[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=0,nCases=0,count=0;
    cin>>count;
    double circuit=0;
    while(count--)
    {
        cin>>nCases;
        circuit=0;
        while(nCases--)
        {
            cin>>str;
            Sum=0;
            int end = 0;
            int pre = 0;
            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
",1.0/circuit);
    }
    
    return 0;
}