Blue Bridge Cup 2013- provincial tournament -C / C ++ - A group of four questions

Topic
Title: upside price tag

     Mike's TV monopoly sample store other store shelves can be called: sample TV stores.

    Its price is four digits (ie thousand dollars).

    Li order price is clear, easy to use and a similar price tag digital prefabricated, as long as the color digital pen painted on it (see p1.jpg).

    This price tag has a feature on some numbers, upside down is a reasonable number. Such as: 1,256,890 can. As a result, if the sign hanging down, it may have turned into another price, for example: 1958 is hanging upside down: 8561, the difference between a few thousand dollars ah !!

    Of course, in most cases fail to read, for example, 1110 can not be reversed, because 0 can not start with a number.

    One day, a tragedy has happened. A store clerk accidentally put a price tag to hang two down. Both price and brand TV sets sold out!

    Fortunately, prices differ significantly, one of which lost more than 2 million price tags, price tags, but earned another eight hundred, together, but earn 558 yuan.

    Based on this information, please calculate: money-losing price tag that the correct price should be how much?

The answer is an integer 4, please submit the numbers directly through the browser.
Note: Do not submit answers to the process, or other assistance in explaining the contents of the class.

answer

9088

Code

 

 1 /*2013-蓝桥杯-省-A-4*/ 
 2 #include<iostream>
 3 #include<map>
 4 #include<vector> 
 5 #include<algorithm>
 6 using namespace std;
 7 map<char ,char> rev_map;
 8 char num[7]={'1','2','5','6','8','9','0'};
 9 vector<int > g800p;
10 map<int , int > g800p_profmap;
11 vector<int > l200p;
12 map<int ,int > l200p_profmap;
13 struct val{
14     char a[4];
15 };
16 
17 void init(){
18     rev_map['1']='1';
19     rev_map['2']='5';
20     rev_map['6']='9';
21     rev_map['8']='8';
22     rev_map['9']='6';
23     rev_map['0']='0'; 
24 }
25 int str2int(char a[]){
26     int temp=0;
27     temp=a[0]-'0'+10*(a[1]-'0')+100*(a[2]-'0')+1000*(a[3]-'0');
28     return temp;
29 }
30 void dfs(val cur,int pos){
31     if(pos==4){
32         if(cur.a[0]=='0'||cur.a[3]=='0'){
33             return;
34         }else{
35             char b[4];
36             int bef_val=str2int(cur.a);
37             reverse_copy(cur.a,cur.a+4,b);
38             b[0]=rev_map[b[0]];
39             b[1]=rev_map[b[1]];
40             b[2]=rev_map[b[2]];
41             b[3]=rev_map[b[3]];
42             int aft_val=str2int(b);
43             int prof=aft_val-bef_val;
44             if(prof>800&&prof<900){
45                 //cout<<"find: (800)  "<<bef_val<<endl;
46                 //cout<<prof<<endl;
47                 g800p.push_back(bef_val);
48                 g800p_profmap[bef_val]=prof;
49             }
50             if(prof<-200&&prof>-300){
51                 //cout<<"find: (-200)  "<<bef_val<<endl;
52                 //cout<<prof<<endl;
53                 l200p.push_back(bef_val);
54                 l200p_profmap[bef_val]=prof;
55             }
56             
57             return;
58         }
59     }else{
60         for(int i=0;i<7;i++){
61             cur.a[pos]=num[i];
62             dfs(cur,pos+1);
63         }
64     }
65 }
66 int main(){
67     init();
68     val a;
69     dfs(a,0);
70     cout<<"dfs ok"<<endl;
71     for(vector<int>::iterator iter1=g800p.begin();iter1!=g800p.end();iter1++){
72         for(vector<int>::iterator iter2=l200p.begin();iter2!=g800p.end();iter2++){
73         if(g800p_profmap[*iter1]+l200p_profmap[*iter2]==558){
74             cout<<*iter2<<" "<<endl;
75             break;
76             }
77         }
78     }    
79     return 0;
80 } 
81