试试你有多聪明——经典逻辑智力题50例

gxqcn · 发表于 2008-8-31 13:03:39

第15题不是早就注明了“已由shshsh_0510在132#、mathe在134#、fmtgy在136#解答”吗？

不过，还是非常感谢你的参与。

gxqcn · 发表于 2008-9-16 20:28:56

我在一本书上看到这到智力题的解题思路也是填写表格形式完成的，
比较直观，也易于推理，类似于有些人用Excel解“数独”问题。

另，我将表格略作编辑，去除了一些冗余标签。
不过，zYr 来论坛不久，帖子排版质量很高，赞一个。

gxqcn · 发表于 2008-9-16 21:12:22

刚刚更新了一下主题帖，
现在仅仅差 No.36 无人解答了。

gxqcn · 发表于 2008-9-16 21:56:20

查了一遍，近期的没找到。
不知是不是老早就有人解答了？

gxqcn · 发表于 2008-9-18 07:34:29

原帖由 zYr 于 2008-9-17 22:14 发表
这“经典逻辑智力题50例”都快做完了，能不能再出个几十例？

你可以试试由 shability 整理的《我整理的一些IQ题目合集》，
我记得里面并未完全解答。

还有无心人发的《数学奥林匹克升级题》，也非常有意思。

gxqcn · 发表于 2008-9-18 19:56:01

刚才在 VC6.0 下编译出现了大量 error，就略作了下修改：

/************
* 关于爱因斯坦98%的人都回答不出来的粗浅程序, 鉴于网络上没有很好的C参考程序, \
* 这里尝试写一个, 聊博一笑! *^_^*, 请在VC6.0或以上环境编译!
* Email: silitex@yeah.net
* 注: 如采用原始枚举: (5!)^5 = 24,883,200,000, 在普通的电脑需运行1分钟以上, \
* 故做一个算法优化, 在枚举时, 把已知的2个条件: 牛奶 = 2, 挪威人 = 0, \
* 直接确定; 这样算法的枚举次数为: (4!)^2*(5!)^3 = 995,328,000.
* 用C++的标准做法应该把所有的内容封装成类, 否则太容易因为下表取错而出错!
***/
#include <stdio.h>
#include <math.h>
#define N 5
#define N_NAME 24 /* Name的枚举次数 */
#define N_COLOR 120
#define N_DRINK 24
#define N_CIG 120
#define N_PET 120
enum E_NAME {ENGLISH, SWEDISH, DANE, NORWEGIAN, GERMAN};
enum E_COLOR {RED, GREEN, WHITE, YELLOW, BLUE};
enum E_DRINK {TEA, COFFEE, MILK, BEER, WATER};
enum E_CIGARETTE {PALLMALL, DUNHILL, BLENDS, BLUEMASTER, PRINCE}; // Cigar类?
enum E_PET {DOG, BIRD, CAT, HORSE, FISH};
const char *NAME[] = {"英国人", "瑞典人", "丹麦人", "挪威人", "德国人"};// 人名
const char *COLOR[] = {"红", "绿", "白", "黄", "蓝"}; // 颜色
const char *DRINK[] = {"茶", "咖啡", "牛奶", "啤酒", "水"}; // 饮料
const char *CIGARETTE[] = {"PallMal", "Dunhill", "混合烟", "BlueMas", \
"Prince"}; // 香烟
const char *PET[] = {"狗", "鸟", "猫", "马", "鱼"}; // 宠物
extern const int ENUM_NAME[][N];
extern const int ENUM_DRINK[][N];
extern const int ENUM_ALL[][N];
int main(void)
{
int tmp_name[N]; // 临时数组, 以上面的枚举变量为索引, 指示在第几栋房子
int tmp_color[N];
int tmp_drink[N];
int tmp_cig[N];
int tmp_pet[N];
int out_name[N]; // 输出数组, 以第几栋房子为位置, 记录上面的枚举变量值
int out_color[N];
int out_drink[N];
int out_cig[N];
int out_pet[N];
int i;
long per;
for (int n = 0; n < N_NAME; n++)
{
for (i = 0; i < N; i++)
tmp_name[ENUM_NAME[n][i]] = i;
