[acm]HDOJ 3082 Simplify The Circuit

题目地址：

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

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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                 else

31                 {

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 }