POJ 3096 Surprising Strings (枚举)

Surprising Strings

 

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 7033 Accepted: 4554

 

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

给出一个字符串,问同样间隔下挑选两个字符组成的字符串是否重复,若是,则 NOT surprising ,否则为 surprising

比如以 AABB 为例

间距为0有: AA、AB、BB

间距为1有: AB、AB (重复)

间距为2有: AB

 

思路

字符串长度并不是很长,因此直接暴力组合出这样所有的字符串,利用STL中的map判断即可。

 

AC 代码

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<iostream>
using namespace std;
#include<queue>
#include<map>
#define INF (1<<25)

int main()
{
    char str[105];
    while(gets(str))
    {
        if(strcmp(str,"*")==0)break;
        int len=strlen(str);
        bool flag=true;
        for(int li=1; li<len; li++) //间距
        {
            map<string,bool>sk;
            for(int i=0; i+li<len; i++)
            {
                string no="";   //组合字符串
                no+=str[i],no+=str[i+li];
                if(sk[no])
                {
                    flag=false;
                    break;
                }
                sk[no]=true;
            }
            if(!flag)break;
        }
        printf("%s is ",str);
        if(!flag)printf("NOT ");
        printf("surprising.\n");
    }
    return 0;
}