[Luogu]P5587 打字练习

题目链接：

https://www.luogu.com.cn/problem/P5587

解题思路：

字符串处理，考虑多种可能情况:

1. 范文和输入均可能存在退格，没见过范文还带退格的艹

2. 计算KPM的时候分母可能为0

3. 输入的行数可能比范文的短或长。

输入输出样例

Input1:

hello world.
aaabbbb
x
EOF
heelo world.
aaacbbbb
y<x
EOF
60

Output1:

18

---

Input2:

<<qwe
EOF
<<qwe
EOF
60

Output2:

3

---

Input3:

<qwe
EOF
qw<e
EOF
60

Output3:

1

代码

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<string>
#include<vector>
#include<queue>
#include<cmath>
#include<set>
#define INF 0x3f3f3f3f typedef long long ll;
using namespace std;int count(string a,string b){int ans=0;int length;vector<char> c;vector<char> d;length=a.length();for(int i=0;i<length;i++){if(a[i]=='<' && !c.empty())  c.pop_back();else{if(a[i]!='<')c.push_back(a[i]);}}length=b.length();for(int i=0;i<length;i++){if(b[i]=='<' && !d.empty())  d.pop_back();else{if(b[i]!='<')d.push_back(b[i]);}}length = min(c.size(),d.size());for(int i=0;i<length;++i)ans+= (c[i]==d[i]);return ans;
}
int main() {int n=0;int i=0;int time = 0;double ans=0.0;string str;string a[100005];while(1){getline(cin,str);if(str!="EOF")  a[n++]=str;else      break;}while(1){getline(cin,str);if(str!="EOF"){if(i<n)  ans+=count(a[i++],str);else     continue;}elsebreak;}cin >> time;cout << (int)((ans*60/time)+0.5);
}