HDU1015 Safecracker

题目地址：http://acm.hdu.edu.cn/showproblem.php?pid=1015

      题目罗嗦了半天，其实意思很简单，就是给定一个目标值target,再给你一个备选字符串(5~12个字符)，要你在这个字符串里选5个出来，满足题中给定的等式，并且你选择的这5个字符组成的字符串必须是所有可能情况中按字典序最大的情况。<?xml:namespace prefix = o ns = "urn:schemas-microsoft-com:office:office" />

      简单分析下就可以看出，就是一个组合问题，问题解的最大规模就是12取5，就是12*11*10*9*8*7，而最小规模是5取5，所以应该用枚举法就可以搞定。

#include  < iostream >
#include  < string >
#include  < vector >
using   namespace  std;

bool  isOk( int  v, int  w, int  x, int  y, int  z, long   int  target)
{//判断是否符合要求
    if((v-w*w+x*x*x-y*y*y*y+z*z*z*z*z)==target)
       return true;
    else
       return false;
}
int  charToInt( char  ch)
{
    return ch-'A'+1;
}
void  mySort( string &  p, int  len)
{//让字符串按字母逆序排列,这样就可以很快找到最大符合要求的情况
    int i,j;
    char temp;
    for(i=0;i<len-1;i++)
     {
      for(j=0;j<len-i-1;j++)
       {
        if(p[j]<p[j+1])
         {
          temp=p[j];
          p[j]=p[j+1];
          p[j+1] = temp;
         }
       }
     }
}

int  main( int  argc, char *  argv[])
{
    long int target;
    int tv,tw,tx,ty,tz;
    string strTmp;
    START:
    while(cin>>target>>strTmp&&!(target==0&&strTmp=="END"))
    {
        int len = strTmp.length();
        mySort(strTmp,len);
        int v,w,x,y,z;
        //下面开始用枚举法检验可能性,各个字符不能取同一个位置
        for(v=0;v<len;++v)
        {
              for(w=0;w<len;++w)
              {
                  if(v!=w)
                  {
                    for(x=0;x<len;++x)
                    {
                          if(x!=v&&x!=w)
                          {
                              for(y=0;y<len;++y)
                              {
                                    if(y!=v&&y!=w&&y!=x)
                                    {
                                        for(z=0;z<len;++z)
                                        {
                                            if(z!=v&&z!=w&&z!=x&&z!=y)
                                            {
                                                tv = charToInt(strTmp[v]);
                                                tw = charToInt(strTmp[w]);
                                                tx = charToInt(strTmp[x]);
                                                ty = charToInt(strTmp[y]);
                                                  tz = charToInt(strTmp[z]);
                                                if(isOk(tv,tw,tx,ty,tz,target))
                                                {
                                                       cout<<strTmp[v]<<strTmp[w]<<strTmp[x]<<strTmp[y]<<strTmp[z]<<endl;
                                                    goto START;//遇到最大的符合要求的值，直接跳出，进入下一个
                                                }
                                            }
                                         }
                                    }
                                }
                           }
                        }
                    }
                }
           }
           cout<<"no solution"<<endl;
    }
    return 0;
}