uva 414

Machined Surfaces 

An imaging device furnishes digital images of two machined surfaces that eventually will be assembled in contact with each other. The roughness of this final contact is to be estimated.

A digital image is composed of the two characters, "X" and " " (space). There are always 25 columns to an image, but the number of rows, N, is variable. Column one (1) will always have an "X" in it and will be part of the left surface. The left surface can extend to the right from column one (1) as contiguous X's.

Similarly, column 25 will always have an "X" in it and will be part of the right surface. The right surface can extend to the left from column 25 as contiguous X's.

Digital-Image View of Surfaces

                  Left                                              Right                 XXXX                                              XXXXX                     XXX                                             XXXXXXX                 XXXXX                                              XXXX                 XX                                               XXXXXX                 .                                                     .                 .                                                     .                 .                                                     .                 XXXX                                               XXXX                 XXX                                              XXXXXX                                                                                            1                                                  25

In each row of the image, there can be zero or more space characters separating the left surface from the right surface. There will never be more than a single blank region in any row.

For each image given, you are to determine the total ``void" that will exist after the left surface has been brought into contact with the right surface. The ``void" is the total count of the spaces that remains between the left and right surfaces after theyhave been brought into contact.

The two surfaces are brought into contact by displacing them strictly horizontally towards each other until a rightmost "X" of the left surface of some row is immediately to the left of the leftmost "X" of the right surface of that row. There is no rotation or twisting of these two surfaces as they are brought into contact; they remain rigid, and only move horizontally.

Note: The original image may show the two surfaces already in contact, in which case no displacement enters into the contact roughness estimation.

Input

The input consists of a series of digital images. Each image data set has the following format:

First line -

A single unsigned integer, N, with value greater than zero (0) and less than 13. The first digit of N will be the first character on a line.

Next N lines -

Each line has exactly 25 characters; one or more X's, then zero or more spaces, then one or more X's.

The end of data is signaled by a null data set having a zero on the first line of an image data set and no further data.

Output

For each image you receive as a data set, you are to reply with the total void (count of spaces remaining after the surfaces are brought into contact). Use the default output for a single integer on a line.

Sample Input (character "B" for ease of reading. The actual input file will use the ASCII-space character, not "B").

4XXXXBBBBBBBBBBBBBBBBXXXXXXXXBBBBBBBBBBBBBBBXXXXXXXXXXXXBBBBBBBBBBBBBBBBXXXXXXBBBBBBBBBBBBBBBBBXXXXXX2XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX1XXXXXXXXXBBBBBBBBBBBBBBXX0

Sample Output

400

#include <stdio.h>#include <string.h>int main(){     int i, j, n;     int max;     while(scanf("%d", &n) == 1)     {          max = 0;          int sum = 0, b[25] = {0};          char a[15][30];          getchar();            if(n == 0) break;          for(i = 0; i < n; i++)          {               //     fgets(a[i], 30, stdin);               gets(a[i]);               for(j = 0; j < strlen(a[i]); j++)               {                    if(a[i][j] == 'X')                    {                         b[i]++;                    }               }               if(max < b[i])               {                    max = b[i];               }          }          for(i = 0; i < n; i++)          {               sum += max - b[i];          }          printf("%d\n", sum);     }     return 0;}

简单题    看题看了半天  英语啊  硬伤啊   思路是把每行的X的个数用数组保存起来  然后找到最大的那个  分别减去比他小的  得到的结果相加就是解答