Glass Beadshttp://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1006


Time Limit: 2 Seconds      Memory Limit: 65536 KB

Once upon a time there was a famous actress. As you may expect, she played mostly Antique Comedies most of all. All the people loved her. But she was not interested in the crowds. Her big hobby were beads of any kind. Many bead makers were working for her and they manufactured new necklaces and bracelets every day. One day she called her main Inspector of Bead Makers (IBM) and told him she wanted a very long and special necklace.

The necklace should be made of glass beads of different sizes connected to each other but without any thread running through the beads, so that means the beads can be disconnected at any point. The actress chose the succession of beads she wants to have and the IBM promised to make the necklace. But then he realized a problem. The joint between two neighbouring beads is not very robust so it is possible that the necklace will get torn by its own weight. The situation becomes even worse when the necklace is disjoined. Moreover, the point of disconnection is very important. If there are small beads at the beginning, the possibility of tearing is much higher than if there were large beads. IBM wants to test the robustness of a necklace so he needs a program that will be able to determine the worst possible point of disjoining the beads.

The description of the necklace is a string A = a1a2 ... am specifying sizes of the particular beads, where the last character am is considered to precede character a1 in circular fashion.

The disjoint point i is said to be worse than the disjoint point j if and only if the string aiai+1 ... ana1 ... ai-1 is lexicografically smaller than the string ajaj+1 ... ana1 ... aj-1. String a1a2 ... an is lexicografically smaller than the string b1b2 ... bn if and only if there exists an integer i, i <= n, so that aj=bj, for each j, 1 <= j < i and ai < bi.

Input

The input consists of N cases. The first line of the
input contains only positive integer N. Then follow the cases. Each case
consists of exactly one line containing necklace description. Maximal length of
each description is 10000 characters. Each bead is represented by a lower-case
character of the english alphabet (a--z), where a < b ... z.

Output

For each case, print exactly one line containing
only one integer -- number of the bead which is the first at the worst possible
disjoining, i.e. such i, that the string A[i] is lexicographically smallest
among all the n possible disjoinings of a necklace. If there are more than one
solution, print the one with the lowest i.

Sample
Input

4
helloworld
amandamanda
dontcallmebfu
aaabaaa

Sample Output

10
11
6
5

题意:输出字典序最小的同构字符串首位置

最小表示法

#include<cstdio>
#include<cstring>
#include<algorithm>
#define N 10001
using namespace std;
char s[N];
int len;
int main()
{
int n,i,j,k;
scanf("%d",&n);
while(n--)
{
scanf("%s",s);
len=strlen(s);
i=; j=;
while(i<len&&j<len)
{
k=;
while(k<len && s[(i+k)%len]==s[(j+k)%len]) k++;
if(k==len) break;
if(s[(i+k)%len]<s[(j+k)%len]) j=max(j+k+,i+);
else i=max(i+k+,j+);
}
printf("%d\n",min(i,j)+);
}
}

最新文章

  1. 每天5分钟 玩转OpenStack 目录列表
  2. grunt任务之seajs模块打包
  3. hadoop fs 命令
  4. PE355
  5. 【GoLang】GoLang 中 make 与 new的区别
  6. Win 8 App开发框架解析
  7. 关于android 将对象写入文件以及从文件读取对象
  8. Caused by: java.lang.ClassNotFoundException: org.apache.commons.io.FileUtils
  9. yii 分页 (ajax)
  10. onActivityResult不执行 或者 onActivityResult的解决方法
  11. SettingsProvider 它SettingsCache
  12. linux Packet socket (1)简单介绍
  13. Microsoft Push Notification Service(MPNS)的最佳体验
  14. mysql之inner join 和left join/right join
  15. 性能秒杀log4net的NLogger日志组件(附测试代码与NLogger源码)
  16. 深入浅出Lua虚拟机
  17. 为什么在Python里推荐使用多进程而不是多线程?
  18. 翻译 Asp.Net Core 2.2.0-preview1已经发布
  19. springboot源码解读01
  20. HDUOJ--点球大战

热门文章

  1. 一:HDFS 用户指导
  2. POJ 3304 Segments(线的相交判断)
  3. 第二次作业 编程题 PAT 1001A+B Format
  4. iOS-修改导航栏文字字体和颜色
  5. YaoLingJump开发者日志(一)
  6. PAT 甲级 1128 N Queens Puzzle
  7. Jmeter系列-webdriver代码范例
  8. 重要的几个按键Tab Ctrl+c Ctrl+d
  9. ajax 请求 后台返回的文件流
  10. BZOJ 1197 花仙子的魔法(递推)