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 = 1000, sum,tmp;
char ch;

while(scanf("%d",&n) != EOF){
printf("n:%d\n",n);
if(n == 0) break;
getchar();
sum = 0; min = 1000;
for(i = 0; i < n;i++){
tmp = 0;
while(1){
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 0;
}