简单枚举类型——植物与颜色 oj

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本文通过一个简单的枚举类型应用实例介绍了如何使用C语言中的枚举类型来匹配特定的输入并输出对应的信息。该实例涉及颜色与花卉之间的映射，展示了枚举类型的实用性和灵活性。

简单枚举类型——植物与颜色

Time Limit: 1000MS Memory Limit: 65536KB

Problem Description

请定义具有red, orange, yellow, green, blue, violet六种颜色的枚举类型color，根据输入的颜色名称，输出以下六种植物花朵的颜色：

Rose(red), Poppies(orange), Sunflower(yellow), Grass(green), Bluebells(blue), Violets(violet)。如果输入的颜色名称不在枚举类型color中，例如输入purple，请输出I don't know about the color purple.

Input

输入数据有多行，每行有一个字符串代表颜色名称，颜色名称最多30个字符，直到文件结束为止。

Output

输出对应颜色的植物名称，例如：Bluebells are blue. 如果输入的颜色名称不在枚举类型color中，例如purple, 请输出I don't know about the color purple.

Example Input

blue
yellow
purple

Example Output

Bluebells are blue.
Sunflower are yellow.
I don't know about the color purple.

Hint

请用枚举类型实现。

Author

lxh

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
enum
{
    red, orange, yellow, green, blue, violet,no
}co;

int main()
{
    char a[30];
    while(scanf("%s",a) != EOF)
          {
    if(strcmp(a,"red") == 0)
        co = red;
    else if(strcmp(a,"orange") == 0)
        co = orange;
    else if(strcmp(a,"yellow") == 0)
        co = yellow;
    else if(strcmp(a,"green") == 0)
        co = green;
    else if(strcmp(a,"blue") == 0)
        co = blue;
    else if(strcmp(a,"violet") == 0)
        co = violet;
    else
        co = no;
    if(co == 0)
        printf("Rose are red.\n");
    else if(co == 1)
        printf("Poppies are orange.\n");
    else if(co == 2)
        printf("Sunflower are yellow.\n");
    else if(co == 3)
        printf("Grass are green.\n");
    else if(co == 4)
        printf("Bluebells are blue.\n");
    else if(co == 5)
        printf("Violets are violet.\n");
    else
        printf("I don't know about the color %s.\n", a);
          }
    return 0;
}