Topic background
D is a small man who will participate in the IO ION players, however, the written exam in the myriad of strange title made him a big headache, come help him!
Title Description
Written exam can be abstracted as \ (n \) item questionnaire that face each question of the questions and answers composition, is a string, to ensure that all surfaces differ topic title.
To test the effect of small D written back, the coach made a mock exam, exam contains \ (q \) item questionnaire that each question has a \ (4 \) option, a small D needs from the \ (4 \) options selected the option that coincides answer.
Now you need to help small D to complete the exam.
Input Format
The first line of two positive integers \ (n-, Q \) .
Next \ (n-\) rows, each row \ (2 \) , separated by a space string representing object of this question and answers questions surface.
Next \ (Q \) rows, each row \ (5 \) , separated by a space character string, the first string showing simulated test questions in this question object surface, the remaining \ (4 \) string by This question object sequence are options options A to D, to ensure that different options.
Output Format
For exams in each question, the output of a character representation corresponding options This question is the answer to ensure that all the problems have solution.
Sample
Entry
3 4
decoak yes
duliuchutiren nonono
csps noiptg
decoak yes no qwq qaq
csps noiptg noippj noi cspj
decoak qwq qaq yesyes yes
duliuchutiren yes no nono nonono
Export
A
A
D
D
text
I am committed to write konjac (like me) can understand the problem solution.
analysis
Simulation, emergent, everyone can do.
method
Pure analog, while being read to lose Dafa is good.
Code
#include<bits/stdc++.h>
int n,q;
struct an{
std::string at;//题面
std::string aa;//答案
}tm[105];//背的题面和答案。
struct zz{
std::string t;//题面
std::string a[5];//选项
}da[105];
char anssc[5]={'1','A','B','C','D'};//直接输出,免得乱搞。
int main()
{
//freopen("in.txt","r",stdin);//测试专用
//freopen("out.txt","w",stdout);
std::cin>>n>>q;//读入
for(int i=1;i<=n;i++)
{
std::cin>>tm[i].at>>tm[i].aa;
}
for(int i=1;i<=q;i++)
{
std::cin>>da[i].t;
int cache=0;//临时存储数据
for(int j=1;j<=n;j++)
{
if(da[i].t==tm[j].at)//找相同的题面
cache=j;
}
for(int j=1;j<=4;j++)
{
std::cin>>da[i].a[j];
if(da[i].a[j]==tm[cache].aa)//找答案
{
std::cout<<anssc[j]<<"\n";//记得回车
}
}
}
return 0;
}