POJ 3096： Surprising Strings (水)

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本文介绍了一个关于字符串的独特性质——surprising特性，并提供了一种通过查找相同字符对及其距离来判断字符串是否拥有此特性的算法实现。代码示例展示了如何遍历字符串并使用集合存储不同距离，最终确定字符串是否令人惊讶。

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Surprising Strings

Time limit:1000 ms Memory limit:65536 kB OS:Linux

Problem Description

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.

题意：

输入一个字符串，相同字母的称为D-pairs，其距离为D-unique，如果只有一种大小的D-unique，则该字符串就为surprising

解题思路：

直接暴力查找所有相同的字符，如果出现就查找该长度有没有出现过，如果有就跳出，not suprising ，否则插入该距离。

Code：

#include <iostream>
#include <string>
#include <set>
using namespace std;

set<int> s;

int main()
{
    string str;
    while(cin>>str)
    {
        if(str=="*")
            break;
        int len=str.length();
        s.clear();
        int flag=1;
        for(int i=0;i<len-1;i++)
        {
            for(int j=i+1;j<len;j++)
            {
                if(str[i]==str[j])
                {
                    int d=j-i;
                    if(s.find(d)!=s.end())
                    {
                        flag=0;
                        break;
                    }
                    s.insert(d);
                }
            }
            if(flag==0)
                break;
        }

        if(flag)
        {
            cout<<str<<" is surprising."<<endl;
        }
        else
        {
            cout<<str<<" is NOT surprising."<<endl;
        }
    }
    return 0;
}