pair解法

#include <bits/stdc++.h>
using namespace std;
int main(){
	pair<string,int> dinary[10001];
	int n,m,ans;
	string look;
	scanf("%d",&n);
	for(int i = 1;i<=n;i++){
		cin>>dinary[i].first;
		scanf("%d",&dinary[i].second);
	}
	scanf("%d",&m);
	for(int i = 1;i<=m;i++){
		cin>>look;
		for(int j = 1;j<=n;j++)
			if(look == dinary[j].first){
				ans = dinary[j].second;
				break;
			}
		printf("%d\n",ans);
	}
	return 0;
}

map解法

#include <bits/stdc++.h>
using namespace std;
int main(){
	map<string,int> dinary;
	int n,m,ans,p;
	string look;
	scanf("%d",&n);
	for(int i = 1;i<=n;i++){
		cin>>look;
		scanf("%d",&p);
		dinary[look] = p;
	}
	scanf("%d",&m);
	for(int i = 1;i<=m;i++){
		cin>>look;
		printf("%d\n",dinary[look]);
	}
	return 0;
}