USACO_Section 1.1_beads

USACO_Section 1.1_beads

#include  < stdio.h >
#include  < stdlib.h >
int  main()
{
    FILE  * fin, * fout;
    fin = fopen( " beads.in " , " r " );
    fout = fopen( " beads.out " , " w " );
     char   * beads;
     int  n;
    fscanf(fin, " %d " , & n);
    beads = ( char   * )malloc( 3 * n * sizeof ( char ));
    fscanf(fin, " %s " ,beads);
     int  i,a,b,left,right,sum = 0 ;
     for (i = n;i < 3 * n; ++ i)
    {
        beads[i] = beads[i - n];
    }
     for (i = n;i < 2 * n; ++ i)
    {
        left = i;
        right = i + 1 ;
         char  ch;

         while (beads[left] == ' w ' && left >= 0 ) -- left;
        ch = beads[left];
         while (left > 0 && (beads[left - 1 ] == ch || beads[left - 1 ] == ' w ' )) -- left;
        a = i - left + 1 ;

         while (beads[right] == ' w ' && right < 3 * n) ++ right;
        ch = beads[right];
         while (right < ( 3 * n - 1 ) && (beads[right + 1 ] == ch || beads[right + 1 ] == ' w ' )) ++ right;
        b = right - i;

         if (a + b > sum)sum = a + b;
         if (a >= n || b >= n || a + b > n)sum = n;
    }
    fprintf(fout, " %d\n " ,sum);
     return   0 ;
}

首先我的想法是从1到n,left=0,right=1,然后往两边数颜色相同的珠子。如果用一个大小为n的数组存字符串，一个很显然的问题就是当left<0或者right>n-1时就要溢出。所以要用到一个取余的函数。

但是这样确实太麻烦了，写的代码也容易出错，我终于决定重写了。新的想法是在字符串两边各复制一份相同的，这样就是大小为3×n的字符串，而循环时只需要从n到2×n-1,解决了溢出的问题。（但是我觉得这并不是一个好方法，因为浪费了三倍的空间）。最终的代码是这样的，虽然AC了，但总不是那么完美。