hdu 1015 Safecracker 暴力dfs

Safecracker

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6937    Accepted Submission(s): 3445

Problem Description

=== Op tech briefing, 2002/11/02 06:42 CST ===
"The item is locked in a Klein safe behind a painting in the second-floor library. Klein safes are extremely rare; most of them, along with Klein and his factory, were destroyed in World War II. Fortunately old Brumbaugh from research knew Klein's secrets and wrote them down before he died. A Klein safe has two distinguishing features: a combination lock that uses letters instead of numbers, and an engraved quotation on the door. A Klein quotation always contains between five and twelve distinct uppercase letters, usually at the beginning of sentences, and mentions one or more numbers. Five of the uppercase letters form the combination that opens the safe. By combining the digits from all the numbers in the appropriate way you get a numeric target. (The details of constructing the target number are classified.) To find the combination you must select five letters v, w, x, y, and z that satisfy the following equation, where each letter is replaced by its ordinal position in the alphabet (A=1, B=2, ..., Z=26). The combination is then vwxyz. If there is more than one solution then the combination is the one that is lexicographically greatest, i.e., the one that would appear last in a dictionary."

v - w^2 + x^3 - y^4 + z^5 = target

"For example, given target 1 and letter set ABCDEFGHIJKL, one possible solution is FIECB, since 6 - 9^2 + 5^3 - 3^4 + 2^5 = 1. There are actually several solutions in this case, and the combination turns out to be LKEBA. Klein thought it was safe to encode the combination within the engraving, because it could take months of effort to try all the possibilities even if you knew the secret. But of course computers didn't exist then."

=== Op tech directive, computer division, 2002/11/02 12:30 CST ===

"Develop a program to find Klein combinations in preparation for field deployment. Use standard test methodology as per departmental regulations. Input consists of one or more lines containing a positive integer target less than twelve million, a space, then at least five and at most twelve distinct uppercase letters. The last line will contain a target of zero and the letters END; this signals the end of the input. For each line output the Klein combination, break ties with lexicographic order, or 'no solution' if there is no correct combination. Use the exact format shown below."

 

 

Sample Input

1 ABCDEFGHIJKL11700519 ZAYEXIWOVU3072997 SOUGHT1234567 THEQUICKFROG0 END

 

 

Sample Output

LKEBAYOXUZGHOSTno solution

 

 

这题没什么好说的 把所有的数据测试一遍就可以了  但是默默地吐槽一个地方  就是第一个测试数据  我让a变量从前向后遍历 得出的数据和 测试答案不同，然后我默默地 让a变量从后向前遍历，，数据就相同了。。。不知道为什么。。。然后就可以AC了。。。。   可能是数据太弱了。。吧

 

#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int cal(char v,char w,char x,char y,char z){
 v-=64;w-=64;x-=64;y-=64;z-=64;
 return v-w*w+x*x*x-y*y*y*y+z*z*z*z*z;
}
int tar;
char str[50];
bool vis[50];
int main()
{
 int len,min,max,flag,i;
 int a,b,c,d,e;
 while(cin>>tar>>str){
  flag=0;
  if(tar==0 && !strcmp(str,"END")) break;
  len=strlen(str);
  sort(str,str+len);
  min=cal(str[2],str[len-2],str[1],str[len-1],str[0]);
  max=cal(str[len-3],str[1],str[len-2],str[0],str[len-1]);
  if(tar>max || tar < min) 
  {
   cout<<"no solution"<<endl;
   continue;
  }
  //if(tar==max) cout<<str[len-3]<<str[1]<<str[len-2]<<str[0]<<str[len-1]<<endl;
  //else if(tar==min) cout<<str[len-3]<<str[1]<<str[len-2]<<str[0]<<str[len-1]<<endl;
  //else
  //{
  for(i=0;i<len;i++)  vis[i]=false;

  //for(a=0;a<len;a++)
  for(a=len-1;a>=0;a--)
  {
   vis[a]=true;
   for(b=0;b<len;b++)
   {
    if(vis[b]==false){
     vis[b]=true;
     for(c=0;c<len;c++)
     {
      if(vis[c]==false){
       vis[c]=true;
       for(d=0;d<len;d++)
       {
        if(vis[d]==false){
         vis[d]=true;
         for(e=0;e<len;e++)
         {
          if(vis[e]==false){
           vis[e]=true;
           if(cal(str[a],str[b],str[c],str[d],str[e])==tar)
           {flag=1;goto End;}
           vis[e]=false;
          }
         }
         vis[d]=false;
        }
       }
       vis[c]=false;
      }
     }
     vis[b]=false;
    }
   }
   vis[a]=false;
  }
  //}
End:if(flag)
  cout<<str[a]<<str[b]<<str[c]<<str[d]<<str[e]<<endl;
 else
  cout<<"no solution"<<endl;
 //}

 }
 return 0;
}