Conversions
Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1674 Accepted Submission(s): 986
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
2008 “Shun Yu Cup” Zhejiang Collegiate Programming Contest - Warm Up（1）
题目大意：国际度量制度和英国度量制度之间的转换。
思路：水题。2015年第一道题，希望新的一年继续努力，得到更多的收获。
#include<iostream>
#include<string>
#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
int N;
double a;
string s;
cin >> N;
for(int i = 1; i <= N; ++i)
{
cin >> a >> s;
printf("%d ",i);
if(s == "kg")
printf("%.4lf lb\n",2.2046*a);
else if(s == "l")
printf("%.4lf g\n",0.2642*a);
else if(s == "lb")
printf("%.4lf kg\n",a*0.4536);
else if(s == "g")
printf("%.4lf l\n",a*3.7854);
}
return 0;
}