1.    简要文字说明与分析

Ø        整体思路：Input à computation àoutput

Ø        逐步精化：

l        对用户输入合法性需要处理：

因为鸡和兔的个数必是正数，由    chicken+rabbit=heads,2*chicken+4*rabbit=feet可知

Feet>=2heads&&feet=<4heads

l        逻辑计算：

将chicken初始化从0开始，每循环一次递增一，相应的rabbit减一，    保持和是定数，即heads  值，当且仅当2*chicken+4*rabbit)==feet 时，问题有解，得问题的解，打印输出

l        漏洞细节：

当rabbit<0时，问题域无解，client须从新输入信息和计算

2.    程序代码与运行结果（注释略）

#include <stdio.h>

#define TRUE 1

main()

{

       int heads,feet,rabbit,chicken;

       printf("please enter the total numbers of heads and feet espectively\n");

while(TRUE)

{

       scanf("%d%d",&heads,&feet);

       while(feet<2heads||feet>4heads||feet<0||heads<0)    /*I fail here*/

       {

              printf("the number you entered is not appropriate,please try again\n");

              scanf("%d%d",&heads,&feet);

       }

       chicken=0;

       rabbit=heads-chicken;

       while(rabbit>=0)

       {

              if((2*chicken+4*rabbit)==feet)

              {

                     printf("chicken=%d,rabbit=%d\n",chicken,rabbit);

                     break;

              }

              chicken++;

              rabbit--;

       }

if(rabbit<0)

{

printf("the number you entered is not appropriate,please try again\n");

}else{

printf("have a good time!\n");

break;

}

}

}

School of Software , NEU

Vincent Wei

23th,Oct,2009