POJ 3096 Surprising Strings

Posted 2020-09-29 西北会法语

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

The input consists of one or more nonempty strings of at most 79 uppercase letters, each string on a line by itself, followed by a line containing only an asterisk that signals the end of the input.

Output

For each string of letters, output whether or not it is surprising using the exact output format shown below.

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;
}