HDU 3082 HDOJ 3082 Simplify The Circuit ACM 3082 IN HDU

 

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题目地址 :

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

题目分析:

     没什么特殊的方法, 简单模拟就可以了  :

 代码 :

/*
Mail to   : miyubai@gamil.com
My Blog   : www.baiyun.me
Link      : http://www.cnblogs.com/MiYu  || http://www.cppblog.com/MiYu
Author By : MiYu
Test      : 1
Complier  : g++ mingw32-3.4.2
Program   : HDU_3082
Doc Name  : Simplify The Circuit
*/
//#pragma warning( disable:4789 )
#include <iostream>
#include <fstream>
#include <sstream>
#include <algorithm>
#include <string>
#include <set>
#include <map>
#include <utility>
#include <queue>
#include <stack>
#include <list>
#include <vector>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
using namespace std;
char str[110];
char *tok;
int main ()
{
    int T;
    scanf ( "%d",&T );
    while ( T -- ) {
        int N;
        double res = 0;
        scanf ( "%d", &N );
        for ( int i = 1; i <= N; ++ i ) {
            scanf ( "%s", str );
            tok = strtok ( str, "-" );
            int t = atoi ( tok );
            int r = t;
            while ( tok = strtok ( NULL, "-" ) ) {
                t = atoi ( tok );
                r += t;
            }
            res += 1.0 / r;
        }
        printf ( "%.2lf\n", 1.0 / res );
    }
    return 0;
}