CF 319D(Have You Ever Heard About the Word?-模拟)

 

D. Have You Ever Heard About the Word?
time limit per test
6 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output

substring of a string is a contiguous subsequence of that string. So, string bca is substring of string abcabc, but string cc is not.

repeating block is a string formed by concatenating some string with itself. So, string abcabc is a repeating block, but strings abcabd,ababab are not.

You've got a sequence of Latin characters (string). At each step you find the shortest substring that is a repeating block, if there exists more than one you must choose the leftmost. As the substring is of form XX (X — some string) you replace this substring with X, in other words you delete one of the X substrings in the substring. You repeat this process until there remains no repeating block in the string.

How would the final string looks like? Look at the sample explanation to understand the statement more precise.

Input

In the first line of input you're given a string of small Latin characters with length between 1 to 50000, inclusive.

Output

Print the final string after applying changes.

Sample test(s)
input
abccabc
output
abc
input
aaaabaaab
output
ab
input
birdbirdbirdistheword
output
birdistheword
Note

At the first sample the string transforms as follows: abccabc  abcabc  abc.

At the second sample the string transforms as follows: aaaabaaab  aaabaaab  aabaaab  abaaab  abaab  abab  ab.


模拟能过……

 

E文是硬伤,要的是最短中最左子串,不是最左中最短子串……

 

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<functional>
#include<cmath>
#include<cctype>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define RepD(i,n) for(int i=n;i>=0;i--)
#define MEM(a) memset(a,0,sizeof(a))
#define MEMI(a) memset(a,127,sizeof(a))
#define MEMi(a) memset(a,128,sizeof(a))
#define MAXN (50000+10)
char s[MAXN];
int main()
{
//	freopen("CF319D.in","r",stdin);
//	freopen(".out","w",stdout);
	int t=1;
	while (t--)
	{
	int n=strlen(gets(s+1));
	For(len,n/2)
	{
		int tot=0;
		For(j,n-len)
		{
			if (s[j]==s[j+len]) tot++;else tot=0;
			if (tot==len)
			{
				Fork(k,j+1,n-len) s[k]=s[k+len];
				n-=len;len=0;break;
			}
		}			
	}
	s[n+1]=0;
	puts(s+1);
	}
	return 0;
}


 


 

posted @ 2013-06-24 20:22  爱生活,爱编程  阅读(318)  评论(0编辑  收藏  举报