414 - Machined Surfaces

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

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题目很难看懂，大意就是计算每次总得空格数，以及获得每行最小的空格数，输出sum - min*linNum。
AC代码：
注意第11行getchar()。后面scanf("%c”，&ch)的话会先获得换行符。

 #include<stdio.h>
 int main(){
     int n;
     int i;
     int min = , sum,tmp;
     char ch;

     while(scanf("%d",&n) != EOF){
         printf("n:%d\n",n);
         if(n == ) break;
         getchar();
         sum = ; min = ;
         for(i = ; i < n;i++){
             tmp = ;
             while(){
                 scanf("%c",&ch);
17
                 if(ch == ' '){
                     tmp++;
                     sum++;
                 }
                 else if(ch == '\n')
                     break;
             }
             min = min > tmp ? tmp:min;
         }
         printf("%d\n",(sum - min*n));

     }
     return ;
 }