| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 5783 | Accepted: 3792 |
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.
Source
解题思路：
推断一个字符串是不是surpring。
条件：该字符串的全部D-pairs,D是字符串中两个字母的距离, 假设全部的D-pairs都不同，那么该字符串是D-unique。
假设对全部的距离D，都满足D-unique，那么字符串就是surpring.
比方ZGBG
0-pairs : ZG GB BG 不同样，是0-unique
1-pairs : ZB GG 不同样，是1-unique
2-pairs: ZG 不同样。是2-unique
综上。ZGBG是surpring
每个D-pairs进行推断。假设在推断一个D-pairs过程中有同样的，那么该字符串肯定不是surpring。
用map<string,bool> 来推断是否字符串已经出现过。要注意其声明的位置，要在每一层D循环里面声明。
代码：
#include <iostream>
#include <stdio.h>
#include <map>
#include <string.h>
using namespace std;
char str[80]; int main()
{
while(scanf("%s",str)!=EOF&&str[0]!='*')
{
int len=strlen(str);
if(len<=2)
{
cout<<str<<" is surprising."<<endl;
continue;
}
bool ok=1;//是surpring
for(int d=0;d<=len-2;d++)//距离d
{
bool dtap=1;//是D-unique
map<string,bool>mp;//注意其声明的位置
for(int s=0;s<=len-2&&s+d+1<len;s++)
{
string temp="";
temp+=str[s];
temp+=str[s+d+1];
if(mp[temp])//该字符串出现过
{
dtap=0;
break;
}
else
mp[temp]=1;
}
if(!dtap)
{
ok=0;
break;
}
}
if(ok)
cout<<str<<" is surprising."<<endl;
else
cout<<str<<" is NOT surprising."<<endl;
}
return 0;
}
版权声明：本文博客原创文章。博客，未经同意，不得转载。