HDU 1985-CSDN博客

本文链接：https://blog.csdn.net/DT_Zhangshuo/article/details/52614604

Conversions

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

在解决这到水题时发现了一些新的解法，就是字符串的判断，用整型去判断字符，那么怎么知道字符串对应的整型是多少呢？提前用%d吧str[0]输出就行了。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int t,str[4];
    double n;
    scanf("%d",&t);
    for(int i=1; i<=t; i++)
    {
        memset(str,0,sizeof(str));
        scanf("%lf",&n);
        scanf("%s",str);
        printf("%d ",i);
        if(str[0]==26475)
        {
            printf("%.4lf lb\n",n*2.2046);
            continue;
        }
        if(str[0]=='g')
        {
            printf("%.4lf l\n",n*3.7854);
            continue;
        }
        if(str[0]==25196)
        {
            printf("%.4lf kg\n",n*0.4536);
            continue;
        }
        else
        {
            printf("%.4lf g\n",n*0.2642);
            continue;
        }
    }
    return 0;
}