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

51 kg2 l7 lb3.5 g0 l

 

Sample Output

1 2.2046 lb2 0.5284 g3 3.1752 kg4 13.2489 l5 0.0000 g

 

Source

2008 “Shun Yu Cup” Zhejiang Collegiate Programming Contest - Warm Up（1）

 

Recommend

linle

 

#include <iostream>#include <string>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>#define LL long long#define maxi 2147483647#define maxl 9223372036854775807#define dg cout << "Here!" << endl;using namespace std;int main(){    int n, t = 1;    double a;    string s;    cin >> n;    while(n--)    {        cin >> a >> s;        if(s == "kg")           printf("%d %.4lf lb\n", t++, a * 2.2046);        else if(s == "lb")            printf("%d %.4lf kg\n", t++, a * 0.4536);        else if(s == "l")            printf("%d %.4lf g\n", t++, a * 0.2642);        else if(s == "g")            printf("%d %.4lf l\n", t++, a * 3.7854);    }    return 0;}