2012 ACM/ICPC Asia Regional Changchun Online-LianLianKan

LianLianKan

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2389 Accepted Submission(s): 382



[align=left]Problem Description[/align]
I like playing game with my friends, although sometimes look pretty naive. Today I invent a new game called LianLianKan. The game is about playing on a number stack.

Now we have a number stack, and we should link and pop the same element pairs from top to bottom. Each time, you can just link the top element with the same-value element. After pop them from stack, all left elements will fall down. Although the game seems
to be interesting, it's really naive indeed.



To prove I am wisdom among my friends, I add an additional rule to the game: for each top element, it can just link with the same-value element whose distance is less than 5.

Before the game, I want to check whether I have a solution to pop all elements in the stack.

[align=left]Input[/align]
There are multiple test cases.

The first line is an integer N indicating the number of elements in the stack initially. (1 <= N <= 1000)

The next line contains N integers ai indicating the elements from bottom to top. (0 <= ai <= 2,000,000,000)

[align=left]Output[/align]
For each test case, output “1” if I can pop all elements; otherwise output “0”.

[align=left]Sample Input[/align]

2
1 1
3
1 1 1
2
1000000 1

[align=left]Sample Output[/align]

1
0
0题意：类似于我们玩的连连看，从上往下，6个水果内如果有相同的2个水果则可以消去，直至水果被消完，输出1，或是找不到可以消去的水果，输出0。思路：数据顶多1000个，直接暴力过。[code]#include<stdio.h>
int main()
{
int a[1111];
int n,i,j;
while(scanf("%d",&n)!=EOF)
{
for(i=0;i<n;i++)
scanf("%d",&a[i]);
if(n%2) //因为每次只能消去2个，所以奇数个水果必然消不完
{
printf("0\n");
continue;
}
int flag=n;
while(flag>0)
{
int hash=0;
for(i=0;i<n;i++)
{
if(a[i]!=-1)
{
int num=0;
for(j=i+1;j<n;j++)
{
if(num>4) //6个水果内找不到2个相同的
break;
if(a[j]==a[i])
{
a[j]=-1; //消去的水果进行标记
a[i]=-1; //消去的水果进行标记
hash=1; //输出标记
flag-=2; //总水果数对应的减去2
break;
}
else
{
if(a[j]!=-1)
num++;
}
}
}
if(hash==1)
break;
}
if(hash==0)
{
printf("0\n");
break;
}
}
if(flag==0)
printf("1\n");
}
return 0;
}

[/code]