POJ 2725
#include <iostream>
#include <string>
#include <algorithm>
#define MAXN 400005
using namespace std; char _m[MAXN];
int next[MAXN];
void give_next(int len);
int ans_len[MAXN];
int ans_len_index;
int ans[MAXN];
bool take_judge(int b,int len);
int main()
{
//freopen("acm.acm","r",stdin);
string s;
int i;
int j;
int len;
int index;
int cur;
//int ans;
while(getline(cin,s))
{
if(s.length() == )
{
break;
}
index = ;
//cout<<s<<endl;
//ans = 0;
//cout<<s<<endl;
for(i = ; i < s.length(); ++ i)
{
_m[i] = s[i];
}
memset(next,-,sizeof(next));
give_next(s.length()); i = s.length();
//cout<<i<<" ";
ans[index ++] = i;
for(; next[i] != ; i = next[i])
{
//cout<<next[i]<<" ";
ans[index++] = next[i];
}
for(i = index-; i >= ; -- i)
{
cout<<ans[i]<<" ";
}
cout<<ans[i]<<endl;
//cout<<endl;
// cout<<endl;
}
return ;
} bool take_judge(int b,int len)
{
int j = ;
int i;
for(i = b; i < len; ++ i,++ j)
{
if(_m[i] != _m[j])
{
break;
}
}
if(i == len)
{
return true;
}
return false;
} void give_next(int len) //此为求next 数组的方法。
{
int i;
int j;
i = ;
j = -;
next[] = -;
while(i < len)
{
if(j == - || _m[i] == _m[j])
{
i ++;
j ++;
next[i] = j;
}
else
j = next[j];
}
}
POJ 2725的更多相关文章
- poj 1279 半平面交核面积
Art Gallery Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 6668 Accepted: 2725 Descr ...
- POJ 3370. Halloween treats 抽屉原理 / 鸽巢原理
Halloween treats Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 7644 Accepted: 2798 ...
- POJ 2356. Find a multiple 抽屉原理 / 鸽巢原理
Find a multiple Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 7192 Accepted: 3138 ...
- POJ 2965. The Pilots Brothers' refrigerator 枚举or爆搜or分治
The Pilots Brothers' refrigerator Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 22286 ...
- POJ 1753. Flip Game 枚举or爆搜+位压缩,或者高斯消元法
Flip Game Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 37427 Accepted: 16288 Descr ...
- POJ 3254. Corn Fields 状态压缩DP (入门级)
Corn Fields Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 9806 Accepted: 5185 Descr ...
- POJ 2739. Sum of Consecutive Prime Numbers
Sum of Consecutive Prime Numbers Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 20050 ...
- POJ 2255. Tree Recovery
Tree Recovery Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 11939 Accepted: 7493 De ...
- POJ 2752 Seek the Name, Seek the Fame [kmp]
Seek the Name, Seek the Fame Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 17898 Ac ...
随机推荐
- hdu6351 2018 Multi-University Training Contest 5 1002 Beautiful Now
题意: 给出一个十进制数,数位两两可以交换,给出最多能交换多少次,以及交换后的数不能有前缀0,问能形成最小和最大的数 * * * 尝试的思路 贪心,将字符串先排出最大以及最小的情况,然后按一定顺序将对 ...
- C++STL priority_queue
priority_queue优先级队列 最大值优先级队列(队头是最大值) 最小值优先级队列(队头是最小值) priority_queue<int> q1;//默认定义为最大值优先级队列 ...
- DDR中寄存器的问题
图中虚线是自动跳转,实线是通过发送命令才能跳转的. 下面是框中对应的命令. ACT = ACTIVATE MPR = Multipurpose register MRS = Mode register ...
- MySQL 安装与使用(一)
操作系统:CentOS release 5.10 (Final) 文件准备: MySQL-server-community-5.1.73-1.rhel5.i386.rpm MySQL-client-c ...
- 1045 Favorite Color Stripe 动态规划
1045 Favorite Color Stripe 1045. Favorite Color Stripe (30)Eva is trying to make her own color strip ...
- Forward团队-爬虫豆瓣top250项目-团队编程项目开发环境搭建过程
本次结对编程和团队项目我都需要用python环境,所以环境的搭建是一样的.(本文部分内容引用自己博客:http://www.cnblogs.com/xingyunqi/p/7527411.html) ...
- (day20)javaEE三大组件之一Servlet (简介(二)servletconfig,servletContext,session,cookie,request,response,out)
javaEE是服务器编程,javaEE提供了服务器的接口让具体的服务器去创建实现的对象 JavaEE是sun公司为了解决企业级开发定义的一套技术,只提供了规范,具体的实现是由服务器完成的 servle ...
- jvm调优的分类
本文部分内容出自https://blog.csdn.net/yang_net/article/details/5830820 调优步骤: 衡量系统现状. 设定调优目标. 寻找性能瓶颈. 性能调优. 衡 ...
- jQuery插件初级练习2
<!DOCTYPE html><html> <head> <meta charset="UTF-8"> <title>& ...
- bzoj网络流
近期看了一些bzoj的网络流,深感智商不够.不过对于网络流又有了进一步的理解. 还是mark一下吧. 献上几篇论文:1)<最小割模型在信息学竞赛中的应用> 2)<浅析一类最小割问题& ...