PAT甲级1006题（打卡）

1.题目介绍

2.考察点，难度

字符串处理类，string转int，map容器存储，难度易

3.解题代码

string本身存在比较函数，可以直接用string来做比较而不需要转换成int。

#include <bits/stdc++.h>
using namespace std;

int main(){
    freopen("D:\\PAT\\Clion\\in.txt","r",stdin);
    int N;
    string in="24:00:00", out="00:00:00",iid="",oid="";
    unordered_map<string,string> I,O;
    cin>>N;
    for(int i=0;i<N;i++){
        string id,a,b;
        cin>>id>>a>>b;
        I.insert({id,a});
        O.insert({id,b});
    }
    for(auto s:I){
        int first = stod(s.second.substr(0,2))-stod(in.substr(0,2));
        int second = stod(s.second.substr(3,2))-stod(in.substr(3,2));
        int third = stod(s.second.substr(6,2))-stod(in.substr(6,2));
        if(first>0 || (first==0 && second>0) || (first==0 && second==0 && third>0)){
            continue;
        }
        else{
            in=s.second;
            iid=s.first;
        }
    }
    for(auto s:O){
        int first = stod(s.second.substr(0,2))-stod(out.substr(0,2));
        int second = stod(s.second.substr(3,2))-stod(out.substr(3,2));
        int third = stod(s.second.substr(6,2))-stod(out.substr(6,2));
        if(first<0 || (first==0 && second<0) || (first==0 && second==0 && third<0)){
            continue;
        }
        else{
            out=s.second;
            oid=s.first;
        }
    }
    cout<<iid<<" "<<oid<<endl;
    return 0;
}

4.原题地址

https://pintia.cn/problem-sets/994805342720868352/problems/994805516654460928