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Uva|Volume 0. Getting Started 414 - Machined Surfaces


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 contiguousX'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 contiguousX's.
Digital-Image View of Surfaces

LeftRight


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 blankregion 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").


4 XXXXBBBBBBBBBBBBBBBBXXXXX XXXBBBBBBBBBBBBBBBXXXXXXX XXXXXBBBBBBBBBBBBBBBBXXXX XXBBBBBBBBBBBBBBBBBXXXXXX 2 XXXXXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXX 1 XXXXXXXXXBBBBBBBBBBBBBBXX 0




Sample Output


4 0 0


题目大意：这个真的很惭愧呀……面对一大堆让人头痛的英文我真的败了……。
关键是这段：
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.

直接引用别人的文字吧：摘自http://blog.sina.com.cn/s/blog_867764f70100svh5.html

本题比较水，没什么技术含量，关键就在于理解题意。一大篇的英文，让人看了就头疼，最悲剧的还有许多单词不认识……
题目的大致意思是：首先输入一个数字N，代表接下来要输入几行字符串；
每行字符串的格式为：由字母X和空格 （“ ”）组成，且两边是字符X，中间是空格。每行的开始有一个或多个X，中间有0个或多个空格，末端有一个或多个X。
要求：将N行中的左边的字符X同时向右移动（或右边的字符X向左移动），直到有一行中间没有空格时停止所有行的移动，此时计算N行中总共有多少个空格。
注意：（1）给出的测试数据XXXXBBBBBBBBBBBBBBBBXXXXX中的字母‘B’是代表空格的意思，因为如果给出空格，你很难看出中间有多少个空格；给出‘B’，一眼就能看出空格的个数。每行的第一个位置没有空格，即左边的X前面没有空格。题目中就给出这句“character "B" for ease of reading. The actual input file will use the ASCII-space character, not "B"”，意思是：字符‘B’是为了阅读的方便，真正的输入文件将会使用标准的ASCII码中的空格字符，不是'B'。
（2）程序中不能使用scanf函数输入字符串，因为scanf遇到空格、Tab、换行就结束读取，而本题中的字符串要求中间有0个或多个字符串。推荐使用fgets函数。
（3）本题要求以N为0结束输出，所以你的程序要有当N=0时退出的语句。
【Uva|Volume 0. Getting Started 414 - Machined Surfaces】解题思想：计算出每个字符串中的空格数，找出一行中含有空格的最小数min，用每行空格数减去min,在求和。
解题方法：用一个一维数组来存储每行的字符，当输入的同时计算出此字符串中含有的空格数，并找出一行中含有空格的最小数。这样的话就不需定义一个二维数组将输入字符串先保存下来在计算了。

MySubmit：
本人比较吝啬，和 494 - Kindergarten Counting Game 一样，我没有开字符数组，这次开了一个 int 数组用于存储每行测试数据里的空格数量。

#include #include typedef struct{//这样定义结构体，就可以直接用 node 定义变量了，否则要用 struct node 去定义。 int v; int i; }node; //如果你不了解，就把这个结构体删了，把下边的 node min; 改为 int min,mini; //min.v 改为 min //min.i 改为 mini int main(){ //freopen("in.txt","r",stdin); //从文件读入测试数据，这样就不用每次测试都输入一遍测试数据了，不会的童鞋，了解一下吧，很简单的。 //但是提交代码的时候要注释掉或者删掉。 int n,i,k,ans,space_count[13]; char ch; node min; while(scanf("%d\n",&n),n){//读入 n 的同时把 n 后面的换行符也读进来。 memset(space_count,0,sizeof(space_count)); //将数组 space_count 清零，也可以用 for 来实现，但没有这个快。 min.v=30; ans=0; for(k=0; k








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