CSU-ACM2017暑期训练1-Debug与STL hdu2736

题目链接:Surprising String

题目

Surprising Strings

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 510 Accepted Submission(s): 348

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

ZGBGXEEAABAABAAABBBCBABCC*

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：
对一个字符串，考虑它的所有只有两个字符（这两个字符相隔dis长）的子串，对于所有合法的dis，每一个情况中出现的所有子串都不应有相同的出现，满足这个条件的字符串称他为surprising。
思路：
map判重，枚举dis。
代码：
#include <iostream>#include <string>#include <cstdio>#include <cstring>#include <map>using namespace std;map<string,int>m;string s;string ss[88];int main(){    while(cin>>s)    {        if(s[0]=='*')            break;        int dis=0;        int flag=1;        int lens=s.length();        while(dis<=lens-2&&flag==1)        {            int k=0;            for(int i=0;i+dis<lens-1;i++)            {                string a,b;                a=s[i];                b=s[i+dis+1];                ss[k]=a+b;                //cout<<ss[k]<<endl;f                k++;            }            for(int j=0;j<k;j++)            {                if(m.find(ss[j])==m.end())                    m[ss[j]]=j;                else                {                    flag=0;                    break;                }            }            dis++;            m.clear();        }        cout<<s;        if(dis==lens-1)            printf(" is surprising.\n");        else            printf(" is NOT surprising.\n");    }    return 0;}