poj-3096-Suprising Strings

poj-3096-Suprising Strings

Surprising Strings

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 7456		Accepted: 4765

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

Mid-Central USA 2006

解法：

　　采用哈希 + 暴力的手段，用双重循环找出所有的 二元string，然后在 hashing vector中找是否存在一样的distance。复杂度为 O(n^2) 

17798509

sunshinestyle

3096

Accepted

436K

0MS

G++

656B

2017-11-04 10:56:31

#include <cstdio> 
#include <cstring> 
#include <cstdlib>  

const int MAXN = 100; 

int vis[MAXN*26 + MAXN]; 

int main(){ 
	freopen("in.txt", "r", stdin); 

	char ch[MAXN]; 

	while(scanf("%s", ch) != EOF){
		getchar();  
		if(strcmp(ch, "*") == 0){
			break; 
		}

		memset(vis, 0, sizeof(vis)); 

		int len = strlen(ch); 
		int flag = 0; 

		for(int i=1; i<len; ++i){
			for(int j=0; j + i <len; ++j){
				int cur = 26 * (ch[ j ] - 'A') + ch[ j+i ] - 'A'; 
				if( vis[ cur ] == i ){
					flag = 1; 
					break;  
				}
				vis[ cur ] = i; 
			}
			if(flag){ break; } 
		} 
		printf("%s is ", ch );  
		if( flag ){
			printf("NOT ");
		}
		printf("surprising.
"); 

	}
	return 0; 
}