源代码 仅供参考:
#include<stdlib.h>
#include <stdio.h>
#include<string.h>
int main(void)
{
FILE *fp;
char data[5][40]={"国家","房子","宠物","饮料","香烟"};
char country[6][40]={" "," "," "," "," "," "};
char house[6][40]={" "," "," "," "," "," "};
char pet[6][40]={" "," "," "," "," "," "};
char drink[6][40]={" "," "," "," "," "," "};
char smoke[6][40]={" "," "," "," "," "," "};
int i, j, k, l, m, n, o, p, q, r, s, t;
bool A1, A2, A3, A4, A5, A6, A7, A8, A9, A10, A11, A12;
if ((fp=fopen("sdfg.txt","w+")) == NULL)
{
fprintf(stderr,"\nError opening file \n");
exit(1);
}
strcpy(country[1],"挪威");//挪威人住在第一个房子里面
strcpy(house[2],"蓝色");
strcpy(drink[3],"牛奶");// 住在中间那个房子里的人喝牛奶
for(i=1; i<6; i++)
for(j=1; j<6; j++)
for(k=1; k<5; k++)
for(l=1; l<6; l++)
for(m=1; m<6; m++)
for(n=1; n<6; n++)
for(o=1; o<6; o++)
{
A1 = (i!=1 && i!=2 );//英国人住在红色的房子里；根据房子和国家判断
A2 = (j!=1 && j!=i ); //瑞典人养狗作为宠物；根据国家和宠物判断
A3 = (k>3 && k!=i);//绿房子的主人喝咖啡；绿房子紧挨着白房子，在白房子的左边；根据房子和国家判断
A4 = (l!=1 && l!=3 && l!=i && l!=j && l!=k); //丹麦人喝茶；根据国家和饮料判断
A5 = (m!=j);//抽Pall Mall牌香烟的人养鸟；根据香烟和宠物判断
A6 = (n!=2 && n!=i && n!=k && n!=m);
//黄色房子里的人抽Dunhill牌香烟；根据房子和香烟判断
A7 = (o!=3 && o!=k && o!=l && o!= m && o!=n
);//抽BlueMaster牌香烟的人和啤酒；根据香烟和饮料判断
if(A1&&A2&&A3&&A4&&A5&&A6&&A7)
{
p = 15-1-i-j-l; //德国人抽Prince牌香烟 ，根据国家判断
q = 15-m-n-o-p; //抽Blends牌香烟的人和养猫的人相邻,根据香烟判断
A8 = (p!=1 && p!=i && p!=j && p!=l && p!=m && p!=n &&
p!=o);//德国人抽Prince牌香烟 ，
A9 = (q!=m && q!=n && q!=o && q!=p); //抽Blends牌香烟的人和养猫的人相邻,
if(A8&&A9)
{
if(q>1 && (q-1)!=j && (q-1)!=m)//抽Blends牌香烟的人和养猫的人相邻,根据宠物判断
{
strcpy(pet[q-1], "猫");
if(n>1 && (n-1)!=j && (n-1)!=m)
{
strcpy(pet[n-1], "马");
t = 15-(q-1)-(n-1)-j-m;
strcpy(pet[t], "鱼");
}
else if(n<5 && (n+1)!=j && (n+1)!=m)
{
strcpy(pet[n+1], "马");
t = 15-(q-1)-(n+1)-j-m;
strcpy(pet[t], "鱼");
}
}
else if(q<5 && (q+1)!=j && (q+1)!=m)
{
strcpy(pet[q+1], "猫");
if(n>1 && (n-1)!=j && (n-1)!=m)
{
strcpy(pet[n-1], "马");
t = 15-(q+1)-(n-1)-j-m;
strcpy(pet[t], "鱼");
}
else if(n<5 && (n+1)!=j && (n+1)!=m)
{
strcpy(pet[n+1], "马");
t = 15-(q+1)-(n+1)-j-m;
strcpy(pet[t], "鱼");
}
}
//抽Blends牌香烟的人和喝矿泉水的人相邻，根据饮料判断
if(q>1 && (q-1)!=3 && (q-1)!=k && (q-1)!=l && (q-1)!=o)
strcpy(drink[q-1],"矿泉水");
else if(q<5 && (q+1)!=3 && (q+1)!=k && (q+1)!=l && (q+1)!=o)
strcpy(drink[q+1],"矿泉水");
//养马的人和抽Dunhill牌香烟的人相邻，根据宠物判断
//抽Blends牌香烟的人和养猫的人相邻,根据宠物判断
if((q>1 && (q-1)!=j && (q-1)!=m) ||(q<5 && (q+1)!=j &&
(q+1)!=m))
//抽Blends牌香烟的人和喝矿泉水的人相邻，根据饮料判断
if((q>1 && (q-1)!=3 && (q-1)!=k && (q-1)!=l && (q-1)!=o) ||(q<5
&& (q+1)!=3 && (q+1)!=k && (q+1)!=l && (q+1)!=o))
//养马的人和抽Dunhill牌香烟的人相邻，根据宠物判断
if((n>1 && (n-1)!=j && (n-1)!=m) || (n<5 && (n+1)!=j &&
(n+1)!=m))
{ strcpy(country,"英国");//英国人住在红色的房子里；
strcpy(house,"红色");
strcpy(country[j],"瑞典"); //瑞典人养狗作为宠物；
strcpy(pet[j], "狗");
strcpy(house[k],"绿色");//绿房子紧挨着白房子，在白房子的左边；
strcpy(house[k+1],"白色") ;
strcpy(drink[k],"咖啡"); //绿房子的主人喝咖啡；
strcpy(country[l],"丹麦");//丹麦人喝茶；
strcpy(drink[l],"茶");
strcpy(smoke[m],"PallMall");//抽Pall Mall牌香烟的人养鸟；
strcpy(pet[m],"鸟");
strcpy(house[n],"黄色"); //黄色房子里的人抽Dunhill牌香烟
strcpy(smoke[n],"Dunhill");
strcpy(smoke[o],"BlueMaster");//抽BlueMaster牌香烟的人和啤酒
strcpy(drink[o],"啤酒");
strcpy(country[p],"德国");//德国人抽Prince牌香烟
strcpy(smoke[p],"Prince");
strcpy(smoke[q],"Blends"); //抽Blends牌香烟的人和养猫的人相邻,
fprintf(fp,"\n");
fprintf(stdout, "\n");
for(r=0; r<5; r++)
{
fprintf(stdout, "%15s", data[r]);
fprintf(fp,"%15s", data[r]);
}
fprintf(fp,"\n");
fprintf(stdout, "\n");
for(s=1; s<6; s++)
{
fprintf(stdout, "%15s", country[s]);
fprintf(stdout, "%15s", house[s]);
fprintf(stdout, "%15s", pet[s]);
fprintf(stdout, "%15s", drink[s]);
fprintf(stdout, "%15s", smoke[s]);
fprintf(stdout, "\n");
fprintf(fp,"%15s", country[s]);
fprintf(fp,"%15s", house[s]);
fprintf(fp,"%15s", pet[s]);
fprintf(fp,"%15s", drink[s]);
fprintf(fp,"%15s", smoke[s]);
fprintf(fp,"\n");
fflush(fp);
}
}
}
}
}
fclose(fp);
system("pause");
return 0;
}
[ 本帖最后由 mathbox 于 2008-6-27 15:16 编辑 ]