for (int c = 0; c < N_COLOR; c++)
{
for (i = 0; i < N; i++)
tmp_color[i] = ENUM_ALL[c][i];
for (int d = 0; d < N_DRINK; d++)
{
for (i = 0; i < N; i++)
tmp_drink[ENUM_DRINK[d][i]] = i;
for (int c2 = 0; c2 < N_CIG; c2++)
{
for (i = 0; i < N; i++)
tmp_cig[i] = ENUM_ALL[c2][i];
for (int p = 0; p < N_PET; p++)
{
for (i = 0; i < N; i++)
tmp_pet[i] = ENUM_ALL[p][i];
/* 前面的就是枚举所有可能行, 下面的工作就根据条件判断\
* 已知我们用掉了5个初始条件, 所以这里判断10个剩余条件.
*/
if (tmp_name[ENGLISH] != tmp_color[RED])
continue;
if (tmp_name[SWEDISH] != tmp_pet[DOG])
continue;
if (tmp_name[DANE] != tmp_drink[TEA])
continue;
if (tmp_color[GREEN] != tmp_drink[COFFEE])
continue;
if (tmp_cig[PALLMALL] != tmp_pet[BIRD])
continue;
if (tmp_color[YELLOW] != tmp_cig[DUNHILL])
continue;
if (tmp_cig[BLUEMASTER] != tmp_drink[BEER])
continue;
if (tmp_name[GERMAN] != tmp_cig[PRINCE])
continue;
#if 1 // 绿在白的左边, 且必须为邻居
if (tmp_color[GREEN]+1 != tmp_color[WHITE])
continue;
#else // 绿在白的左边, 不必为邻居
if (tmp_color[GREEN] >= tmp_color[WHITE])
continue;
#endif
if (abs(tmp_cig[BLENDS]-tmp_pet[CAT]) != 1)
continue;
if (abs(tmp_pet[HORSE]-tmp_cig[DUNHILL]) != 1)
continue;
if (abs(tmp_name[NORWEGIAN]-tmp_color[BLUE]) != 1)
continue;
#if 1 // 这个条件多余, 可以不用.
if (abs(tmp_cig[BLENDS]-tmp_drink[WATER]) != 1)
continue;
#endif
// 如果上面的条件都满足, 则显示出来
for (i = 0; i < N; i++)
{
out_name[tmp_name[i]] = i;
out_color[tmp_color[i]] = i;
out_drink[tmp_drink[i]] = i;
out_cig[tmp_cig[i]] = i;
out_pet[tmp_pet[i]] = i;
}
printf("\nposi: 0 1 2 3 4");
printf("\nname: %-7s %-7s %-7s %-7s %-7s\n", \
NAME[out_name[0]], NAME[out_name[1]], \
NAME[out_name[2]], NAME[out_name[3]], \
NAME[out_name[4]]);
printf("color: %-7s %-7s %-7s %-7s %-7s\n", \
COLOR[out_color[0]], COLOR[out_color[1]], \
COLOR[out_color[2]], COLOR[out_color[3]], \
COLOR[out_color[4]]);
printf("drink: %-7s %-7s %-7s %-7s %-7s\n", \
DRINK[out_drink[0]], DRINK[out_drink[1]], \
DRINK[out_drink[2]], DRINK[out_drink[3]], \
DRINK[out_drink[4]]);
printf("cig: %-7s %-7s %-7s %-7s %-7s\n", \
CIGARETTE[out_cig[0]], CIGARETTE[out_cig[1]], \
CIGARETTE[out_cig[2]], CIGARETTE[out_cig[3]], \
CIGARETTE[out_cig[4]]);
printf("pet: %-7s %-7s %-7s %-7s %-7s\n", \
PET[out_pet[0]], PET[out_pet[1]], \
PET[out_pet[2]], PET[out_pet[3]], \
