hdu2087剪花布条+hdu3746Cyclic Nacklace 【kmp复习】

剪花布条

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22868    Accepted Submission(s): 14289

Problem Description

一块花布条，里面有些图案，另有一块直接可用的小饰条，里面也有一些图案。对于给定的花布条和小饰条，计算一下能从花布条中尽可能剪出几块小饰条来呢？

 

Input

输入中含有一些数据，分别是成对出现的花布条和小饰条，其布条都是用可见ASCII字符表示的，可见的ASCII字符有多少个，布条的花纹也有多少种花样。花纹条和小饰条不会超过1000个字符长。如果遇见#字符，则不再进行工作。

 

Output

输出能从花纹布中剪出的最多小饰条个数，如果一块都没有，那就老老实实输出0，每个结果之间应换行。

 

Sample Input

abcde a3aaaaaa  aa#

 

Sample Output

03

 

Author

qianneng

 

Source

冬练三九之二  

#include<stdio.h>#include<string.h>#define N 1010int next[N],count ;void GetNext(char *s){    int  k,j;    next[0] = k = -1;    j = 0;    while(s[j]!='\0')    {        if( k == -1||s[j]==s[k])            next[++j] = ++k;        else            k = next[k];    }    return ;}void GetKmp(char *s1,char *s2){    int i,j,l;    i = j = 0;    l = strlen(s1);//    printf("l=%d\n",l);    while(i <= l)    {    //    printf("i=%d j=%d\n",i,j);        if(s2[j] == '\0')        {            j = 0;            ++count;        }        else         {            while(j!=-1&&s1[i]!=s2[j])                j = next[j];            ++i;            ++j;        }    }    return;}int main(){    char s1[N],s2[N];    while(scanf("%s%s",s1,s2),s1[0]!='#')    {        count = 0;        GetNext(s2);        GetKmp(s1,s2);        printf("%d\n",count);    }    return 0;}

 

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10844    Accepted Submission(s): 4637

Problem Description

CC always becomes very depressed at the end of this month, he has checked his credit card yesterday, without any surprise, there are only 99.9 yuan left. he is too distressed and thinking about how to tide over the last days. Being inspired by the entrepreneurial spirit of "HDU CakeMan", he wants to sell some little things to make money. Of course, this is not an easy task.

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.

 

Input

The first line of the input is a single integer T ( 0 < 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 ).

 

Output

For each case, you are required to output the minimum count of pearls added to make a CharmBracelet.

 

Sample Input

3aaaabcaabcde

 

Sample Output

025

 

Author

possessor WC

 

#include<stdio.h>#include<string.h>#define N 100010int next[N],l;char s[N];void GetNext(char *s){    int i,j,k,l,min;    j = i = 0;    next[0] = k = -1;    l = strlen(s);    while(j<= l)    {        if(k == -1||s[k]==s[j])            next[++j] = ++k;        else            k = next[k];    }    min = l-next[l];    if(l%min==0&&l!=min)        printf("0\n");    else    {        k = l%min;        printf("%d\n",min-k);    }    return;}int main(){    int t;    scanf("%d",&t);    getchar();    while(t--)    {        scanf("%s",s);        GetNext(s);    }    return 0;}