moves between cells sharing common edges and passes only through cells containing the same flowers.

You are to help Kanno depict such a grid of flowers, with $n$ and $m$ arbitrarily chosen under the constraints given below. It can be shown that at least one solution exists under the constraints of this problem.

Note that you can choose arbitrary $n$ and $m$ under the constraints below, they are not given in the input.

Input

The first and only line of input contains four space-separated integers $a$, $b$, $c$ and $d$ ($1\le a,b,c,d\le 100$) — the required number of connected components of Amaranths, Begonias, Centaureas and Dianthuses, respectively.

Output

In the first line, output two space-separated integers $n$ and $m$ ($1\le n,m\le 50$) — the number of rows and the number of columns in the grid respectively.

Then output $n$ lines each consisting of $m$ consecutive English letters, representing one row of the grid. Each letter should be among 'A', 'B', 'C' and 'D', representing Amaranths, Begonias, Centaureas and Dianthuses, respectively.

In case there are multiple solutions, print any. You can output each letter in either case (upper or lower).

Examples

input

Copy

5 3 2 1

output

Copy

4 7
DDDDDDD
DABACAD
DBABACD
DDDDDDD

input

Copy

50 50 1 1

output

Copy

4 50
CCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCCC
ABABABABABABABABABABABABABABABABABABABABABABABABAB
BABABABABABABABABABABABABABABABABABABABABABABABABA
DDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDDD

input

Copy

1 6 4 5

output

Copy

7 7
DDDDDDD
DDDBDBD
DDCDCDD
DBDADBD
DDCDCDD
DBDBDDD
DDDDDDD

Note

In the first example, each cell of Amaranths, Begonias and Centaureas forms a connected component, while all the Dianthuses form one.

题意：问是否能构成一个图，有a个‘A’联通块，b个‘B’联通块，c个‘C’联通块，d个‘D’

联通块，且图的不大于50*50。

题解：直接把图分为两部分上面全填上‘A’，下面全填上‘B’，然后a--，b--，隔空填A，B，C，D即可

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
char s[55][55];
int check(int i,int j,char ch)
{
	if(s[i][j-1]==ch) return 0;
	if(s[i][j+1]==ch) return 0;
	if(s[i+1][j]==ch) return 0;
	if(s[i-1][j]==ch) return 0;
	return 1;
}
int main()
{
	int a,b,c,d;
	int n=50;
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)
		{
			if(i<=25)
			s[i][j]='A';
			else s[i][j]='B';
		}
	}
	cin>>a>>b>>c>>d;
	a--;b--;
	
	for(int i=1;i<=25;i++)
	{
		for(int j=1;j<=n;j+=2)
		{
			if(b>0&&check(i,j,'B')) s[i][j]='B',b--;
			else if(c>0&&check(i,j,'C')) s[i][j]='C',c--;
			else if(d>0&&check(i,j,'D')) s[i][j]='D',d--;
		}
	}
	
	for(int i=26;i<=n;i++)
	{
		for(int j=1;j<=n;j+=2)
		{
			if(a>0&&check(i,j,'A')) s[i][j]='A',a--;
			else if(c>0&&check(i,j,'C')) s[i][j]='C',c--;
			else if(d>0&&check(i,j,'D')) s[i][j]='D',d--;
		}
	}
	
	cout<<50<<" 50\n";
	
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=n;j++)cout<<s[i][j];
		cout<<endl;
	}
	return 0;
}