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

,否则输出surprising。



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.

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
char s[500];
char t[500];
int main()
{
int i,j;
int n;
while(scanf("%s",s)!=EOF)
{
if(s[0]=='*')
break;
n = strlen(s);
bool flag = true;
for(i=0;i<n;i++)
{
if(!flag)
break;
map<string,int>mapp;
for(j=0;j+i+1<n;j++)
{
if(!flag)
break;
t[0] = s[j];
t[1] = s[j+i+1];
t[2] = '\0';
if(mapp[t]>0)
{
printf("%s is NOT surprising.\n",s);
flag = false;
}
else
mapp[t]++;
}
}
if(flag)
cout<<s<<" is surprising."<<endl;
}
return 0;
}

C - Surprising Strings的更多相关文章

  1. [POJ3096]Surprising Strings

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

  2. HDOJ 2736 Surprising Strings

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

  3. POJ 3096 Surprising Strings

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

  4. HDU 2736 Surprising Strings

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

  5. 【字符串题目】poj 3096 Surprising Strings

    Surprising Strings Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 6193   Accepted: 403 ...

  6. Surprising Strings

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

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

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

  8. POJ 3096:Surprising Strings

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

  9. hdu 2736 Surprising Strings(类似哈希,字符串处理)

    重点在判重的方法,嘻嘻 题目 #define _CRT_SECURE_NO_WARNINGS #include<stdio.h> #include<string.h> int ...

随机推荐

  1. T-SQL语言基础

    1.T-SQL语言 CREATE:创建新对象,包括数据库.表.视图.过程.触发器和函数等常见数据库对象. ALTER:修改已有对象的结构. DROP:用来删除已有的对象.有些对象是无法删除的,因为它们 ...

  2. apple iphone 3gs 有锁机 刷机 越狱 解锁 全教程(报错3194,3014,1600,短信发不出去等问题可参考)

    以自身经历列步骤如下:(基本思路就是刷6.1.6,越狱,降级基带,解锁) 一.准备工作 1.下载3gs 6.1.6官方固件.地址:http://act.feng.com/wetools/index.p ...

  3. Linux ./configure && make && make install 编译安装和卸载

    正常的编译安装/卸载: 源码的安装一般由3个步骤组成:配置(configure).编译(make).安装(make install).   configure文件是一个可执行的脚本文件,它有很多选项, ...

  4. c++ string用法

    首先,为了在我们的程序中使用string类型,我们必须包含头文件 .如下: #include  //注意这里不是string.h string.h是C字符串头文件 1.声明一个C++字符串 声明一个字 ...

  5. cetos 6.3 安装 apache+mysql+php

      1.安装 apache 服务器 yum install httpd 启动服务 service httpd start or /etc/init.d/httpd start 2.安装 mysql 数 ...

  6. spring 构造注入 异常 Ambiguous constructor argument types - did you specify the correct bean references as constructor arguments

    你可能在做项目的时候,需要在项目启动时初始化一个自定义的类,这个类中包含着一个有参的构造方法,这个构造方法中需要传入一些参数. spring提供的这个功能叫“构造注入”, applicationCon ...

  7. dictionary(字典)

    dictionary(字典):   字典对象   字典是一种key - value 的数据类型,使用就像我们上学用的字典,通过笔划.字母来查对应页的详细内容. 1.      dic={"n ...

  8. sort 命令

    sort sort -t': ' -k 2n -t 可以自定义分隔符 -k 可以自定义分割后取第几个字符串作为排序值 2n表示第二个值,并作为数字来排序

  9. [BZOJ 1046] [HAOI2007] 上升序列 【DP】

    题目链接:BZOJ - 1046 题目分析 先倒着做最长下降子序列,求出 f[i],即以 i 为起点向后的最长上升子序列长度. 注意题目要求的是 xi 的字典序最小,不是数值! 如果输入的 l 大于最 ...

  10. hdu 5072 Coprime

    http://acm.hdu.edu.cn/showproblem.php?pid=5072 题意:给出 n 个互不相同的数,求满足以下条件的三元无序组的个数:要么两两互质要么两两不互质. 思路:根据 ...