POJ 3096 Surprising Strings

标签：统一   get   out   because   uppercase   issue   letter   red   sam   

The D-pairs of a string of letters are the ordered pairs of letters that are distance D from each other. A string is D-unique if all of its D-pairs are different. A string is surprising if it is D-unique for every possible distance D.

Consider the string ZGBG. Its 0-pairs are ZG, GB, and BG. Since these three pairs are all different, ZGBG is 0-unique. Similarly, the 1-pairs of ZGBG are ZB and GG, and since these two pairs are different, ZGBG is 1-unique. Finally, the only 2-pair of ZGBG is ZG, so ZGBG is 2-unique. Thus ZGBG is surprising. (Note that the fact that ZG is both a 0-pair and a 2-pair of ZGBG is irrelevant, because 0 and 2 are different distances.)

Acknowledgement: This problem is inspired by the "Puzzling Adventures" column in the December 2003 issue of Scientific American.

Input

Output

Sample Input

ZGBG
X
EE
AAB
AABA
AABB
BCBABCC
*

Sample Output

ZGBG is surprising.
X is surprising.
EE is surprising.
AAB is surprising.
AABA is surprising.
AABB is NOT surprising.
BCBABCC is NOT surprising.

题目大意是在一个字符串中取任意间距的两个字符，如果在统一间距情况下，没有一组字符是重复的，那么这个数就是surprising。 枚举比较即可。

#include<iostream>
#include<string>

using namespace std;

int main()
{
	string s; 
	while(getline(cin,s))
	{
		if(s[0]==‘*‘)
			return 0;
		int x=s.length();

		int flag=0;
		for(int i=0;i<x;i++)
		{
			for(int j=i+1;j<x;j++)
			{
				if(s[i]==s[j])
				{
					//cout<<s[i]<<s[j]<<endl; 
					for(int k=1;k<x;k++)
					{
						if(s[i+k]==s[j+k]&&i+k<x&&j+k<x)
						{
							
							flag=1;
							}	
					}
				}
			}
		}
		if(flag==0)
			cout<<s<<" is surprising."<<endl;
			else
				cout<<s<<" is NOT surprising."<<endl;
	}
	
	return 0;
}

POJ 3096 Surprising Strings

标签：统一   get   out   because   uppercase   issue   letter   red   sam   

原文地址：http://www.cnblogs.com/xibeiw/p/7237713.html