ZOJ Problem Set 2988 Conversions

Conversions

Time Limit: 1 Second      Memory Limit: 32768 KB

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:

Type Metric English equivalent
Weight 1.000 kilograms 2.2046 pounds
Weight 0.4536 kilograms 1.0000 pound
Volume 1.0000 liter 0.2642 gallons
Volume 3.7854 liters 1.0000 gallon

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: Greater New York Region 2007

SOURCE CODE:

#include<iostream>
#include
<iomanip>
#include
<string>
using namespace std;

int main()
{
int cases;cin>>cases;
for(int i = 1; i <= cases; i++)
{
double num;cin>>num;
string s;cin>>s;
cout
<<i << " "<<fixed<<setprecision(4);
if(s == "kg")
{
cout
<<2.2046 * num<<" lb"<<endl;
}
else if( s == "lb")
{
cout
<<0.4536 * num<<" kg"<<endl;
}
else if(s == "l")
{
cout
<<0.2642 * num<<" g"<<endl;
}
else if(s == "g")
{
cout
<<3.7854 * num<<" l"<<endl;
}
}
return 0;
}

这是一道考察输出格式控制的题目,利用标准库的iomanip很容易就做到了输出保留四位小数的要求。

原文地址:https://www.cnblogs.com/malloc/p/2101788.html