爱因斯坦谜题的c语言解答
[旧文搬迁]
题目如下,
1、在一条街上，有5座房子，喷了5种颜色。 2、每个房里住着不同国籍的人 3、每个人喝不同的饮料，抽不同品牌的香烟，养不同的宠物 另外, 1、英国人住红色房子 2、瑞典人养狗 3、丹麦人喝茶 4、绿色房子在白色房子左面 5、绿色房子主人喝咖啡 6、抽Pall Mall香烟的人养鸟 7、黄色房子主人抽Dunhill香烟 8、住在中间房子的人喝牛奶 9、挪威人住第一间房 10、抽Blends香烟的人住在养猫的人隔壁 11、养马的人住抽Dunhill香烟的人隔壁 12、抽Blue Master的人喝啤酒 13、德国人抽Prince香烟 14、挪威人住蓝色房子隔壁 15、抽Blends香烟的人有一个喝水的邻居
问题是：谁养鱼？
以下是c语言的解答
#include <stdio.h>
#include <stdlib.h>
enum Nat
{
ENG,SWE,DEN,NOR,GER
};
enum Hou
{
RED,GREEN,WHITE,YELLOW,BLUE
};
enum Pet
{
DOG,BIRD,CAT,HORSE,FISH
};
enum Dri
{
TEA,COFFEE,MILK,BEER,WATER
};
enum Cig
{
PALLMALL,DUNHILL,BLENDS,BLUEMASTER,PRINCE
};
struct S//scence
{
int *nat; //nationality
int *hou; //house
int *pet; //pet
int *dri; //drink
int *cig; //cigar
} ;
void Permute(int a[120][5]) //permutation process
{
int i=0,j=0,r1,r2,r3,r4,r5;
for (r1=0;r1<5;r1++)
{
for (r2=0;r2<5;r2++)
{
if (r2!=r1)
for (r3=0;r3<5;r3++)
{
if (r3!=r2&&r3!=r1)
for (r4=0;r4<5;r4++)
{
if (r4!=r3&&r4!=r2&&r4!=r1)
for (r5=0;r5<5;r5++)
{
if (r5!=r4&&r5!=r3&&r5!=r2&&r5!=r1)
{
a[i][j++]=r1;
a[i][j++]=r2;
a[i][j++]=r3;
a[i][j++]=r4;
a[i++][j]=r5;
j=0;
}
}
}
}
}
}
//printf("\n***枚举过程完成***\n\n");
return ;
}
int NatPos(struct S s,enum Nat t) //get position by nationality
{
int i;
for (i=0;i<5;i++)
if (s.nat[i]==t)
return i;
return -1;
}
int HouPos(struct S s,enum Hou t) //get position by house
{
int i;
for (i=0;i<5;i++)
if (s.hou[i]==t)
return i;
return -2;
}
int PetPos(struct S s,enum Pet t) //get position by pet
{
int i;
for (i=0;i<5;i++)
if (s.pet[i]==t)
return i;
return -3;
}
int DriPos(struct S s,enum Dri t) //get position by drink
{
int i;
for (i=0;i<5;i++)
if (s.dri[i]==t)
return i;
return -4;
}
int CigPos(struct S s,enum Cig t) //get position by cigar
{
int i;
for (i=0;i<5;i++)
if (s.cig[i]==t)
return i;
return -5;
}
int main()
{
int a[120][5];
struct S s;
int r1,r2,r3,r4,r5,i;
Permute(a); //get permuted array
for (s.dri=a[0],r1=0;r1<120;s.dri+=5,r1+=1) //permute 5 rounds for scene
{
if (s.dri[2]==MILK) //condition required
for (s.nat=a[0],r2=0;r2<120;s.nat+=5,r2+=1)
{
if (s.nat[0]==NOR&&s.dri[NatPos(s,DEN)]==TEA) //condition required
for (s.hou=a[0],r3=0;r3<120;s.hou+=5,r3+=1)
{
if (s.hou[NatPos(s,ENG)]==RED&&s.dri[HouPos(s,GREEN)]==COFFEE&&
(HouPos(s,GREEN)+1==HouPos(s,WHITE))&&
((NatPos(s,NOR)==HouPos(s,BLUE)+1)||(NatPos(s,NOR)==HouPos(s,BLUE)-1)))
for (s.pet=a[0],r4=0;r4<120;s.pet+=5,r4+=1)
