Codeforces 430 B. Balls Game

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B. Balls Game

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

Iahub is training for the IOI. What is a better way to train than playing a Zuma-like game?

There are n balls put in a row. Each ball is colored in one of k colors.
Initially the row doesn't contain three or more contiguous balls with the same color. Iahub has a single ball of color x. He can insert his ball at any
position in the row (probably, between two other balls). If at any moment there are three or more contiguous balls of the same color in the row, they are destroyed immediately. This rule is applied multiple times, until there are no more sets of 3 or more
contiguous balls of the same color.

For example, if Iahub has the row of balls [black, black, white, white, black, black] and a white ball, he can insert the ball between two white balls. Thus three white balls are destroyed, and then four black balls become contiguous, so all four balls are
destroyed. The row will not contain any ball in the end, so Iahub can destroy all 6 balls.

Iahub wants to destroy as many balls as possible. You are given the description of the row of balls, and the color of Iahub's ball. Help Iahub train for the IOI by telling him the maximum number of balls from the row he can destroy.

Input

The first line of input contains three integers: n (1 ≤ n ≤ 100), k (1 ≤ k ≤ 100)
and x (1 ≤ x ≤ k). The next line contains n space-separated
integers c1, c2, ..., cn (1 ≤ ci ≤ k).
Number ci means
that the i-th ball in the row has color ci.

It is guaranteed that the initial row of balls will never contain three or more contiguous balls of the same color.

Output

Print a single integer — the maximum number of balls Iahub can destroy.

Sample test(s)

input

6 2 2
1 1 2 2 1 1

output

6

input

1 1 1
1

output

0

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int Stack[110],top;
int n,k,c;
int a[110];

int doit(int x)
{
int ans=0,nt=top;
if(Stack[top]!=x)
{
for(int i=top;i>0;i--)
{
if(Stack[i]==Stack[top]) ans++,nt=i;
else break;
}
}
if(ans<3)
{
Stack[++top]=x;
ans=0;
}
else
{
top=nt;
Stack[top]=x;
}
return ans;
}

int ck(int pos,int x)
{
int ret=0;
top=0;
for(int i=0;i<n;i++)
{
ret+=doit(a[i]);
if(i==pos)
{
ret+=doit(x);
}
}
ret+=doit(999);
return ret;
}

int main()
{
scanf("%d%d%d",&n,&k,&c);
for(int i=0;i<n;i++)
{
scanf("%d",a+i);
}
int ans=0;
for(int i=0;i<n-1;i++)
{
if(a[i]==a[i+1]&&a[i]==c)
ans=max(ans,ck(i,c)-1);
}
printf("%d\n",ans);
return 0;
}