[acm]HDOJ 3082 Simplify The Circuit

题目地址：

http://acm.hdu.edu.cn/showproblem.php?pid=3082

---

字符串处理+并联电阻公式

 //11481261    2014-08-18 16:52:47    Accepted    3082    0MS    384K    733 B    G++    空信高手
 #include<string>
 #include<iostream>
 #include<cstdio>
 #include<cstdlib>

 using namespace std;

 int main()
 {
     //freopen("input.txt","r",stdin);
     string str;
     int Sum=,nCases=,count=;
     cin>>count;
     double circuit=;
     while(count--)
     {
         cin>>nCases;
         circuit=;
         while(nCases--)
         {
             cin>>str;
             Sum=;
             int end = ;
             int pre = ;
             while(end<str.length())
             {
                 if(str[end]=='-')
                     end++;
                 else
                 {
                     pre=end;
                     while((str[end]!='-')&&(end<str.length()))
                     {
                         end++;
                     }
                     //字符串分割
                     string str1 = str.substr(pre,end-pre);
                     int i = atoi(str1.c_str());
                     Sum+=i;
                 }
             }
             circuit+=1.0/Sum;  //并联电阻公式
         }
         printf("%.2lf\n",1.0/circuit);
     }

     return ;
 }