{
if (s.pet[NatPos(s,SWE)]==DOG) //condition required
for (s.cig=a[0],r5=0;r5<120;s.cig+=5,r5+=1)
{
//<----
//////////////////////////////////////////////////////////////////
if (((CigPos(s,BLENDS)==DriPos(s,WATER)+1)||(CigPos(s,BLENDS)==DriPos(s,WATER)-1))&&
((PetPos(s,HORSE)==CigPos(s,DUNHILL)+1)||(PetPos(s,HORSE)==CigPos(s,DUNHILL)-1))&&
((CigPos(s,BLENDS)==PetPos(s,CAT)+1)||(CigPos(s,BLENDS)==PetPos(s,CAT)-1))&&
s.dri[CigPos(s,BLUEMASTER)]==BEER&&s.pet[CigPos(s,PALLMALL)]==BIRD&&
s.cig[HouPos(s,YELLOW)]==DUNHILL&&s.cig[NatPos(s,GER)]==PRINCE)
{
switch (s.nat[PetPos(s,FISH)])
{
case ENG:printf("***英国人养鱼***\n");break;
case SWE:printf("***瑞典人养鱼***\n");break;
case DEN:printf("***丹麦人养鱼***\n");break;
case NOR:printf("***挪威人养鱼***\n");break;
case GER:printf("***德国人养鱼***\n");break;
default:printf("***解不存在***\n");break;
}
printf("\n此时具体情况如下");
printf("\n\n*****************\n");
printf(" 标号 1 2 3 4 5 \n");
printf("\n 国籍 ");
for (i=0;i<5;i++)
{
switch (s.nat[i])
{
case ENG:printf("英国 ");break;
case SWE:printf("瑞典 ");break;
case DEN:printf("丹麦 ");break;
case NOR:printf("挪威 ");break;
case GER:printf("德国 ");break;
default:printf("错误 ");break;
};
}
printf("\n\n 房屋颜色 ");
for (i=0;i<5;i++)
{
switch (s.hou[i])
{
case RED:printf("红色 ");break;
case GREEN:printf("绿色 ");break;
case WHITE:printf("白色 ");break;
case YELLOW:printf("黄色 ");break;
case BLUE:printf("蓝色 ");break;
default:printf("错误 ");break;
};
}
printf("\n\n 宠物 ");
for (i=0;i<5;i++)
{
switch (s.pet[i])
{
case DOG:printf("狗 ");break;
case BIRD:printf("鸟 ");break;
case CAT:printf("猫 ");break;
case HORSE:printf("马 ");break;
case FISH:printf("鱼 ");break;
default:printf("错误 ");break;
};
}
printf("\n\n 饮品 ");
for (i=0;i<5;i++)
{
switch (s.dri[i])
{
case TEA:printf("茶 ");break;
case COFFEE:printf("咖啡 ");break;
case MILK:printf("牛奶 ");break;
case BEER:printf("啤酒 ");break;
case WATER:printf("水 ");break;
default:printf("错误 ");break;
};
}
printf("\n\n 香烟 ");
for (i=0;i<5;i++)
{
switch (s.cig[i])
{
case PALLMALL:printf("帕玛 ");break;
case DUNHILL:printf("杜希尔 ");break;
case BLENDS:printf("布莱登 ");break;
case BLUEMASTER:printf("布鲁玛 ");break;
case PRINCE:printf("布里斯 ");break;
default:printf("错误 ");break;
};
}
printf("\n\n*****************\n\n");
}
}
}
}
}
}
return 0;
}
Thu, 25 Aug 2022 04:06:10 -1100
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