题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3746
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
As Christmas is around the corner, Boys are busy in choosing
christmas presents to send to their girlfriends. It is believed that chain
bracelet is a good choice. However, Things are not always so simple, as is known
to everyone, girl's fond of the colorful decoration to make bracelet appears
vivid and lively, meanwhile they want to display their mature side as college
students. after CC understands the girls demands, he intends to sell the chain
bracelet called CharmBracelet. The CharmBracelet is made up with colorful pearls
to show girls' lively, and the most important thing is that it must be connected
by a cyclic chain which means the color of pearls are cyclic connected from the
left to right. And the cyclic count must be more than one. If you connect the
leftmost pearl and the rightmost pearl of such chain, you can make a
CharmBracelet. Just like the pictrue below, this CharmBracelet's cycle is 9 and
its cyclic count is 2:
Now CC has brought in
some ordinary bracelet chains, he wants to buy minimum number of pearls to make
CharmBracelets so that he can save more money. but when remaking the bracelet,
he can only add color pearls to the left end and right end of the chain, that is
to say, adding to the middle is forbidden.
CC is satisfied with his ideas and
ask you for help.
< T <= 100 ) which means the number of test cases.
Each test case
contains only one line describe the original ordinary chain to be remade. Each
character in the string stands for one pearl and there are 26 kinds of pearls
being described by 'a' ~'z' characters. The length of the string Len: ( 3 <=
Len <= 100000 ).
count of pearls added to make a CharmBracelet.
#include<cstdio>
#include<cstring>
#define MAX 100000+5
using namespace std;
int Next[MAX];
char pat[MAX];
int len,cycle;
void getnext()
{
int i=, j=-;
len=strlen(pat);
Next[]=-;
while(i<len)
{
if(j == - || pat[i] == pat[j]) Next[++i]=++j;
else j=Next[j];
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%s",pat);
getnext();
cycle=len-Next[len];
if(len%cycle==)
{
if(len/cycle==) printf("%d\n",cycle);
else printf("0\n");
}
else printf("%d\n",cycle-(len-len/cycle*cycle));
}
}
题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1358
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
题目大意：
给出一个长度为N的字符串，对其每个前缀串（满足长度≥2，且可以为字符串本身），考察是否存在循环节循环（即存在循环节，且循环节个数≥2）；
若存在，则输出该前缀串的长度，并且输出这个前缀串中循环节循环次数；
例如字符串：“aaa”：
前缀串①：“aa”，长度为2，循环节个数为2；
前缀串②：“aaa”，长度为3，循环节个数为3；
例如字符串：“aabaabaabaab”：
前缀串①：“aa”，长度为2，循环节个数为2；
前缀串②：“aabaab”，长度为6，循环节个数为2；
前缀串③：“aabaabaab”，长度为9，循环节个数为3；
前缀串④：“aabaabaabaab”，长度为12，循环节个数为4；
解题思路：
有了上面那题len-Next[len]的铺垫，本题就比较简单了，进行暴力枚举，判断之后输出即可。
AC代码：
#include<cstdio>
#include<cstring>
#define MAX 1000000+5
int Next[MAX],len;
char pat[MAX];
void getNext()
{
int i=, j=-;
Next[]=-;
while(i<len)
{
if(j == - || pat[i] == pat[j]) Next[++i]=++j;
else j=Next[j];
}
}
int main()
{
int kase=;
while(scanf("%d",&len) && len!=)
{
scanf("%s",pat);
getNext();
printf("Test case #%d\n",++kase);
for(int i=,cycle;i<=len;i++)
{
cycle=i-Next[i];
if(i%cycle == && i/cycle >= ) printf("%d %d\n",i,i/cycle);
}
printf("\n");
}
}