PAT甲级1157 Anniversary

思路

用两个map存储输出的两部分id，mapp1存储校友的id，mapp2存储出席的id。当输入出席id时，用mapp1.find()函数搜索是否为校友，若是则sum++，并将该id记为1；否则记为0。
重写mapp2的compare函数，根据id的第7-14位从小到大排，即年纪越大的越靠前。
输出sum和排序后mapp2中第一个值为1的id。

代码

#include<iostream>
#include<map>
#include<string>
using namespace std;
struct MyCmp{bool operator()(const string& str1,const string& str2)const{string s1 = str1.substr(6,8);string s2 = str2.substr(6,8);return s1 < s2;}
};
int main(){int n,m,i,sum = 0;string str;cin>>n;map<string,int> mapp1;map<string,int,MyCmp> mapp2;for(i = 0;i < n; ++ i){cin>>str;mapp1[str] = 0;}cin>>m;for(i = 0;i < n; ++ i){cin>>str;if(mapp1.find(str) != mapp1.end()){sum ++;mapp2[str] = 1;}else mapp2[str] = 0;}cout<<sum<<endl;map<string,int,MyCmp>::iterator it = mapp2.begin();if(sum > 0){for(;it != mapp2.end(); ++ it){if(it->second == 1)cout<<it->first<<endl;break;}}else cout<<it->first<<endl;return 0;
}

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PAT甲级1157 Anniversary

思路

代码

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