zoukankan      html  css  js  c++  java
  • 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

    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; 
    }
    

      

  • 相关阅读:
    Azure产品目录
    AWS产品目录
    BD
    Cloud Resource
    do-release-upgrade升级笔记
    Gluster vs Ceph:开源存储领域的正面较量
    OpenStack大规模部署详解
    SECURITY ONION:防御领域的kali
    vue非父子组件间传参问题
    vue源码之响应式数据
  • 原文地址:https://www.cnblogs.com/zhang-yd/p/7782743.html
Copyright © 2011-2022 走看看