hdoj 3746 Cyclic Nacklace 【kmp 求至少在结尾添加多少个字符可以凑够至少两次循环】

Cyclic Nacklace

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3515    Accepted Submission(s): 1606

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

因为变量写错了，错了两次 郁闷。。。

设字符串长度为len

最小循环节minn = len - f[len]；f为失配函数。

那么 minn - (f[len] % minn)就是我们需要的答案了。

 

mp：202ms

 

#include <cstdio>#include <cstring>#define MAX 100000+10using namespace std;int f[MAX];char P[MAX];void getfail(){    int i, j;    int len = strlen(P);    f[0] = f[1] = 0;    for(i = 1; i < len; i++)    {        j = f[i];//由前推后         while(j && P[i] != P[j])//继续         j = f[j];        f[i+1] = P[i]==P[j]?j+1:0;     }}int main(){    int t;    int minn;//最小循环节     scanf("%d", &t);    while(t--)    {        scanf("%s", P);        getfail();        int len = strlen(P);        minn = len - f[len];        if(f[len] && len % minn == 0)//因为要循环两次及其以上 所以f[len] == 0不符合         printf("0\n");        else        printf("%d\n", minn-(f[len]%minn));//最小循环节长度 减去已有长度     }    return 0;}

kmp：140ms

 

#include <cstdio>#include <cstring>#define MAX 100000+10using namespace std;int f[MAX];char P[MAX];void getfail(){    int len = strlen(P);    int i = 0;    int j = -1;    f[0] = -1;    while(i < len)    {        if(j == -1 || P[i] == P[j])        f[++i] = ++j;        else        j = f[j];    }}int main(){    int t;    int minn;    scanf("%d", &t);    while(t--)    {        scanf("%s", P);        getfail();        int len = strlen(P);        minn = len - f[len];        if(f[len] && len % minn == 0)        printf("0\n");        else        printf("%d\n", minn-(f[len]%minn));    }    return 0;}