Surprising Strings
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 6193   Accepted: 4036

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. 题目分析:一个字符串例如:AABB,字符的两两组合为:间距为0时:AA AB BB
间距为1时:AB AB (出现相同情况,可以判断该串为 not surprise )
间距为2时: AB
代码:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <string>
#include <iostream>
#include <iomanip>
#include <algorithm> using namespace std; struct node
{
char a;
char b;
bool operator == (const node &x)const{
return a==x.a&&b==x.b; //重载定义==符号 判断两个结构体相等
}
}q[10000]; int main()
{
char s[100];
int i, j, k, ll;
int len;
while(scanf("%s", s) && strcmp(s, "*")!=0 ){
len = strlen(s);
if(len<=2){
printf("%s is surprising.\n", s);
}
else{
bool flag=false;
for(i=0; i<=len-2; i++)
{
int e=0;
for(j=0; j+i<len; j++)
{
q[e].a=s[j]; q[e++].b=s[j+i+1];
}
for(k=0; k<e; k++)
{
for(ll=k+1; ll<e; ll++)
{
if(q[k] == q[ll]){
flag=true; break;
}
}
}
if(flag==true){
break;
}
}
if(flag==true)
printf("%s is NOT surprising.\n", s);
else
printf("%s is surprising.\n", s);
}
}
return 0;
}

【字符串题目】poj 3096 Surprising Strings的更多相关文章

  1. POJ 3096 Surprising Strings

    Surprising Strings Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5081   Accepted: 333 ...

  2. [ACM] POJ 3096 Surprising Strings (map使用)

    Surprising Strings Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 5783   Accepted: 379 ...

  3. POJ 3096 Surprising Strings(STL map string set vector)

    题目:http://poj.org/problem?id=3096 题意:给定一个字符串S,从中找出所有有两个字符组成的子串,每当组成子串的字符之间隔着n字符时,如果没有相同的子串出现,则输出 &qu ...

  4. POJ 3096:Surprising Strings

    Surprising Strings Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6258   Accepted: 407 ...

  5. C - Surprising Strings

                                   C - Surprising Strings 题意:输入一段字符串,假设在同一距离下有两个字符串同样输出Not surprising ,否 ...

  6. KMP POJ 2406 Power Strings

    题目传送门 /* 题意:一个串有字串重复n次产生,求最大的n KMP:nex[]的性质应用,感觉对nex加深了理解 */ /************************************** ...

  7. [POJ3096]Surprising Strings

    [POJ3096]Surprising Strings 试题描述 The D-pairs of a string of letters are the ordered pairs of letters ...

  8. HDOJ 2736 Surprising Strings

    Surprising Strings Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Other ...

  9. HDU 2736 Surprising Strings

                                    Surprising Strings Time Limit:1000MS     Memory Limit:65536KB     64 ...

随机推荐

  1. PhoneNumber

    项目地址:PhoneNumber 简介:一个获取号码归属地和其他信息(诈骗.骚扰等)的开源库   一个获取号码归属地和其他信息(诈骗.骚扰等)的开源库.支持本地离线(含归属地.骚扰.常用号码)和网络( ...

  2. 工厂方法模式之C++实现

    说明:本文仅供学习交流,转载请标明出处.欢迎转载. 工厂方法模式与简单工厂模式的差别在于:在简单工厂模式中.全部的产品都是有一个工厂创造,这样使得工厂承担了太大的造产品的压力,工厂内部必须考虑所以的产 ...

  3. MFC——9.多线程与线程同步

    Lesson9:多线程与线程同步 程序.进程和线程是操作系统的重点,在计算机编程中.多线程技术是提高程序性能的重要手段. 本文主要解说操作系统中程序.进程和线程之间的关系,并通过相互排斥对象和事件对象 ...

  4. android之Context对各种服务的管理

    经常,当我们须要用到服务的时候能够通果Context来获取:Context.getSystemService(name):比方:当我们想知道当前电话状态(来电/去电/sim卡状态等)时候,我们能够通过 ...

  5. selenium3 踩坑--move_to_element()报错

    问题:selenium3 使用move_to_element()报错,报错信息如下图所示: 网上没有找到合适的解决办法,回退到稳定的selenium2可以解决. pip install seleniu ...

  6. 【转】php和java之间rsa加密互通

    以下是php封装好的类,引入即可使用 <?php /** * 作者:pjp * 邮箱:vippjp@163.com */ class RSA{ private $privateKey='';// ...

  7. windows 2008配置运行PHP5.5.X

    1.安装web5.0平台安装程序.web5.0平台安装程序:http://www.iis.net/downloads (实际上更方便的是用WebPlalform安装PHP:http://www.mic ...

  8. EasyNVR无插件IPC摄像机直播方案前端构建之:区分页面是自跳转还是分享依据

    区分分享还是跳转 对于前端一些页面的展示,通常有两种方式:通过入口链接一步步进入,或是通过分享链接直接进入:对于这两种方式的区别是什么?在进行前端书写时又应该如何处理? 以EasyNVR为例来进行说明 ...

  9. android菜鸟学习笔记1----环境搭建

    Step1 JDK安装及配置: 1.下载并安装JDK: 根据自己系统情况,选择安装相应的JDK版本 当前系统:64位WIN8,内存8G 选择了Java SE 8u45 即JDK 1.8.0_45,可以 ...

  10. SpringMVC @ResponseBody和@RequestBody使用

    @ResponseBody用法 作用: 该注解用于将Controller的方法返回的对象,根据HTTP Request Header的Accept的内容,通过适当的HttpMessageConvert ...