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关闭窗口 有A、B、C、D、E五人，每人额头上都帖了一张黑或白的纸。五人对坐，每人都可以看到其它人额头上的纸的颜色。五人相互观察后，
A说：“我看见有三人额头上帖的是白纸，一人额头上帖的是黑纸。”
B说：“我看见其它四人额头上帖的都是黑纸。”
C说：“我看见一人额头上帖的是白纸，其它三人额头上帖的是黑纸。”
D说：“我看见四人额头上帖的都是白纸。”
E什么也没说。
现在已知额头上帖黑纸的人说的都是谎话，额头帖白纸的人说的都是实话。问这五人谁的额头是帖白纸，谁的额头是帖黑纸？
*问题分析与算法设计
假如变量A、B、C、D、E表示每个人额头上所帖纸的颜色，0 代表是黑色，1 代表是白色。根据题目中A、B、C、D四人所说的话可以总结出下列关系：
A说： a&&b+c+d+e==3||!a&&b+c+d+e!=3
B说： b&&a+c+d+e==0||!b&&a+c+d+e!=0
C说： c&&a+b+d+e==1||!c&&a+b+d+e!=1
D说： d&&a+b+c+e==4||!d&&a+b+c+e!=4
穷举每个人额头所帖纸的颜色的所有可能的情况，代入上述表达式中进行推理运算，使上述表达式为“真”的情况就是正确的结果。
*程序与程序注释
#include<stdio.h>
void main()
{
int a,b,c,d,e;
for(a=0;a<=1;a++) /*黑色：0 白色：1*/
for(b=0;b<=1;b++) /*穷举五个人额头帖纸的全部可能*/
for(c=0;c<=1;c++)
for(d=0;d<=1;d++)
for(e=0;e<=1;e++)
if((a&&b+c+d+e==3||!a&&b+c+d+e!=3)
&&(b&&a+c+d+e==0||!b&&a+c+d+e!=0)
&&(c&&a+b+d+e==1||!c&&a+b+d+e!=1)
&&(d&&a+b+c+e==4||!d&&a+b+c+e!=4))
{
printf("A is pasted a piece of %s paper on his forehead.\n",
a?"white":"black");
printf("B is pasted a piece of %s paper on his forehead.\n",
b?"white":"black");
printf("C is pasted a piece of %s paper on his forehead.\n",
c?"white":"black");
printf("D is pasted a piece of %s paper on his forehead.\n",
d?"white":"black");
printf("E is pasted a piece of %s paper on his forehead.\n",
e?"white":"black");
}
}
*运行结果
A is pasted a paper of black paper on his forehead. (黑)
B is pasted a paper of black paper on his forehead. (黑)
C is pasted a paper of white paper on his forehead. (白)
D is pasted a paper of black paper on his forehead. (黑)
E is pasted a paper of white paper on his forehead. (白)
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