题解 | #简单错误记录#

#include <iostream>
#include <unordered_set>
#include <deque>
#include <math.h>
using namespace std;

struct Ans{
    string name;
    int line;
    int count;
};

bool Check(Ans Ans1_Temp, deque<Ans>&ans){//检查Ans1_Temp是否已经在ans中出现过。
    deque<Ans>::iterator m1_Iter;
    m1_Iter = ans.begin();
    while(m1_Iter != ans.end()){
        if(m1_Iter->name == Ans1_Temp.name && m1_Iter->line == Ans1_Temp.line){//重复了
            m1_Iter->count++;
            return false;
        }
        m1_Iter++;
    }
    return true;
}

int main() {
    unordered_set<string> trash;
    deque<Ans> ans;
    Ans Ans1_Temp;
    string m1_Str;
    string str1_Temp;
    int num = 0;
    while(getline(cin, m1_Str)){
        Ans1_Temp.count = 0;
        Ans1_Temp.line = 0;
        int i = m1_Str.length() - 1, j;
        while(m1_Str[i] != '\\') i--;
        j = i;
        while(m1_Str[j] != ' ') j++;
        if(j - i -1 <= 16){
            str1_Temp = m1_Str.substr(i + 1, j - i -1);
        }
        else{
            str1_Temp = m1_Str.substr( j - 16, 16);
        }
        Ans1_Temp.name = str1_Temp;
        i = m1_Str.length() - 1;
        while(m1_Str[i] != ' ') i--;
        str1_Temp = m1_Str.substr(i + 1, m1_Str.length()-1);
        j = 0;
        while(j <= str1_Temp.length() - 1){
            Ans1_Temp.line += ((int)str1_Temp[j] - 48) * pow(10, str1_Temp.length() - j - 1);
            j++;
        }
        Ans1_Temp.count++;
        // cout << Ans1_Temp.name << " " << Ans1_Temp.line << " " << Ans1_Temp.count << endl;
        str1_Temp = Ans1_Temp.name + str1_Temp;
        if(trash.count(str1_Temp) == 0){//在垃圾回收站中的不要了。
            if(Check(Ans1_Temp, ans)){//Check是否已经出现过了。
                if(num >= 8){//没出现过，此时若已经记录了8条信息，头pop，尾push
                    str1_Temp = ans.front().name + to_string(ans.front().line);
                    trash.insert(str1_Temp);
                    ans.pop_front();
                    ans.push_back(Ans1_Temp);
                }else{//没出现过，此时记录未满8条，直接尾push
                    ans.push_back(Ans1_Temp);
                    num++;
                }
            }
        }
        // cout << num << endl;
    }
    deque<Ans>::iterator ans_iter;
    ans_iter = ans.begin();
    while(ans_iter != ans.end()){
        cout << ans_iter->name << " " << ans_iter->line  << " " << ans_iter->count << endl;
        ans_iter++;
    }
    return 0;
}

这道题目理解上不难，就是流程上麻烦点，对字符处理要细致