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简单枚举类型——植物与颜色

2018-03-08 23:58 323 查看
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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.  

Sample Input

blue
yellow
purple

Sample Output

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

Hint

 请用枚举类型实现。
#include <stdio.h>
#include <string.h>
int main()
{
enum color {red, orange, yellow, green, blue, violet, no};
enum color a;
char st[44];
while(scanf("%s", st) != EOF)
{
if(strcmp(st, "red") == 0)
a = red;
else if(strcmp(st, "orange") == 0)
a = orange;
else if(strcmp(st, "yellow") == 0)
a = yellow;
else if(strcmp(st, "green") == 0)
a = green;
else if(strcmp(st, "blue") == 0)
a = blue;
else if(strcmp(st, "violet") == 0)
a = violet;
else
a = no;
//printf("a:%d\n", a);
switch(a)
{
case 0: printf("Rose are red.\n"); break;
case 1: printf("Poppies are orange.\n"); break;
case 2: printf("Sunflower are yellow.\n"); break;
case 3: printf("Grass are green.\n"); break;
case 4: printf("Bluebells are blue.\n"); break;
case 5: printf("Violets are violet.\n"); break;
case 6: printf("I don't know about the color %s.\n", st); break;
}
}
return 0;
}

/***************************************************
User name: jk170602王子原
Result: Accepted
Take time: 0ms
Take Memory: 148KB
Submit time: 2018-03-08 11:42:21
****************************************************/
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