UVA 414 - Machined Surfaces
2012-04-09 22:21
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再一次感叹英语不好的孩子伤不起。看两几次,终于看懂了题目问什么。
大意就是:左边的图案碰到右边的后,中间还剩下多少个空格。值得一提的是,输入例子里用了B代替空格,而实际的输入用例是用空格的。记住这个特点。
思路:统计所有空格数r,然后记录最小空格那一行的空格数min,结果为r-min*n
code:
#include<iostream>
#include<stdio.h>
#include<string>
#include<cstring>
using namespace std;
int main()
{
string s;
int a[110],i,j,result,n,min;
while(cin>>n&&n!=0)
{
getchar();
memset(a,0,sizeof(a));
min=26;
result=0;
for(i=0;i<n;i++)
{
getline(cin,s);
for(j=0;j<25;j++)if(s[j]==' ')a[i]++;
if(min>a[i])min=a[i];
result+=a[i];
}
result-=min*n;
cout<<result<<endl;
}
return 0;
}
大意就是:左边的图案碰到右边的后,中间还剩下多少个空格。值得一提的是,输入例子里用了B代替空格,而实际的输入用例是用空格的。记住这个特点。
思路:统计所有空格数r,然后记录最小空格那一行的空格数min,结果为r-min*n
code:
#include<iostream>
#include<stdio.h>
#include<string>
#include<cstring>
using namespace std;
int main()
{
string s;
int a[110],i,j,result,n,min;
while(cin>>n&&n!=0)
{
getchar();
memset(a,0,sizeof(a));
min=26;
result=0;
for(i=0;i<n;i++)
{
getline(cin,s);
for(j=0;j<25;j++)if(s[j]==' ')a[i]++;
if(min>a[i])min=a[i];
result+=a[i];
}
result-=min*n;
cout<<result<<endl;
}
return 0;
}
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