我拒绝5人俗称黑5人都有谁

我拒绝5人俗称黑5人都有谁

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!=3B说： b&&a+c+d+e==0||!b&&a+c+d+e!=0C说： c&&a+b+d+e==1||!c&&a+b+d+e!=1D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。*程序说明与注释#includeint in(){int a,b,c,d,e;for(a=0;a<=1;a++) for(b=0;b<=1;b++) 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)

俗称黑5人