| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 6017 | Accepted: 3934 |
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：给你个字符串例如abcd，距离为0的字符串有ab,bc,cd，距离为1的字符串有ac,bc，距离为2的字符串有ad，如果不存在在相同距离下的两个字符串相同，则这个字符串是surprising，否则是NOT surprising；
#include <iostream>
#include <string.h>
using namespace std;
int main(){
char s[100];
while (cin>>s){
if (strcmp(s,"*")==0)
break;
int len=strlen(s),flag=0;
for (int i=0;i<len;i++){
for (int j=i+1;j<len;j++){
if (s[i]==s[j]){
for (int k=1;k<len-j;k++){
if (s[k+i]==s[k+j]){
flag=1;
goto A;
}
}
}
}
}
A:
if (flag==0)
cout<<s<<" is surprising."<<endl;
else
cout<<s<<" is NOT surprising."<<endl;
}
return 0;
}原文地址：http://blog.csdn.net/codeforcer/article/details/40786021