PET[out_pet[4]]);
}
}
}
// 显示当前的进度
per = ((N_COLOR*(long)n)+(long)c+1)*100/(N_NAME*N_COLOR);
printf("\rProgress: %02d%% ", per);
}
}
return 0;
}
const int ENUM_NAME[][N] = {
{NORWEGIAN, ENGLISH, SWEDISH, DANE , GERMAN },
{NORWEGIAN, ENGLISH, SWEDISH, GERMAN , DANE },
{NORWEGIAN, ENGLISH, DANE , SWEDISH, GERMAN },
{NORWEGIAN, ENGLISH, DANE , GERMAN , SWEDISH},
{NORWEGIAN, ENGLISH, GERMAN , SWEDISH, DANE },
{NORWEGIAN, ENGLISH, GERMAN , DANE , SWEDISH},
{NORWEGIAN, SWEDISH, ENGLISH, DANE , GERMAN },
{NORWEGIAN, SWEDISH, ENGLISH, GERMAN , DANE },
{NORWEGIAN, SWEDISH, DANE , ENGLISH, GERMAN },
{NORWEGIAN, SWEDISH, DANE , GERMAN , ENGLISH},
{NORWEGIAN, SWEDISH, GERMAN , ENGLISH, DANE },
{NORWEGIAN, SWEDISH, GERMAN , DANE , ENGLISH},
{NORWEGIAN, DANE , ENGLISH, SWEDISH, GERMAN },
{NORWEGIAN, DANE , ENGLISH, GERMAN , SWEDISH},
{NORWEGIAN, DANE , SWEDISH, ENGLISH, GERMAN },
{NORWEGIAN, DANE , SWEDISH, GERMAN , ENGLISH},
{NORWEGIAN, DANE , GERMAN , ENGLISH, SWEDISH},
{NORWEGIAN, DANE , GERMAN , SWEDISH, ENGLISH},
{NORWEGIAN, GERMAN , ENGLISH, SWEDISH, DANE },
{NORWEGIAN, GERMAN , ENGLISH, DANE , SWEDISH},
{NORWEGIAN, GERMAN , SWEDISH, ENGLISH, DANE },
{NORWEGIAN, GERMAN , SWEDISH, DANE , ENGLISH},
{NORWEGIAN, GERMAN , DANE , ENGLISH, SWEDISH},
{NORWEGIAN, GERMAN , DANE , SWEDISH, ENGLISH},
};
const int ENUM_DRINK[][N] = {
{TEA , COFFEE , MILK , BEER , WATER },
{TEA , COFFEE , MILK , WATER , BEER },
{TEA , BEER , MILK , COFFEE , WATER },
{TEA , BEER , MILK , WATER , COFFEE },
{TEA , WATER , MILK , COFFEE , BEER },
{TEA , WATER , MILK , BEER , COFFEE },
{COFFEE , TEA , MILK , BEER , WATER },
{COFFEE , TEA , MILK , WATER , BEER },
{COFFEE , BEER , MILK , TEA , WATER },
{COFFEE , BEER , MILK , WATER , TEA },
{COFFEE , WATER , MILK , TEA , BEER },
{COFFEE , WATER , MILK , BEER , TEA },
{BEER , TEA , MILK , COFFEE , WATER },
{BEER , TEA , MILK , WATER , COFFEE },
{BEER , COFFEE , MILK , TEA , WATER },
{BEER , COFFEE , MILK , WATER , TEA },
{BEER , WATER , MILK , TEA , COFFEE },
{BEER , WATER , MILK , COFFEE , TEA },
{WATER , TEA , MILK , COFFEE , BEER },
{WATER , TEA , MILK , BEER , COFFEE },
{WATER , COFFEE , MILK , TEA , BEER },
