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  • Conversions(水)

    Conversions

    Time Limit: 1000/500 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
    Total Submission(s): 1054    Accepted Submission(s): 618

    Problem Description
    Conversion between the metric and English measurement systems is relatively simple. Often, it involves either multiplying or dividing by a constant. You must write a program that converts between the following units:

     
    Input
    The first line of input contains a single integer N, (1 ≤ N ≤ 1000) which is the number of datasets that follow.
    Each dataset consists of a single line of input containing a floating point (double precision) number, a space and the unit specification for the measurement to be converted. The unit specification is one of kg, lb, l, or g referring to kilograms, pounds, liters and gallons respectively.
     
    Output
    For each dataset, you should generate one line of output with the following values: The dataset number as a decimal integer (start counting at one), a space, and the appropriately converted value rounded to 4 decimal places, a space and the unit specification for the converted value.

     
    Sample Input
    5 1 kg 2 l 7 lb 3.5 g 0 l
     
    Sample Output
    1 2.2046 lb 2 0.5284 g 3 3.1752 kg 4 13.2489 l 5 0.0000 g
     
    Source
     
    Recommend
    linle
     
     1 #include <iostream>
     2 #include <string>
     3 #include <cstdio>
     4 #include <cmath>
     5 #include <cstring>
     6 #include <algorithm>
     7 #define LL long long
     8 #define maxi 2147483647
     9 #define maxl 9223372036854775807
    10 #define dg cout << "Here!" << endl;
    11 using namespace std;
    12 
    13 int main()
    14 {
    15     int n, t = 1;
    16     double a;
    17     string s;
    18     cin >> n;
    19     while(n--)
    20     {
    21         cin >> a >> s;
    22         if(s == "kg")
    23            printf("%d %.4lf lb\n", t++, a * 2.2046);
    24         else if(s == "lb")
    25             printf("%d %.4lf kg\n", t++, a * 0.4536);
    26         else if(s == "l")
    27             printf("%d %.4lf g\n", t++, a * 0.2642);
    28         else if(s == "g")
    29             printf("%d %.4lf l\n", t++, a * 3.7854);
    30     }
    31     return 0;
    32 }
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  • 原文地址:https://www.cnblogs.com/cszlg/p/2910477.html
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