来自不同国家的4位留学生A，B，C，D在一起交谈，他们每人只会中、英、法、日4种语言中的2种，情况是：没有人既能讲日语又能讲法语；A能讲日语，D不会日语，但A和D能互相交谈，B不会英语，但A和C交谈时却要B当翻译；B，C，D3人想互相交谈，但找不到共同的语言；只有一种语言3人都会。请编程确定这四人分别会哪两种语言。
请用穷举法 谢谢
请用C语言编写 急急急！！
#include <stdio.h>
void deal(void)
{
int a[4][2];
int i, j, k, n, m, x, y, z, t, k1, k2, k3, k4, k5;
a[0][0] = 3;
// 用 0 1 2 3 分别代表 中 英 法 日
// 数组a[4][2] 代表有四个人 存放两种语言 这个 方法 真的 不是很好 但思想 差不多 楼主 自己 改改
for (i = 0; i < 3; i++)
{
a[0][1] = i; //第一个的赋值
for (j = 0; j < 3; j++)
{
if (j == 1)
{
j = j + 1;
}
a[1][0] = j;
for (k = j + 1; k < 4; k++)
{
if (k == 1)
{
k = k + 1;
}
a[1][1] = k;
for (m = 0; m < 3; m++)
{
a[2][0] = m;
for (n = m + 1; n < 4; n++)
{
a[2][1] = n;// 第三个的赋值
for (t = 0; t < 2; t++)
{
a[3][0] = t;
for (k1 = t + 1; k1 < 3; k1++)
{
z = 0;
x = 0;
a[3][1] = k1;
k5 = 0;
if (a[0][0] == a[3][0] || a[0][0] == a[3][1] || a[0][1] == a[3][0] || a[0][1] == a[3][1])// A D 能够讲话
{
x++;
}
if (a[0][0] != a[2][0] && a[0][0] != a[2][1] && a[0][1] != a[2][0] && a[0][1] != a[2][1]) // A C 不能交谈
{
if (a[1][0] == a[0][0] || a[1][0] == a[0][1] || a[1][1] == a[0][0] || a[1][1] == a[0][1]) // A B 翻译
{
if (a[1][0] == a[2][0] || a[1][0] == a[2][1] || a[1][1] == a[2][0] || a[1][1] == a[2][1]) // B C翻译
{
x++;
}
}
}
for (k2 = 0; k2 < 2; k2++) //b c d 不能相互交流 不是 注意理解
{
for (k3 = 0; k3 < 2; k3++)
{
for (k4 = 0; k4 < 2; k4++)
{
if (a[1][k3] == a[2][k4] == a[3][k4])
{
k5 = 1;
}
}
}
}
if (x == 2 && k5 == 0)
{
for (x = 0; x < 4; x++)
{
for ( y = 0; y < 2; y++)
{
printf("%d ", a[x][y]);
}
printf("\n");
}
printf("\n\n");
}
}
}
}
}
}
}
}
}
int main(void)
{
deal();
}
// 这个方法不是很好 可以用循环来判断只是写了 就没管了 我没实际检验 如有问题 改改就行
void deal(void)
{
int a[4][2];
int i, j, k, n, m, x, y, z, t, k1, k2, k3, k4, k5;
a[0][0] = 3;
// 用 0 1 2 3 分别代表 中 英 法 日
// 数组a[4][2] 代表有四个人 存放两种语言 这个 方法 真的 不是很好 但思想 差不多 楼主 自己 改改
for (i = 0; i < 3; i++)
{
a[0][1] = i; //第一个的赋值
for (j = 0; j < 3; j++)
{
if (j == 1)
{
j = j + 1;
}
a[1][0] = j;
for (k = j + 1; k < 4; k++)
{
if (k == 1)
{
k = k + 1;
}
a[1][1] = k;
for (m = 0; m < 3; m++)
{
a[2][0] = m;
for (n = m + 1; n < 4; n++)
{
a[2][1] = n;// 第三个的赋值
for (t = 0; t < 2; t++)
{
a[3][0] = t;
for (k1 = t + 1; k1 < 3; k1++)
{
z = 0;
x = 0;
a[3][1] = k1;
k5 = 0;
if (a[0][0] == a[3][0] || a[0][0] == a[3][1] || a[0][1] == a[3][0] || a[0][1] == a[3][1])// A D 能够讲话
{
x++;
}
if (a[0][0] != a[2][0] && a[0][0] != a[2][1] && a[0][1] != a[2][0] && a[0][1] != a[2][1]) // A C 不能交谈
{
if (a[1][0] == a[0][0] || a[1][0] == a[0][1] || a[1][1] == a[0][0] || a[1][1] == a[0][1]) // A B 翻译
{
if (a[1][0] == a[2][0] || a[1][0] == a[2][1] || a[1][1] == a[2][0] || a[1][1] == a[2][1]) // B C翻译
{
x++;
}
}
}
for (k2 = 0; k2 < 2; k2++) //b c d 不能相互交流 不是 注意理解
{
for (k3 = 0; k3 < 2; k3++)
{
for (k4 = 0; k4 < 2; k4++)
{
if (a[1][k3] == a[2][k4] == a[3][k4])
{
k5 = 1;
}
}
}
}
if (x == 2 && k5 == 0)
{
for (x = 0; x < 4; x++)
{
for ( y = 0; y < 2; y++)
{
printf("%d ", a[x][y]);
}
printf("\n");
}
printf("\n\n");
}
}
}
}
}
}
}
}
}
int main(void)
{
deal();
}
// 这个方法不是很好 可以用循环来判断只是写了 就没管了 我没实际检验 如有问题 改改就行
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第1个回答 2011-06-21
10分...估计没那个动力
第2个回答 2011-06-20
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第3个回答 2011-06-21
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