首页 > 代码库 > [acm]HDOJ 3082 Simplify The Circuit

[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

 

 1 //11481261    2014-08-18 16:52:47    Accepted    3082    0MS    384K    733 B    G++    空信高手 2 #include<string> 3 #include<iostream> 4 #include<cstdio> 5 #include<cstdlib> 6  7 using namespace std; 8  9 int main()10 {11     //freopen("input.txt","r",stdin);12     string str;13     int Sum=0,nCases=0,count=0;14     cin>>count;15     double circuit=0;16     while(count--)17     {18         cin>>nCases;19         circuit=0;20         while(nCases--)21         {22             cin>>str;23             Sum=0;24             int end = 0;25             int pre = 0;26             while(end<str.length())27             {28                 if(str[end]==-)29                     end++;30                 else31                 {32                     pre=end;    33                     while((str[end]!=-)&&(end<str.length()))34                     {35                         end++;36                     }37                     //字符串分割38                     string str1 = str.substr(pre,end-pre);39                     int i = atoi(str1.c_str());40                     Sum+=i;41                 }42             }43             circuit+=1.0/Sum;  //并联电阻公式44         }45         printf("%.2lf\n",1.0/circuit);46     }47     48     return 0;49 }