HDU1985 ZOJ2988 UVALive3911 Conversions【水题】

Conversions

Time Limit: 1000/500 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2723 Accepted Submission(s): 1558

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）

问题链接：HDU1985 ZOJ2988 UVALive3911 Conversions
问题简述：（略）
问题分析：
    单位转换问题，大水题。
程序说明：（略）
参考链接：（略）
题记：（略）

AC的C语言程序如下：

/* HDU1985 ZOJ2988 UVALive3911 Conversions */

#include <stdio.h>

int main(void)
{
    int n, k;
    double a;
    char s[8];
    scanf("%d", &n);
    for(k = 1; k <= n; k++) {
        scanf("%lf%s", &a, s);
        if(s[0] == 'k')
            printf("%d %.4lf lb\n", k, a * 2.2046);
        else if(s[0] == 'g')
            printf("%d %.4lf l\n", k, a * 3.7854);
        else if(s[0] == 'l' && s[1] == 'b')
            printf("%d %.4lf kg\n", k, a * 0.4536);
        else if(s[0] == 'l')
            printf("%d %.4lf g\n", k, a * 0.2642);
    }

    return 0;
}