题解 | #查找学生信息#
查找学生信息
https://www.nowcoder.com/practice/fe8bff0750c8448081759f3ee0d86bb4
可以直接看成两个字符串,然后利用map的at()返回值与count()看是否存在。
#include <iostream>
#include <string>
#include <map>
using namespace std;
map<string,string> mymap;
int main() {
int n;
while (cin >> n) {
string num,information;
for(int i = 0; i< n;i++){
cin >> num;
getline(cin,information);
mymap[num] = information;
}
cin >> n;
for(int i = 0;i < n;i++){
cin >> num;
if(mymap.count(num) == 0) cout << "No Answer!" << endl;
else cout << num << mymap.at(num) << ' ' << endl;
}
}
return 0;
}
