首尾相连的珠子

这道题是从跑黑同学的博客上看到的
runningdark.blog.chinaunix.net

看了之后在gcc上跑了一下，觉得好像少了一个从0开始的最小计数，而且getnextoccur这个子函数没有看懂，于是在他的思路之上修改了一下，去掉了getnextoccur函数，全部用getfirstvalue函数实现得到以下的代码：

/*

 * =====================================================================================

 *

 * Filename: bead.c

 *

 * Description:

 *

 * Version: 1.0

 * Created: 10/20/2012 11:46:48 AM

 * Revision: none

 * Compiler: gcc

 *

 * Author: [e-mail] (),

 * Organization:

 *

 * =====================================================================================

 */

#include <stdlib.h>

#include <stdio.h>



#define MAXK 10

#define MAXM 100 /* */

int getfirstvalue(int i,int *input,int n,int size){

    int match = (1<<n)-1;

    int end = i-1;

    int v = 0;

    do{

        end++;

        v |= 1<<input[end%size];

    }while(v!=match);

    return end;

}

/*
int getnextoccur(int tarindex, int *input, int size){

    int i;

    for(i = tarindex+1;i<size*2;i++){

        if( input[i%size] == input[tarindex])

            return i%size;

    }

    return -1;

}
*/


/*

 * === FUNCTION ======================================================================

 * Name: main

 * Description:

 * =====================================================================================

 */

    int

main ( int argc, char *argv[] )

{

    int input[] = {0,1,2,3,4,4,5,4,2,3,1,0,2,3,3,1,1,0,3,2,1,4,5,3,4,2,4,1,0,5};

    int n = 6;

    int size = sizeof(input)/sizeof(int);

    int end[size];

    int endnum;

    int i = 0;

    int minlen = size;

    for(i = 0; i<size;i++){

        end[i] = getfirstvalue(i,input,n,size);

    endnum=end[i]%size;

        printf ( "start = %d end = %d\n", i,endnum );

        minlen = (end[i]-i+1)>minlen? minlen : end[i]-i+1;

    }

    printf ( "min length = %d\n", minlen );

    return EXIT_SUCCESS;

} /* ---------- end of function main ---------- */

我的思路呢，就是把每一颗珠子遍历一遍，求出每一个的最小值，然后比较minlen

如果小于minlen就修改minlen的值

求最小值的方法借用了跑黑同学左移填满数据位的方法