【字符串题目】poj 3096 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
- POJ 3096 Surprising Strings
Surprising Strings Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5081 Accepted: 333 ...
- [ACM] POJ 3096 Surprising Strings (map使用)
Surprising Strings Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5783 Accepted: 379 ...
- 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 ...
- POJ 3096:Surprising Strings
Surprising Strings Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 6258 Accepted: 407 ...
- 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 ...
- 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 */ /************************************** ...
- [POJ3096]Surprising Strings
[POJ3096]Surprising Strings 试题描述 The D-pairs of a string of letters are the ordered pairs of letters ...
- HDOJ 2736 Surprising Strings
Surprising Strings Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Other ...
- HDU 2736 Surprising Strings
Surprising Strings Time Limit:1000MS Memory Limit:65536KB 64 ...
随机推荐
- IO流中文件和文件夹的删除程序举例
/* * 删除功能(无论是文件夹还是文件都是用delete方法,仅仅能一级一级的删除.):public boolean delete() * * 注意: * A:假设你创建文件或者目录忘了写盘符路径, ...
- AngularJS中,<span class="bluetext" ng-bind="ctrl.user.name|uppercase"></span>和{{ctrl.user.name|uppercase}}是等价的,但不等于<span class="bluetext" ng-bind="ctrl.user.name|uppercase"/>
代码下载:https://files.cnblogs.com/files/xiandedanteng/angularjsAttenSpan.rar AngularJS中,<span class= ...
- Java内存泄漏及分析
对于内存泄漏,首先想到的是C语言,其实不然,java中也有各种的内存泄漏.对于java程序员,在虚拟即中,不需要为每一个新建对象去delete/free内存,不容易出现内存泄漏.但是,正 是由于这种机 ...
- struts2实现文件查看、下载
CreateTime--2017年9月7日10:25:33 Author:Marydon struts2实现文件查看.下载 1.界面展示 <a style="color: #199 ...
- 在linux oracle 10g/11g x64bit环境中,goldengate随os启动而自己主动启动的脚本
在linux.oracle 10g/11g x64bit环境中,goldengate随os启动而自己主动启动的脚本 背景描写叙述: goldengate安装于/u01/ggs文件夹下 rhel5.5 ...
- 【Python】分析文本split()
分析单个文本 split()方法,是以空格为分隔符将字符串拆分成多个部分,并将这些部分存储到一个列表中 title = 'My name is oliver!' list = title.split( ...
- Effective C++ 35,36,37
35.使公有继承体现 "是一个" 的含义. 共同拥有继承意味着 "是一个".如 class B:public A. 说明类型B的每个对象都是一个类型A的对象, ...
- hadoop生态系统学习之路(六)hive的简单使用
一.hive的基本概念与原理 Hive是基于Hadoop之上的数据仓库,能够存储.查询和分析存储在 Hadoop 中的大规模数据. Hive 定义了简单的类 SQL 查询语言,称为 HQL.它同意熟悉 ...
- 使用python处理实验数据-yechen_pro_20171231
整体思路 1.观察文档结构: - 工况之一 - 流量一28 - 测点位置=0 -测点纵断面深度-1 -该点数据Speedxxxxxxxx.txt -测点纵断面深度-2 -测点纵断面深度-3 -... ...
- Dubbo(一)Dubbo资料
这个资料绝对权威了:http://dubbo.io/user-guide/