UVa - 414 - Machined Surfaces(AC)

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 Nwill 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

题意：首先输入整数N，表明要输入的digital images行数， 注意 0<N<13, 当输入0时应该退出， 此题难以看懂的地方是如何contact的问题，刚开始我以为是从左右同时向中间直到相遇，后来才发现是从左或者右侧向另一侧extend, 也就是从一侧，向另一侧扩展，当有一行contact时，则停止，计算此时整个图所有的空格数，其实只需要整个图的空格数减去最小的空格数乘N即可

注意：  (1)while循环输入n, 按回车，结果回车也作为一个字符

(2)变量初始化的位置

(3)fgets(buf, sizeof(buf)-1, stdin)遇到回车符则结束输入，然后自动在末尾添加结束符'\0'

#include <stdio.h> #include <string.h>#include <ctype.h>#define column 30char buf[column];int main(){int n, i, j;int min, sum;int count;while(scanf("%d", &n) == 1){if(n == 0)break;sum = 0; //注意sum和min的初始化位置，在while循环里面min = 25;//while读进去按回车后，回车也算一个字符进入循环中，所以i<n+1, 多一次for(i = 0; i<n+1; i++){count = -1;//初值为-1是因为,后面count会把字符串末尾的0算进去，所以这里要-1//fgets()当遇到回车符时输入结束，并自动在末尾添加'\0'结束符fgets(buf, sizeof(buf), stdin);int len = strlen(buf);for(j = 0; j < len; j++){if(!isalpha(buf[j])) //判断是否为字母count++;}sum = sum+count;if(count < min && i>0)//与上i>0，是因为i为0时，回车符给算进来了min = count;}printf("%d\n",sum-n*min);}return 0;}