Surprising Strings

本文介绍了一种判断字符串是否具有Surprising特性的方法。Surprising特性是指字符串在所有可能的距离D下都是D-unique。文章给出了一个示例,并提供了一个C++实现方案,用于检查输入的字符串是否符合此特性。

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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.


第一天的集训题。。水题
#include<iostream>
#include<cstring>
using namespace std;
char s[80];
int sum,k;
int main()
{
    while(cin>>s,strcmp(s,"*")!=0)
    {
        k=0;
        if (strlen(s)<=2) cout<<s<<" is surprising."<<endl;
        else
        {
            for(int i=1;i<strlen(s)-1;i++)
            {
                sum=0;
                for(int j=i;j<strlen(s);j++)
                    if (s[j]==s[j-i]) sum++;
                if (sum>=2)
                {
                    k=1;
                    break;
                }
            }
            if (k==1) cout<<s<<" is NOT surprising."<<endl;
            else cout<<s<<" is surprising."<<endl;
        }
    }
    return 0;
}

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