{WATER , COFFEE , MILK , BEER , TEA },
{WATER , BEER , MILK , TEA , COFFEE },
{WATER , BEER , MILK , COFFEE , TEA },
};
const int ENUM_ALL[][N] = {
{0,1,2,3,4}, {0,1,2,4,3}, {0,1,3,2,4}, {0,1,3,4,2}, {0,1,4,2,3},
{0,1,4,3,2}, {0,2,1,3,4}, {0,2,1,4,3}, {0,2,3,1,4}, {0,2,3,4,1},
{0,2,4,1,3}, {0,2,4,3,1}, {0,3,1,2,4}, {0,3,1,4,2}, {0,3,2,1,4},
{0,3,2,4,1}, {0,3,4,1,2}, {0,3,4,2,1}, {0,4,1,2,3}, {0,4,1,3,2},
{0,4,2,1,3}, {0,4,2,3,1}, {0,4,3,1,2}, {0,4,3,2,1}, {1,0,2,3,4},
{1,0,2,4,3}, {1,0,3,2,4}, {1,0,3,4,2}, {1,0,4,2,3}, {1,0,4,3,2},
{1,2,0,3,4}, {1,2,0,4,3}, {1,2,3,0,4}, {1,2,3,4,0}, {1,2,4,0,3},
{1,2,4,3,0}, {1,3,0,2,4}, {1,3,0,4,2}, {1,3,2,0,4}, {1,3,2,4,0},
{1,3,4,0,2}, {1,3,4,2,0}, {1,4,0,2,3}, {1,4,0,3,2}, {1,4,2,0,3},
{1,4,2,3,0}, {1,4,3,0,2}, {1,4,3,2,0}, {2,0,1,3,4}, {2,0,1,4,3},
{2,0,3,1,4}, {2,0,3,4,1}, {2,0,4,1,3}, {2,0,4,3,1}, {2,1,0,3,4},
{2,1,0,4,3}, {2,1,3,0,4}, {2,1,3,4,0}, {2,1,4,0,3}, {2,1,4,3,0},
{2,3,0,1,4}, {2,3,0,4,1}, {2,3,1,0,4}, {2,3,1,4,0}, {2,3,4,0,1},
{2,3,4,1,0}, {2,4,0,1,3}, {2,4,0,3,1}, {2,4,1,0,3}, {2,4,1,3,0},
{2,4,3,0,1}, {2,4,3,1,0}, {3,0,1,2,4}, {3,0,1,4,2}, {3,0,2,1,4},
{3,0,2,4,1}, {3,0,4,1,2}, {3,0,4,2,1}, {3,1,0,2,4}, {3,1,0,4,2},
{3,1,2,0,4}, {3,1,2,4,0}, {3,1,4,0,2}, {3,1,4,2,0}, {3,2,0,1,4},
{3,2,0,4,1}, {3,2,1,0,4}, {3,2,1,4,0}, {3,2,4,0,1}, {3,2,4,1,0},
{3,4,0,1,2}, {3,4,0,2,1}, {3,4,1,0,2}, {3,4,1,2,0}, {3,4,2,0,1},
{3,4,2,1,0}, {4,0,1,2,3}, {4,0,1,3,2}, {4,0,2,1,3}, {4,0,2,3,1},
{4,0,3,1,2}, {4,0,3,2,1}, {4,1,0,2,3}, {4,1,0,3,2}, {4,1,2,0,3},
{4,1,2,3,0}, {4,1,3,0,2}, {4,1,3,2,0}, {4,2,0,1,3}, {4,2,0,3,1},
{4,2,1,0,3}, {4,2,1,3,0}, {4,2,3,0,1}, {4,2,3,1,0}, {4,3,0,1,2},
{4,3,0,2,1}, {4,3,1,0,2}, {4,3,1,2,0}, {4,3,2,0,1}, {4,3,2,1,0},
};

复制代码

gxqcn · 发表于 2008-9-19 08:37:24

估计主要是 include 的同名头文件的干扰吧。

gxqcn · 发表于 2008-11-12 07:40:05

原帖由疯猪于 2008-11-12 05:28 发表
两个圆环，半径分别是1和2，小圆在大圆内部绕大圆圆周一周，问小圆自身转了几周？如果在大圆的外部，小圆自身转几周呢？

其实把大圆间断后拉直就可以了
自转数是周长的比是2周
内外都一样

错！

不信，你自己拿两个相同的硬币在桌子上模拟下，
绕固定硬币滚动一周的硬币，在此过程中自转了两周！（而不是一周）

gxqcn · 发表于 2009-12-2 17:07:48

有没有详细解析啊
这样很难懂
jmyhyu 发表于 2009-12-2 16:58

参见图示：http://bbs.emath.ac.cn/viewthrea ... p;fromuid=8#pid5717

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